(*) Distance between two points:
In general for the points A(x1, 0), B(x2, 0) on the X-axis, the distance between A and B is
|x2 - x1|
example : A=(-2, 0) and B=(-6, 0)
|x2 - x1 | = -6 -(-2) | = |-6+2| = |-4| = 4 units
Similarly, if two points lie on Y-axis, then the distance between the points A and B would be the difference between their y-coordinates of the points.
The distance between two points (0, y1) ( 0, y2) would be |y2 - y1|
example: A = (0,2) B=(0,7)
|y2 - y1| = | 7-2| = 5 units.
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Example-1. What is the distance between A=(4,0) and B=(8,0)
sol) Here the points lie on x-axis
The distance between the points will be The difference in the x-coordinates
|x^2 - x1| = | 8 - 4| = 4 units.
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Example-2. A and B are two points given by (8,3) (-4,3). Find the distance between A and B.
sol) Since the y-coordinates(y1 = 3 and y2=3) are equal, points lie on a line, parallel to x-axis.
Here x1 and x2 are lying in two different quadrants and y-coordinates are equal.
Distance AB = |x2 - x1|
= |-4 - 8| = |-12| = 12 units
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Example-3. Let's find the distance between two points A(4,2) and B(8,6)
Sol) Compare these points with
x1 = 4 x2 = 8
y1 = 3 y2 = 6
Using distance formula
Distance AB = d
= √(8-4)^2 + (6-3)^2
= √16 + 9
= √25 = 5 units
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Example 4. Show that the points
A(4,2),
B(7,5) and
C(9,7) are three points lie on a same line.
Sol) Now, we find the distances AB, BC, AC
AB = (4,2) (7,5)
x1 = 4 x2 = 7
y1 = 2 y2 = 5
= √(7-4)^2 + (5-2)^2
= √9 + 9
=√18
=√9 *2
= 3√2 units
BC = √(9-7)^2 + ( 7-5)^2
= √4 + 4
= √8
= 2√2 units
AC = √(9-4)^2 + (7-2)^2
= √25 + 25
= √50
= 5√2 units
Now AB + BC = 3√2 + 2√2
= 5√2 = AC.
Therefore, that the three points (4,2), (7,5) and (9,7) lie on a straight line.
(Points that lie on the same line are called collinear points).
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Example-5. Are the points (3,2), (-2,-3) and (2,3) form a triangle?
sol) Let us apply the distance formula to find the distances PQ, QR and PR, where
P(3,2), Q(-2,-3) and R(2,3) are the given points. We have
PQ = √(-2-3)^2 + (-3-2)^2
= √25 + 25
= √50
= 5√2 = 7.07 units (approx)
QR = √(2-(-2)^2 + (3-(-3)^2
= √(4)^2 + (6)^2
= √52
QR = 7.21 units (approx)
PR =√(2-3)^2 + (3-2)^2
= √(-1)^2 + 1^2
= √2
PR = 1.41 units (approx)
Since the sum of any two of these distances is greater than the third distance, therefore, the points P,Q and R form a triangle and all the sides of triangle is unequal
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Example-6. Show that the points
(1,7),
(4,2),
(-1,-1) and
(-4,4) are the vertices of a square.
sol) One way of showing that ABCD is a square is to use the property that all sides should be equal and both its diagonals should also be equal.
Now.
AB = √(1-4)^2 + (7-2)^2
= √9+25
= √34 units
BC = √(4+1)^2 + (2+1)^2
= √34 units
CD = √(-1+4)^2 + (-1-4)^2
= √34 units
DA = √(-4-1)^2 + (4+7)^2
= √34 units
Diagonals are
AC = √(1+1)^2 + (7+1)^2
= √68 units
BD = √(4+4)^2 + (2-4)^2
= √68 units
Since AB=BC=CD=DA and AC=BD.
So all the four sides of the quadrilateral ABCD are equal and its diagonals AC and BD are also equal.
Therefore, ABCD is a square.
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Example-7. Madhuri, Meena, pallavi are seated at
A(3,1)
B(6,4)
C(8,6) respectively.
Do you think they are seated in a line?
Give reasons to your answer.
sol) Using distance formula, we have
AB = √(6-3)^2 + (4-1)^2
= √18
= √9 * 2
= 3√2 units.
BC = √(18-6)^2 + (6-4)^2
= √4 + 4
=2√2 units.
AC =√(8-3)^2 = (6-1)^2
= √50
=5√2 units
Since, AB + BC = 3√2 + 2√2
= 5√2 = AC,
we can say that the points A,B and C are collinear.
Therefore , they are seated in a line.
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Example-8. Find a relation between x and y such that the point (x,y) is equidistant from the points (7,1) and (3,5).
sol) Let P(x,y) be equidistant from the points
A(7,1) and B(3,5).
Given that AP=BP
√(x-7)^2 + (y-1)^2 = √(x-3)^2 + (y-5)^2
√(x^2 - 14x + 49) + (y^2 - 2y +1) = √(x^2 -6x+9) + (y^2 - 10y + 25)
√(x^2 +y^2 -14x-2y+50) - (x^2+y^2-6x-10y+34) = 0
So, AP^2 = BP^2
-8x + 8y = -16
divide by 8
we get,
i.e., x- y = 2 which is the required relation.
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Example-9. Find the point on the y-axis which is equidistant from both the points A(6,5) and B(-4,3).
sol) We know that a point on the Y-axis is of the form (0,y).
So, let the point P(0,y) be equidistant from A and B. Then
PA = √(6-0)^2 + (5-y)^2
PB = √(-4-0)^2 + (3-4)^2
PA^2 = PB^2
So,
(6-0)^2 + (5-y)^2 = (-4-0)^2 + (3-y)^2
i.e., 36 + 25 + y^2 - 10y = 16 + 9 + y^2 - 6y
i.e., 4y = 36
i.e., y = 9
So, the required point is (0,9)
Let us check our solution :
AP =√(6-0)^2 + (5-9)^2
= √52
BP =√(-4-0)^2 + (3-9)^2
= √52
So, (0,9) is equidistant from (6,5) and (4,3)
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Exercise - 7.1
(1) Find the distance between the following pairs of points
1) (2,3) and (4,1)
sol) Let A=(2,3) and B= (4,1)
(x1, y1) = (2,3) and (x2,y2) = (4,1)
x1 = 2
x2 = 4
y1= 3
y2 = 1
Using distance formula :-
AB = _/(4-2)^2 + ( 1-3)^2
AB = _/(2)^2 + (-2)^2
AB = _/ 4 + 4
AB = _/8
AB = _/4*2
AB = 2_/2
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2) (-5,7) and ( -1,3)
sol) Let A = (-5,7) and B = (-1, 3)
(x1, y1) = ( -5, 7) and (x2,y2) = ( -1, 3)
x1 = -5
x2 = -1
y1 = 7
y2 = 3
Using Formula :-
AB=_/ (-1-(-5)^2 + ( 3-7)^2
AB = _/ (-1+5)^2 + (-4)^2
AB = _/ (4)^2 + 16
AB = _/ 16 +16
AB = _/32
AB = _/ 16*2
AB = 4_/2 units
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3) (-2, -3) and (3, 2)
sol) Let A = (-2, -3) and B =(3,2)
x1 = -2
x2 = 3
y1= -3
y2 = 2
using Formula :-
AB= _/(3-(-2))^2 + ( 2-(-3))^2
AB = _/ (3+2)^2 + (2+3)^2
AB = _/ (5)^2 + (5)^2
AB = _/ 25 + 25
AB = _/ 50
AB = _/25*2
AB = 5_/2 units.
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4) (a,b) and (-a, -b)
sol) Let A = (a-b) and B =(-a, -b)
(x1, y1) = ( a , b) (x2,y2) = (-a, -b)
x1 = a
x2 = -a
y1 = b
y2 = -b
Using Formula :-
AB = _/(-a -a)^2 + (-b - b)^2
AB = _/ (-2a)^2 + ( -2b)^2
AB = _/ 4a^2 + 4b^2
AB = _/4(a^2 + b^2)
AB = 2_/(a^2 + b^2)
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(2) Find the distance between the points (0,0) and (36,15).
sol) Let A= (0,0) and B =(36,15)
(x1,y1) = (0,0) and (x2, y2) = (36,15).
x1 = 0
x2 = 36
y1 = 0
y2 = 15
Using Formula :-
AB = _/(36-0)^2 + (15-0)^2
AB = _/ 1296 + 225
AB = _/1521
AB = _/(39)^2
AB = 39
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(3) Verify that the point (1,5) (2,3) and (-2, -1) are collinear or not.
sol) Collinear :- Points that lie on the same line are called collinear points.
If AB + BC = AC . We say that points lie on a straight line.
Now,
Let A = (1,5) B = (2,3) C = ( -2, -1)
Now we find the distances AB, BC, AC
Distance between AB:-
(x1,y1) = (1, 5) and ( x2, y2) = ( 2,3)
x1= 1
x2 = 2
y1= 5
y2= 3
Using Formula :-
AB = _/ (2-1)^2 + (3 - 5) ^2
AB = _/ (1)^2 + (-2)^2
AB = _/1 + 4
AB = _/5 -----------(1)
Distance between BC :-
(x1, y1) = ( 2,3) and (x2,y2) = ( -2, -1)
x1 = 2
x2 = -2
y1 = 3
y2 = -1
Using Formula :-
BC = _/ (-2 -2)^2 + ( -1-3)^2
BC = _/ (-4)^2 + (-4)^2
BC = _/ 16 + 16
BC = _/ 32
BC = _/ 16*2
BC = 4_/2 ------------------------(2)
Distance between AC :-
(x1, y1) = (1,5) and (x2, y2) = (-2, -1)
x1 = 1
x2 = -2
y1 = 5
y2= -1
Using Formula :-
AC = _/ (-2-1)^2 + ( -1-5)^2
AC = _/ (-3)^2 + (-6)^2
AC = _/ (9 + 36)
AC = _/ 45
AC = _/ 9 * 5
AC = 3_/5 ----------------(3)
Verifying
AB + BC = AC
_/5 + 4_/2 =/= 3_/5
Here, the sum of any two of these distances is not equal to the third distance.
.^. the three given points do not lie on a straight line
Hence, they are not collinear.
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(4) Check whether (5,-2) , (6,4) and (7, -2) are the vertices of an isosceles triangle.
sol) Let A = (5, -2) B =( 6,4) C= ( 7, -2)
Isosceles triangle :- Any two of the three sides are equal
i.e, AB = BC or BC = CA or CA = AB
Distance between AB :-
(x1, y1) = ( 5, -2) and (x2, y2) = ( 6, 4)
x1 = 5
x2 = 6
y1 = -2
y2 = 4
Using Formula :-
AB = _/ (6 - 5) ^2 + ( 4- (-2)^2
AB = _/ (1)^2 + (6)^2
AB = _/ 1 + 36
AB = _/ 37 --------------------(1)
Distance between BC
(x1, y1) = ( 6 , 4 ) and (x2, y2) = ( 7, -2)
x1 = 6
x2 = 7
y1 = 4
y2 = -2
Using Formula :-
BC = _/ ( 7 - 6)^2 + ( -2 - 4) ^2
BC = _/ (1)^2 + (-6)^2
BC = _/ 1 + 36
BC = _/ 37 --------------------(2)
Distance between AC :-
(x1, y1) = ( 5, -2) and ( x2, y2) = ( 7, -2)
x1 = 5
x2 = 7
y1 = -2
y2 = -2
Using Formula :-
AC = _/ (7 -5)^2 + ( -2 -(-2)^2
AC = _/(2)^2 + ( -2+2)^2
AC = _/4 + 0
AC = 2 ------------------(3)
Since AB = BC
two sides are equal . Hence it is an isosceles triangle.
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(5) In a class room, 4 friends are seated at the points A,B,C and D as shown in figure. Jarina and Phani walk into the class and after observing for a few minutes Jarina asks phani " Don't you think ABCD is a square? " Phani disagrees.
using distance formula, find which of them is correct. why?
sol) One way of showing that ABCD is a square is to use the property that all its sides should be equal and both its diagonal should also be equal.
i.e, sides = AB = BC = CD = DA
diagonals = AC = BD
Now from the figure.
A(3, 4), B( 6, 7), C (9, 4) and D( 6, 1)
Distance between AB :-
(x1, y1) = (3, 4) and (x2, y2) = ( 6, 7)
x1 = 3
x2 = 6
y1 = 4
y2 = 7
Using Formula :-
AB=_/(6 -3)^2 + ( 7 - 4)^2
AB = _/ (3)^2 + ( 3)^2
AB = _/ 9 + 9
AB = _/18
AB = _/2*9
AB = 3_/2 ----------------------(1)
Distance between BC :-
(x1, y1) = ( 6, 7) and (x2,y2) = ( 9,4)
x1 = 6
x2 = 9
y1 = 7
y2 = 4
Using Formula :-
BC = _/ ( 9 -6)^2 + ( 4 - 7)^2
BC = _/ (3)^2 + (-3)^2
BC = _/9 + 9
BC = _/18
BC =_/ 2*9
BC = 3_/2 ------------(2)
Distance between CD :-
(x1, y1) = ( 9, 4) and (x2, y2) = ( 6, 1)
x1 = 9
x2 = 6
y1 = 4
y2 = 1
CD = _/(6 - 9)^2 + ( 1-4)^2
CD = _/ (-3)^2 + (-3)^2
CD = _/ (9) + (9)
CD = _/ 18
CD = _/ 2*9
CD =3 _/2---------------(3)
Distance between DA :-
(x1, y1) = ( 6, 1) and ( x2, y2) = ( 3, 4)
x1 = 6
x2 = 3
y1 = 1
y2 = 4
DA = _/ ( (3 -6) ^2 + ( 4 -1)^2
DA = _/ (-3)^2 + (3)^2
DA = _/ 9 + 9
DA = _/18
DA = _/ 2 * 9
DA = 3_/2 --------------(4)
Diagonal AC :-
A =( 3,4) and C = ( 9, 4)
x1 = 3
x2 = 9
y1 = 4
y2 = 4
AC = _/ ( 9-3)^2 + ( 4 -4)^2
AC = _/ (6)^2 + ( 0)^2
AC = _/ 36 + 0
AC = 6
Diagonal BD :-
B( 6,7) and D(6,1)
x1 = 6
x2 = 6
y1 = 7
y2 = 1
BD = _/ (6-6)^2 + ( 1-7)^2
BD = _/ (0)^2 + ( -6)^2
BD = _/ 0+ 36
BD = 6
Here, sides =AB = BC = CD = DA
Diagonals = AC = BD
Hence, ABCD is a square.
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(6) Show that the following points form a equilateral triangle A(a,0), B(-a,0), C(0, a_/3)
sol) If the distance between all the sides are equal . Then we say that the triangle is equilateral triangle.
Distance between AB :-
(x1,y1) = (a , 0) and (x2, y2) = ( -a, 0)
x1 = a
x2 = -a
y1 = 0
y2 = 0
Using Formula :-
AB = _/(-a -a)^2 + ( 0 -0)^2
AB = _/ (-2a)^2 + 0
AB = _/ 4a^2
AB = 2_/a^2 -----(1)
Distance between BC :-
(x1, y1) = ( -a , 0) and (x2,y2) = ( 0, a_/3)
x1 = -a
x2 = 0
y1 = 0
y2 = a_/3
Using Formula :-
BC = _/(0 + a)^2 + (a_/3 - 0)^2
BC = _/ a^2 + 3a^2
BC = _/4a^2
BC = 2_/a^2 ----------(2)
Distance Between AC :-
(x1, y1) = (a, 0) and (x2, y2) = (0, a_/3)
x1 = a
x2 = 0
y1 = 0
y2 = a_/3
Using Formula :-
AC = _/(0 -a) ^2 + (a_/3 - 0 )^2
AC = _/ a^2 + 3a^2
AC =_/4a^2
Ac =2_/a^2 ----(3)
Here (1), (2), and (3) have same value. Hence they form the equilateral triangle.
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( Hint : Area of rhombus = 1/2 * product of its diagonals)
sol) Let A= ( -7, -3), B =( 5, 10), C = ( 15, 8), D =( 3, -5) are the given points.
Proving points taken in order are the corners of a parallelogram :-
In parallellogram opposite sides are equal
Distance between AB :-
(x1, y1) = ( -7, -3) and (x2, y2) = (5, 10)
x1 = -7
x2 = 5
y1 = -3
y2 = 10
Using Formula :-
AB = _/ ( 5 -(-7)^2 + (10-(-3)^2
AB = _/ (5 + 7)^2 + ( 10 + 3)^2
AB = _/ (12)^2 + ( 13)^2
AB = _/ 144 + 169
AB = _/ 313 -----------------(1)
Distance between BC :-
(x1, y1) = (5, 10) and (x2, y2) = ( 15, 8)
x1 = 5
x2 = 15
y1 = 10
y2 = 8
Using Formula :-
BC = _/ (15 - 5)^2 + ( 8 - 10)^2
BC = _/ (10)^2 + ( -2)^2
BC = _/ 100 + 4
BC = _/ 104 --------(2)
Distance between CD :-
(x1, y1) = ( 15, 8) and (x2, y2) = ( 3, -5)
x1 = 15
x2 = 3
y1 = 8
y2 = -5
CD = _/ (3 - 15)^2 + ( -5 -8)^2
CD = _/ 144 + 169
CD = _/ 313 --------------(3)
Distance between DA :-
(x1, y1) = ( 3, -5) and (x2, y2) = ( -7, -3)
x1 = 3
x2 = -7
y1 = -5
y2 = -3
DA = _/ (-7 -3)^2 + ( -3 +5)^2
DA = _/ (-10)^2 + (2)^2
DA = _/ 104 ----------------------(4)
Here AB = CD and BC = DA. Opposite sides are equal. Hence it is a parallelogram.
Finding area :-
Diagonal of a parallelogram divides it into two congruent triangles.
Divide the parallelogram. ABCD into two triangles ABC and ACD
Area (ABCD) = Area ( /_\ ABC) + Area ( /_\ACD)
We have formula for finding area of triangle with three points:-
(x1,y1) = (-7, -3) (x2,y2) = ( 5, 10) (x3,y3) = ( 15, 8)
x1 = -7
x2 = 5
x3 = 15
y1 = -3
y2 = 10
y3 = 8
Area(/_\ ABC) = 1/2 [ x1(y2 -y3) + x2(y3 - y1) + x3(y1-y2)]
=> 1 [ -7 ( 10-8) + 5 ( 8+3) + 15 (-3-10)]
2
=> 1 [ -7(2) + 5(11) + 15(-13)]
2
=> 1 [ -14 + 55 -195]
2
=> 1 [ -154 ]
2
Since the area of two(2) triangles formed inside the parallelogram are equal.
As we got area one triangle we can just simply multiply with "2"
i.e,2 * 1 [ -154 ]
2
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(8) Show that the points ( -4,-7), (-1,2), ( 8,5) and (5,-4) taken in order are the vertices of a rhombus.
sol) In rhombus all sides are equal but diagonals are not equal
let A ( -4, -7) , B(-1, 2) , C ( 8, 5) and D ( 5, -4) are the given points,
We have,
AB = _/ (-1 + 4) ^2 + ( 2 +7)^2
AB = _/ 9 + 81
AB = _/ 90
AB = _/9 * 10
AB = 3_/10 ------------------(1)
BC = _/ (8 +1)^2 + ( 5-2)^2 => _/ 81 + 9 => 3_/10
CD = _/(5-8)^2 + ( -4 -5)^2 => _/ (9 + 81) => 3_/10
DA = _/ ( 5+4)^2 + ( -4 +7)^2 => _/( 9 +81) = > 3_/10
Here AB = BC = CD = DA, so all four sides of the quadrilateral ABCD are equal.
Now let us find distances of the diagonals AC and BD
AC = _/(8+4)^2 + (5 +7)^2 => _/ (12)^2 + (12)^2 => 12_/2
BD = _/ ( 5 + 1)^2 + ( -4-2)^2 => _/ (6)^2 + (6)^2 => 6_/2
Here , AC =/= BD
.^. AB, BC, CD, DA are the vertices of a rhombus.
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(9) Name the type of quadrilateral formed, if any, by the following points, and give reason for your answer.
(1) (-1, -2) , (1,0),(-1,2), (-3,0)
(2) (-3,5), (3,1), (0,3), (-1, -4)
(3) (4,5), (7,6), (4,3), (1,2)
sol) Let A =(-1, -2) B =(1,0) C =(-1, 2) D =(-3,0)
AB =_/(1+1)^2 + ( 0-2)^2 = _/ (4 + 4) = _/8 = 2_/2
BC = _/(-1-1)^2 + ( 2-0)^2 = _/ (4 + 4) = _/8 = 2_/2
CD = _/(-3+1)^2 + ( 0-2)^2 = _/8 = 2_/2
DA = _/(-3+1)^2 + ( 0 +2)^2 = _/8 = 2_/2
Diagonal :-
AC = _/(-1 +1)^2 + (2 +2)^2 = _/16 = 4
BD = _/(-3-1)^2 + (0+0)^2 = 4
Here, AB =BC = CD = DA and AC =BD
so all the four sides of the quadrilateral ABCD are equal and its diagonal are also equal.
.^. , ABCD is a square.
---------------------------------
Let A = (-3,5) B =(3,1) C = (0,3) D=(-1, -4)
Findng Distance AB :- A=( -3,5) B =(3,1)
x1 = -3
x2 = 3
y1 = 5
y2 = 1
Using Formula :-
AB = _/(3(-3)^2 + (1-5)^2
AB = _/(3+3)^2 + (-4)^2
AB = _/ 36 + 16
AB = _/ 52
AB = _/ 4* 13
AB = 2_/13
Finding Distance BC :- B =( (3, 1) C=(0,3)
x1 = 3
x2 = 0
y1 = 1
y2 = 3
Using Formula :-
BC = _/ (0-3)^2 + (3-1)^2
BC = _/9 + 4
BC = _/ 13
Finding distance CD :- C=( 0, 3) D =(-1, -4)
x1 = 0
x2 = -1
y1= 3
y2 = -4
CD = _/ (-1-0)^2 + ( -4-3)^2
CD = _/ 1 + 49
CD = _/50
Finding Distance DA :- D =( -1, -4) A=(-3, 5)
x1 = -1
x2 = - 3
y1 = -4
y2 = 5
DA = _/(-3-(-1)^2 + ( 5-(-4)^2
DA = _/ (-2)^2 + (9)^2
DA = _/4 + 81
DA = _/85
Since AB=/= BC =/=CD =/= DA
So, it is not a quadrilateral
We can find out what actual figure this points formed by using graph
A=( -3, 5) B=(3,1) C=(0,3) D=(-1, -4)
Here points A,B,C are collinear.
We check if AC + BC = AB
Finding Distance AC :- A =(-3,5) C =( 0, 3)
x1 = -3
x2 = 0
y1 = 5
y2 = 3
AC = _/ (0 -(-3)^2 + ( 3 -5)^2
AC = _/ (9) + ( -2)^2
AC = _/ 9 + 4
AC = _/13
We check if
AC + BC = AB
_/13 + _/13 = 2_/13
Hence, ABCD is a triangle, not a quadrilateral.
------------------------------------------------------------
(3) Let A=( 4,5) B=(7,6) C=(4,3) D=(1,2)
Finding distance AB :- A = ( 4,5) B= ( 7,6)
x1=4
x2=7
y1= 5
y2 =6
AB = _/ ( 7-4)^2 + ( 6 - 5)^2
AB = _/ 9 + 1
AB = _/10
Finding distance BC :- B=(7,6) C=( 4,3)
x1 = 7
x2 = 4
y1= 6
y2 =3
BC = _/((4-7)^2 + ( 3-6)^2
BC = _/ 9 + 9
BC = _/ 18
Finding distance CD :- C= (4,3) D=( 1,2)
x1 = 4
x2 = 1
y1 = 3
y2 = 2
CD = _/(1-4)^2 + ( 2-3)^2
CD = _/9 + 1
CD = _/10
*****************************************************
(10) Find the point on the x-axis which is equidistant from (2, -5) and (-2,9).
sol) Let A=( 2, -5) B=(-2, 9)
RTP :- Point on x-axis...
As we know on x-axis , y =0
Hence the required point to find on x-axis is C = (x, 0)
Given :- point on x-axis C= (x,0) is equidistant from "A" and "B"
i.e, AC = BC
Now,
Finding distance AC :- A=( 2, -5) C=( x, 0)
x1 = 2
x2 = x
y1 = -5
y2 = 0
AC = _/ ( x -2)^2 + ( 0 - (-5)^2
AC = _/ x^2 + 2^2 - 2(x)(2) + (5)^2
AC = _/ x^2 + 4 - 4x + 25
AC = _/ x^2 - 4x + 29 --------(1)
Finding distance BC :- B = ( -2, 9) C =(x,0)
x1 = -2
x2 = x
y1 = 9
y2 = 0
BC = _/(x -(-2)^2 + ( 0 -9)^2
BC = _/ (x +2)^2 + 81
BC = _/x^2 + 2^2 + 2(2)(x) + 81
BC = _/x^2 + 4 +4x + 81
BC = _/ x^2 + 4x + 85 -------------------(2)
Now AC = BC
_/(x^2 -4x +29) = _/ x^2 + 4x + 85
squaring on both sides
(_/(x^2 -4x +29))^2 = (_/ x^2 + 4x + 85 )^2
x^2 - 4x +29 = x^2 + 4x + 85
-4x - 4x = 85 - 29
-8x = 56
s = - 56
8
x = -7
Hence the required point is C = (-7, 0)
point C = (-7,0) is equidistant from A(2, -5) B(-2. 9)
*********************************************
(11) If the distance between two points (x,7) and (1, 15) is10, find the value of "x"
sol) Let A = (x,7) B=(1,15)
x1 = x
x2 = 1
y1 = 7
y2 = 15
Using Formula :-
AB =_/(1-x)^2 + ( 15-7)^2
10 =_/(1)^2 + (x)^2 - 2(1)(x) + (8)^2
10 = _/ 1 + x^2 - 2x + 64
10 = _/ x^2 -2x + 65
Now, squaring on both sides
(10)^2 = (_/x^2 -2x + 65)^2
100 = x^2 -2x + 65
x^2 - 2x = 100 - 65
x^2 - 2x = 35
x^2 - 2x - 35 = 0
x^2 -7x + 5x - 35 = 0
x(x - 7) + 5(x - 7) = 0
x-7 = 0 or x +5 = 0
x = 7 or x = -5
**************************
(12) Find the values of "y" for which the distance between the points P(2,-3) and Q(10,y) is 10 units.
sol) Given :-P = ( -2, 3) and Q =( 10, y)
Distance between PQ = 10 units
Finding Distance PQ :-
x1 = -2
x2 = 10
y1 = 3
y2 = y
_/( 10-2)^2 + (y-(-3)^2 = 10
_/ 64 + (y +3)^2 = 10
_/ 64 + y^2 + (3)^2 + 2(y)(3) = 10
-/ 64 + y^2 + 9 +6y = 10
_/ y^2 + 6y + 73 = 10
squaring on both sides, we get
y^2 + 6y + 73 = 100
y^2 + 6y + 73 - 100 = 0
y^2 + 6y - 27 = 0
y^2 + 9y - 3y - 27 = 0
y(y + 9) -3 ( y +9) = 0
(y-3) (y+9) = 0
y = 3 or y = -9 is the solution
*************************
(13) Find the radius of the circle whose center is (3,2) and passes through ( -5,6).
sol) Let 'r' be the radius of the given circle.
Radius(r) :- Distance between the center of the circle and any point on the circumference
Given :- Center (p) = ( 3,2) and passing through point Q=(-5,6)
Finding distance PQ :-
x1 = 3
x2 = -5
y1 = 2
y2 = 6
PQ= _/ (-5-3)^2 + ( 6 -2)^2
=> _/ 64 + 16
=> _/ 80
=> _/ 16 * 5
=> 4_/5
r = 4_/5 units
****************************
(14) Can you draw a triangle with vertices (1,5), (5,8) and (13, 14)? Give reason.
sol) Let A = ( 1,5) B = ( 5, 8) C = ( 13, 14)
Finding distance AB :- A=( 1, 5) B =( 5, 8)
x1 = 1
x2 = 5
y1 = 5
y2 = 8
AB = _/ ( 5-1)^2 + (8-5)^2
AB = _/ 16 + 9
AB = _/ 25
AB = 5
finding distance BC :- B=( 5, 8) C = ( 13, 14)
x1 = 5
x2 = 13
y1 = 8
y2 = 14
BC = _/(13 - 5)^2 + ( 14-8)^2
BC = _/ 64 + 36
BC = _/ 100
BC = 10
Finding distance AC :- A = ( 1,5) C = ( 13, 14)
x1 = 1
x2 = 13
y1 = 5
y2 = 14
AC = _/ (13-1)^2 + (14-5)^2
AC = _/ 144 + 81
AC= _/ 225
AC = 15
If A, B, C are collinear triangle is not possible
Lets check
AB = 5
BC = 10
AC = 15
AB + BC = AC
5 + 10 = 15
15 = 15
All points lie on the same line.
Thus, we cannot draw a triangle with the given vertices
*************************************************
(15) Find a relation between "x" and "y" such that the point (x,y) is equidistant from the points (-2,8) and (-3,-5).
sol) Let A = ( -2,8) B=( -3, 5) and P =(x,y)
Given :- point "P" is equidistant from A and B
i.e AP = BP
Distance between AP :- A = ( -2, 8) P = ( x,y)
x1= -2
x2 = x
y1 = 8
y2 = y
AP = _/ (x-(-2)^2 + ( y-8)^2
AP = _/ (x +2)^2 + (y-8)^2
AP = _/(x)^2 + (2)^2 + 2(x)(2) + (y)^2 + (8)^2 - 2(y)(8)
AP = _/ x^2 + 4 + 4x + y^2 + 64 - 16y
AP = _/x^2 + y^2 + 4x - 16y + 68
Distance between BP :- B =( -3, -5) P =(x,y)
x1 = -3
x2 = x
y1 = -5
y2 = y
BP = _/ x-(-3)^2 + ( y - (-5)^2
BP = _/ (x +3)^2 + ( y+5)^2
BP = _/ x^2 + (3)^2 + 2(x)(3) + y^2 + (5)^2 + 2(y)(5)
BP = _/ x^2 + 9 + 6x + y^2 + 25 + 10y
BP = _/ x^2 + y^2 + 6x + 10y + 34
Since AP = BP
_/x^2 + y^2 + 4x - 16y + 68 = _/ x^2 + y^2 + 6x + 10y + 34
4x - 6x -16y - 10y + 68 - 34 = 0
-2x - 26y + 34 = 0
divide by "2"
-x -13y + 17 = 0
x + 13y - 17 = 0
x + 13y = 17 is the required equation
*********************************************************
Section formula
The coordinates of the points P(x,y) which divides the line segment joining the points A(x1,y1) and
B(x2,y2), internally in the ratio m1 : m2 are
(m1x2 + m2x1 , m1y2 + m2y1 )
m1 + m2 m1 + m2
If the ratio in which P divides AB is k:1, then the coordinated of the point P are
(kx2 + x1 , ky2 + y1)
k+1 k+1
*****************************************
Example-10. Find the coordinates of the point which divides the line segment joining the points (4, -3) and (8,5) in the ratio 3:1 internally.
sol) Let P(x,y) be the required point. using the section formula
P(x,y) = (m1x2 + m2x1 , m1y2 + m2y1 )
m1 + m2 m1 + m2
we get,
x = 3(8) + 1(4) = 24 + 4 = 28 = 7
3 + 1 4 4
y = 3(5) + 1(-3) = 15 - 3 = 12 = 3
3+1 4 4
P(x,y) = (7,3) is the required point.
*****************************************
Example-11. Find the mid point of the line segment joining the points (3,0) and (-1,4)
sol) The mid point M(x,y) of the line segment joining the points (x1, y1) and (x2,y2)
M(x,y) = ( x1 + x2, y1 + y2 )
2 2
.^. The mid point of the line segment joining the points (3,0) and (-1,4) is
M(x,y) = (3 +(-1), 0+4) = ( 2, 4 ) = (1,2)
2 2 2 2
*****************************************
(*) Centroid of a Triangle
The coordinates of the centroid are given by
[ x1 + x2 + x3, y1 + y2 + y3 ]
3 3
Example-12. Find the centroid of the traingle whose vertices are
(3,-5)
(-7,4)
(10,-2) respectively.
sol) The coordinated of the centroid are
= [ x1 + x2 + x3, y1+y2+y3]
3 3
= [ 3 + (-7) + 10, (-5) + 4 + (-2) ]
3 3
= (2, -1) is the centroid
*****************************************
Example-13. In what ratio does the point (-4,6) divide the line segment joining the points A(-6,10) and B(3, -8)?
sol) Let (-4,6) divide AB internally in the ratio m1:m2.
Using the section formula, we get
(-4, 6) = (m1x2 + m2x1 , m1y2 + m2y1 )
m1 + m2 m1 + m2
(-4,6) = [3m1 -6m2, -8m1 + 10m2] --(!)
m1+m2 m1+m2
We know that if(x,y)= (a,b) then
x= a and y = b
So,
-4 = 3m1 - 6m2
m1 + m2
and
6 = -8m1 + 10m2
m1+m2
Now,
-4 = 3m1 -6m2 gives us
m1+ m2
-4m1 - 4m2 = 3m1 -6m2
i.e., 7m1 = 2m2
m1 = 2
m2 7
i.e., m1 : m2 = 2:7
We should verify that the ratio satisfies the y-coordinate also.
Now,
-8m1 + 10m2 (Dividing throughout by m2)
m1 + m2
-8 m1 + 10
= m2
m1 + 1
m2
= -8 * (2/7) + 10
(2/7) + 1
= -16/7 + 10
(9/7)
= -16 + 70
9
= 54/9 = 6
Therefore, the point (-4,6) divides the line segment joining the points A(-6,10) B(3,-8) in the ratio 2:7
*****************************************
(*) Trisectional Points of a Line
Example-14. Find the coordinates of the points of trisection(The points which divide a line segment into 3 equal parts are said to be the sectional points) of the line segment joining the points A(2,-2) and B(-7,4).
Sol) Let P and Q be the points of trisection of AB
i.e., AP = PQ = QB
A---------P----------Q---------B
(2,-2) (-7,4)
Therefore, P divides AB internally in the ration 1:2
Therefore, the coordinates of P are ( by applying the section formula)
P(x,y) = (m1x2 + m2x1 , m1y2 + m2y1 )
m1 + m2 m1 + m2
[ 1(-7) + 2(2), 1(4) + 2(-2) ]
1 + 2 1 + 2
i.e., [ -7 + 4 , 4 -4 ]
3 3
= [ -3 , 0 ]
3 3
= ( -1 , 0)
Now, Q also divides AB internally in the ratio 2:1
So, the coordinates of Q are
= [2(-7) + 1(2), 2(4) + 1(-2) ]
2 + 1 2 + 1
i.e., [ -14 + 2 , 8 -2 ]
3 3
= [ -12 , 6 ]
3 3
= [ -4, 2 ]
Therefore, the coordinates of the points of trisection of the line segment are P(-1,0) and Q(-4, 2)
*****************************************
Example-15. Find the ratio in which the y-axis divides the line segment joining the points (5, -6) and (1, -4) . Also find the point of intersection.
sol) Let the ratio be K : 1. Then by the section formula, the coordinates of the point which divides AB in the ratio K : 1 are
[ K(-1) + 1(5), K(-4) + 1(-6) ]
K + 1 K + 1
i.e., [ -K + 5, -4K -6 ]
K + 1 K + 1
This point lies on the y-axis, and we know that on the y-axis the abscissa is 0 .
Therefore,
-K + 5 = 0
K + 1
-K + 5 = 0
K = 5.
So, the ratio is K:1 = 5 : 1
Putting the value of K=5,
we get the point of intersection as
= (-5 + 5, -4(5) -6)
5 + 1 5 + 1
= (0, -20-6)
6
= ( 0, -26 )
6
= (0, -13 )
3
*****************************************
Example -16. Show that the points
A(7,3)
B(6,1)
C(8,2) and
D(9,4)
taken in that order are vertices of parallelogram.
sol) We know that the diagonals of a parallelogram bisect each other.
.^. So the midpoints of the diagonals AC and DB should be equal.
Now, we find the mid points of AC and DB by using
( x1 + x2, y1 + y2 )
2 2
Midpoint of AC = (7 +8, 3+2)
2 2
= ( 15, 5 )
2 2
Midpoint of DB = ( 9 + 6, 4+1 )
2 2
= ( 15, 5 )
2 2
Hence,
midpoint of AC = midpoint of DB
Therefore, the points A,B,C,D are vertices of a parallelogram.
*****************************************
Example-17. If the points
A(6,1)
B(8,2)
C(9,4) and
D(p,3)
are the vertices of a parallelogram, taken in order, find the value of P.
sol) We know that diagonals of a //gram bisect each other.
So,
the coordinates of the midpoint of AC = Coordinates of the midpoint BD
i.e., ( 6 + 9, 1+4) = ( 8 + p , 5 )
2 2 2 2
( 15, 5 ) = ( 8 + p, 5 )
2 2 2 2
15 = 8 +P
2 2
15 = 8 + p
p = 7.
*****************************************
Exercise - 7.2
(1) Find the co-ordinates of the point which divides the join of (-1, 7) and (4, -3) in the ratio 2:3.
sol) Let A= ( -1, 7) B=(4, -3) and m1:m2 = 2; 3
x1 = -1
x2 = 4
y1 = 7
y2 = -3
Let P(x,y) be the required point. Using the section formula.
Finding "x" :-
x = m1x2 + m2x1
m1 + m2
x1 = -1
x2 = 4
m1 :m2 = 2 : 3
x = 2(4) + 3(-1)
2 + 3
x = 8 - 3
5
x = 5
5
x = 1
Finding "y" :-
y = m1y2 + m2y1
m1 + m2
y1 = 7
y2 = -3
m1 : m2 = 2 : 3
y = 2(-3) + 3(7)
2 + 3
y = -6 + 21
5
y = 15
5
y = 3
Hence the required p(x,y) = p( 1, 3)
*********************************
(2) Find the coordinates of the points of trisection of the line segment joining (4, -1) and (-2, -3).
sol) Trisection :- It means dividing a line segment in three equal parts (OR) dividing a line segment in the ratio 1:2 and 2 :1 internally
A-------------P--------Q---------B
(4, -1) ( -2, -3)
P & Q are two points on AB such that
AP = PQ = QB (equal parts)
Finding "P":-
A----------P--------Q--------B
----1--- --------2--------
Case (1) :- When m1 : m2 = 1 :2
A=( 4,-1) and B=( -2, -3)
x1= 4
x2 = -2
y1 = -1
y2 = -3
Let P(x,y) be the required point, using section formula.
x = m1x2 + m2x1
m1 + m2
x = 1(-2) + 2(4)
1 + 2
x = -2 + 8
3
x = 6
3
x = 2
Finding "y":-
y = m1y2 + m2y1
m1 + m2
y = 1(-3) + 2(-1)
1 + 2
y = -3 -2
3
y = -5
3
Hence, point P(x,y) = P ( 2, -5/3)
Finding Q :-
A------P----Q--------B
------2---- ---1---
case(2) :- When m1 : m2 = 2 ;1
A=( 4,-1) and B=( -2, -3)
x1= 4
x2 = -2
y1 = -1
y2 = -3
Let Q(x,y) be the required point . Using section formula
x = m1x2 + m2x1
m1 + m2
x = 2(-2) + 1(4)
2 + 1
x = -4 + 4
3
x = 0
Finding "y":-
y = m1y2 + m2y1
m1 + m2
y = 2(-3) + 1(-1)
2 + 1
y = -6 -1
3
y = -7
3
Hence , Point Q(x,y) = Q( 0, -7/3)
A --------P -------------Q-------------B
(4, -1) (2, -5/3) (0, -7/3) (-2, -3)
*************************************
(3) Find the ratio in which the line segment joining the points ( -3, 10) and (6, -8) is divided by (-1, 6).
sol) Given :- Point P(-1,6) divide the line segment joining the points (-3,10) and (6,3)
A------------------P--------------------B
(-3,10) (-1,6) (6, -8)
RTP :- m1:m2=??
Using section formula, we get
x1 = -3
x2 = 6
y1 = 10
y2 = -8
P(-1, 6) = m1(6) + m2(-3) , m1(-8) + m2(10)
m1 + m2 m1 + m2
Now,
-1 = 6m1 - 3m2
m1 + m2
-1(m1 + m2) = 6m1 - 3m2
-m1 - m2 = 6m1 - 3m2
-m1 - 6m1 = -3m2 + m2
-7m1 = - 2m2
m1 = 2
m2 7
m1:m2 = 2:7
.^. , the point (-1,6) divides the line segment joining the points A(-3, 10) and B(6, -8) in the ratio 2 :7
*********************************************************************************
(4) If (1,2) , (4,y) , (x ,6) and (3,5) are the vertices of a parallelogram taken in order, find "x" and "y".
sol) Let A=(1,2) B=(4,y) C=(x,6) D=(3,5) are the vertices of the parallelogram
In parallelogram diagonals bisect each other.
.^. , the mid-points of the diagonals AC and BD should be equal.
Let us now find the mid-points of diagonal(AC) and diagonal(BD)
mid-point of AC :- A(1,2) C(x,6)
x1 = 1
x2 = x
y1 = 2
y2 = 6
x1 + x2 = 1 +x
2 2
y1 + y2 = 2 + 6 = 4
2 2
AC = ( 1 + x , 4) -------------(1)
2
Mid-point BD :- B( 4,y) D(3,5)
x1 = 4
x2 = 3
y1 = y
y2 = 5
x1 + x2 = 4 + 3 = 7
2 2 2
y1 + y2 = y + 5
2 2
BD = ( 7 , y +5 ) -------(2)
2 2
Since diagonals in parallelogram is equal we can equate AC and BD
equating (1) and (2)
(1+x , 4) = ( 7 , y +5 )
2 2 2
Finding "x"
1 + x = 7
2 2
1+x = 7
x = 6
Finding "y" :-
4 = y + 5
2
8 = y + 5
y = 3
******************************************************
(5) Find the coordinates of a point A, where AB is the diameter of a circle whose centee is (2, -3) and B is ( 1,4).
sol) Let point A=(x,y)
Given :- center(C) = (2, -3) and B =(1, 4)
We know that the mid-point M(x,y) of the line segment joining the points ( x1, y1) and (x2, y2) is
M(x,y) = (x1 + x2 , y1 + y2 )
2 2
Here
M(x,y) = C(2,-3)
A(x,y) B( 1, 4)
x1 = x
x2 = 1
y1 = y
y2 = 4
Now,
C(2,-3) = (x + 1 , y + 4 )
2 2
2 = x +1
2
4 = x +1
3 = x
- 3 = y + 4
2
-6 = y +4
y = -10
Hence, A( 3, -10)
**********************
(6) If A and B are ( -2, -2) and (2, -4) respectively. Find the coordinates of "P" such that AP = 3/7 AB and "P" lies on the segment AB.
sol) Let P=(x,y)
Given :- AP = 3 AB
7
A--------P-------------------------------B
(-2, -2) (x,y) (2, -4)
=> 7AP = 3AB
-----------------------
But AB = AP +PB
-----------------------
=> 7AP = 3(AP + PB)
=> 7AP = 3AP + 3PB
=> 7AP - 3AP = 3BP
=> 4AP = 3PB
AP = 3
PB 4
Here the point "P" divides "AB" in the ratio of 3:4
Finding Point P(x,y) :-
We have, m1 :m2 = 3 :4
A(-2, -2) B(2, -4)
x1= -2
x2 = 2
y1 = -2
y2 = -4
using section formula :-
Finding "x" :-
x = m1x2 + m2 x1
m1 + m2
x = 3(2) + 4(-2)
3 + 4
x = 6 - 8
7
x = -2
7
Finding "y" :-
y = m1y2 + m2y1
m1 + m2
y = 3(-4) + 4(-2)
3 + 4
y = -12 - 8
7
y = -20
7
Thus, the co-ordinates of "P" are P ( -2/7, -20/7)
A ------------------P---------------------------------B
(-2, -2) (-2/7, -20/7) (2, -4)
*******************************************
(7) Find the coordinates of points which divide the line segment joining A(-4,0) and B (0,6) into four equal parts.
sol) Given :- A(-4, 0) B(0,6)
Given that lines segment divided into 4 equal parts which have A(-4,0) starting point and B(0,6) ending point
A---------P-------Q-------R--------B
(-4,0) (x1,y1) (x2,y2) (x3,y3) (0,6)
Given all 4 parts are equal
i.e, AP = PQ= QR= RB
Finding co-ordinates of "P" :- A (-4, 0) B(0,6)
A-----------P-------Q------R------B
|------1----| |-------------3----------}
Here m1 :m2 = 1 ; 3
x1 = -4
x2 = 0
y1 = 0
y2 = 6
Using section formula :-
p(x,y) = ( 1(0) + 3(-4) , 1(6) + 3(0) }
1 + 3 1 + 3
p(x,y) = -12 , 6
4 4
p(x,y) = (-3, 3/2)
Finding co-ordinates of "Q" :- A( -4, 0) B(0,6)
A------P-----Q------R------B
------2------ ------2------
m1 :m2 = 2: 2
x1 = -4
x2 = 0
y1 = 0
y2 = 6
Using section formula:-
Q(x,y) = ( 2(0) + 2(-4) , 1(6) + 3(0) )
4 4
Q(x,y) = (-2, 3)
Finding co-ordinates of "R" :- A( -4, 0) B(0,6)
A------P------Q--------R-----B
|---------------3-------| |----1--|
m1:m2 = 3 : 1
x1 = -4
x2 = 0
y1 = 0
y2 = 6
Using section Formula :-
R(x,y) = (3(0) + 1(-4) , 3(6) + 1(0)
4 4
R(x,y) = (-1, 9/2)
A---------------P----------Q------------R--------B
(-4,0) (-3, 3/2) (-2,3) (-1, 9/2) (0,6)
******************************************
(8) Find the coordinates of the points which divides the line segment joining A(-2, 2) and B (2,8) into four equal parts.
sol) Given :- A (-2, 2) B(2,8)
Given line segment is divided into 4 equal parts
A----------P-------Q-----R-----B
AP = PQ = QR = RB
Finding co-ordinates of "P" :- A (-2, 2) B(2,8)
A-----------P-------Q------R------B
|------1----| |-------------3----------}
Here m1 :m2 = 1 ; 3
x1 = -2
x2 = 2
y1 = 2
y2 = 8
Using section formula :-
P(x,y) = {(1(2) + 3(-2) , 1(8) + 3(2)}
4 4
P(x,y) = ( -1, 7/2)
Finding co-ordinates of "Q" :- A (-2, 2) B(2,8)
A-----------P-------Q------R------B
|------2 ------------| |-----2-------}
Here m1 :m2 = 2 :2
x1 = -2
x2 = 2
y1 = 2
y2 = 8
Using section formula :-
Q(x,y) = ( 2(2) + 2(-2) , 2(8) + 2(2) }
4 4
Q(x,y) = (0,5)
Finding co-ordinates of "R" :- A (-2, 2) B(2,8)
A-----------P-------Q------R------B
|------3--------------------| |---1--|
Here m1 :m2 = 3 : 1
x1 = -2
x2 = 2
y1 = 2
y2 = 8
Using section formula :-
R(x,y) = ( 3(2) + 1(-2) , 3(8) + 1(2) )
4 4
R(x,y) = (1, 13/2 )
A---------------P----------Q------------R--------B
(-2,2) (-1, 7/2) (0,5) (1, 13/2) (2,8)
*******************************************
(9) Find the coordinates of the point which divides the line segment joining the points ( a+b, a-b) and (a-b, a+b) in the ratio 3 :2 internally.
sol) Let P(x,y) be the required point.
Given :- m1 m2 = 3 :2
A(a+b, a-h) B(a-b, a+b)
x1 = a+b
x2 = a-b
y1 = a-h
y2 = a +b
using Section Formula :-
x = 3(a-b) + 2(a+b)
3 + 2
x = 3a - 3b + 2a + 2b
5
x = 5a - b
5
y = 3(a+b) + 2(a-b)
3 + 2
y = 3a + 3b + 2a - 2b
5
y = 5a + b
5
P(x,y) = ( 5a - b, 5a +b ) is the required point
5 5
******************************************
(10) Find the coordinates of centriod of the following
1) (-1, 3), (6, -3) and (-3, 6)
2) (6,2), (0,0) and (4, -7)
3) (1,-1), (0,6) and (-3,0)
sol) Let A ( -1, 3) B(6,3) C( -3,6)
x1 = -1
x2 = 6
x3 = -3
y1 = 3
y2 = -3
y3 = 6
The co-ordinates of the centroid are
G(x,y) = ( x1 + x2 + x3, y1+y2+y3)
3 3
G(x,y) = ( -1 + 6 +(-3). 3 + (-3) +6)
3 3
G(x,y) = (2, 2 )
3
2) (6,2), (0,0) and (4, -7)
x1 = 6
x2 = 0
x3 = 4
y1 = 2
y2 = 0
y3 = -7
The co-ordinates of the centroid are
G(x,y) = ( x1 + x2 + x3, y1+y2+y3)
3 3
G(x,y) = ( 6+0+4). 2+0-7)
3 3
G(x,y) = (10, -5 )
3 3
3) (1,-1), (0,6) and (-3,0)
x1 = 1
x2 = 0
x3 = -3
y1 = -1
y2 = 6
y3 = 0
The co-ordinates of the centroid are
G(x,y) = ( x1 + x2 + x3, y1+y2+y3)
3 3
G(x,y) = ( 1 +0-3). -1+6+0)
3 3
G(x,y) = (-2, 5 )
3 3
*****************************************
Example 18. Find the area of a triangle whose vertices are (1, -1), (-4, 6) and (-3, -5)
sol) The area of the triangle formed by the vertices
A(1,-1) B(-4, 6) C(-3,-5)
Using Formula
=1 |x1(y2 - y3) + x2(y3 - y1) + x3 (y1-y2) |
2
= 1 |1(6+5) + (-4)(-5+1) + (-3)(-1-6) |
2
= 1 | 11 + 16 + 21 |
2
= 24
So the area of the triangle is 24 square units.
*****************************************
Example-19. Find the area of a traingle formed by the points
A(5,2)
B(4,7) and
C(7,-4)
sol) The area of the triangle formed by the vertices is given by
1 |5(7+4) + 4(-4-2) + 7(2-7)|
2
= 1 | 55 -24 -35|
2
= |-4| = |-2|
2
Since area is a measure, which cannot be negative, we will take the numberical value 2 or absolute value
i.e., |-2| = 2
Therefore,
the area of the triangle = 2 square units.
*****************************************
Example-20. If A(-5,7), B(-4,-5), C(-1,-6) and D(4,5) are the vertices of a quadrilateral. Then, find the area of the quadrilateral ABCD.
sol) By joining B to D, you will get two triangles ABD and BCD.
The area of △ABD
= 1 |-5(-5-5) + (-4)(5-7) + 4(7+5)|
2
= 1 |50 + 8 + 48|
2
= 106 = 53 square units
2
Also,
The area of △BCD
= 1 |-4(-6-5) - 1(5+5) + 4(-5+6)|
2
= 1 | 44 - 10 + 4|
2
= 19 square units.
Area of △ABD + Area of △BCD
So, the area of quadrilateral
ABCD = 53 + 19= 72 square units.
*****************************************
(*) Collinearity
Example-21. The points (3,-2)(-2,8) and (0,4) are three points in a plane.Show that these points are collinear.
sol) By using area of the triangle formula
△ = 1 | 3(8-4) + (-2)(4-(-2) + 0(-2-8)|
2
= 1 | 12-12| = 0
2
The area of the triangle is 0.
Hence the three points are collinear or lie on the same line.
****************************************
(*) Area of Triangle-'Heron's Formula'
A triangle whose lengths of sides are a,b, and c.
______________
A = √S(S-a)(S-b)(S-c)
where
S = a + b + c
2
Example-22. Find the value of 'b' for which the points are collineary.
sol). Let given points
A(1,2), B(-1,b), C(-3, -4)
Then
x1=1 y1=2
x2 = -1 y2 = b
x3 = -3 y3 = -4
We know, area of
△ABC = 1 |x1(y2-y3) + x2(y3-y1) +x3(y1-y2)|
2
.^. 1 |1(b +4) + (-1)(-4-2) + (-3)(2-b)| = 0
2
(= 0 because The given points are collinear)
|b + 4 + 6 -6 + 36 | = 0
|4b + 4| = 0
4b + 4 = 0
.^. b = -1
*****************************************
Exercise- 7.3
(1) Find the area of the triangle whose vertices are
1) (2,3)(-1, 0), (2,-4)
2) (-5, -1), (3,-5), (5,2)
3) (0,0), (3,0) and (0,2)
sol) Area of triangle formed by the vertices
A( 2,3) B(-1,0) C(2, -4)
x1 = 2
x2 = -1
x3 = 2
y1 = 3
y2 =0
y3 = -4
Formula :-
Area= 1/2 | x1(y2 - y3) +x2(y3 - y1) + x3(y1 - y2) |
=> 1/2 |2(0 +4) + (-1)( -4-3) + 2(3-0) |
=> 1/2 |8 + 7 +6 |
=> 21 sq.units
2
2) (-5, -1), (3,-5), (5,2)
A( -5, -1) B(3, -5) C(5, 2)
x1 = -5
x2 = 3
x3 = 5
y1 = -1
y2 = -5
y3 = 2
Formula :-
Area= 1/2 | x1(y2 - y3) +x2(y3 - y1) + x3(y1 - y2) |
=> 1/2 |-5(-5-2) + 3( 2+1) + 5(-1+5) |
=> 1/2 |35+9+20 |
=> 64 sq.units
2
=> 32 sq.units
(3) (0,0) (3,0) and ( 0,2)
sol) Area of triangle formed by the vertices
A( 0,0) B(3, 0) C(0, 2)
x1 = 0
x2 = 3
x3 = 0
y1 = 0
y2 = 0
y3 = 2
Formula :-
Area= 1/2 | x1(y2 - y3) +x2(y3 - y1) + x3(y1 - y2) |
=> 1/2 |0(0-2) + 3( 2-0) + 0(0-0) |
=> 1/2 |6 |
=> 3 sq.units.
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(2) Find the value of 'K' for which the points are collinear.
1) (7,-2) (5,1) (3,k)
2) (8,1) (k, -4) (2, -5)
3) K,K), (2,3) and (4, -1).
Sol) When points are collinear they lie on same line. It means it cannot form Triangle and the area /_\ = 0
1) (7,-2) (5,1) (3,k)
Formula :-
Area= 1/2 | x1(y2 - y3) +x2(y3 - y1) + x3(y1 - y2) |
0 = 1/2 | 7(1-k) + 5 (k-(-2) + 3(-2-1) |
0 = 1/2 |7-7k + 5k + 10 -9 |
0 = 1/2 | 8 - 2k |
0 = 1/2 |2(4-k|
4-k = 0
4=k
--------------------------
2) (8,1) (k, -4) (2, -5)
----------------------------
sol) x1 = 8
Formula :-
Area= 1/2 | x1(y2 - y3) +x2(y3 - y1) + x3(y1 - y2) |
0 = 1/2 |8(-4+5) + k(-5-1) + 2 (1+4)|
0 = 1/2 |8 -6k +10|
0= 1/2 |18-6k|
0 = 1/2 ||6(3-k)|
0 = 3(3-k)
k = 3
Formula :-
Area= 1/2 | x1(y2 - y3) +x2(y3 - y1) + x3(y1 - y2) |
0 = 1/2 | k(3+1) + 2(-1-k) +4(k-3) |
0 = 1/2 |4k -2 -2k + 4k -12|
0 = 1/2 |6k - 14|
0 = 3k - 7
3k = 7
k = 7/3
********************************************
(3) Find the area of the triangle formed by joining the mid-points of the sides of the triangle w hose vertices are (0, -1), (2,1) and (0,3) . find the ratio of this area to the area of the given triangle.
sol) let A=(0,-1) B =(2,1) C=(0,3) are the vertices of the triangle
RTP :- triangle formed by joining the mid-points of the sides of the traingle.
The mid-point of AB be D :-
A=(0,-1) B =(2,1)
D(x,y)
x1= 0
x2 =2
y1= -1
y2= 1
D(x,y) =( x1 +x2, y1+y2)
2 2
=> (0+2, -1+1) => (1,0)
2 2
The mid-point of BC be E :-
B =(2,1) C=(0,3)
E(x,y)
x1= 2
x2 =0
y1= 1
y2= 3
E(x,y) =( x1 +x2, y1+y2)
2 2
E(x,y) = (2+0, 1+3 ) => (1,2)
2 2
The mid-point of AC be F :-
A=(0,-1) C =(0,3)
F(x,y)
x1= 0
x2 =0
y1= -1
y2= 3
F(x,y) =( x1 +x2, y1+y2)
2 2
F(x,y) =( 0+0, -1+3)
2 2
F(x,y) = (0,1)
Finding area of triangle ABC
A=(0,-1) B =(2,1) C=(0,3)
x1=0
x2 =2
x3=0
y1=-1
y2=1
y3=3
Area= 1/2 | x1(y2 - y3) +x2(y3 - y1) + x3(y1 - y2) |
=> 1/2 |0(1-3) + 2(3+1) + 0(-1-1)|
=> 1/2 |8|
=> 4 sq.units
Finding area of triangle DEF
D=(1,0) E =(1,2) F=(0,1)
x1=1
x2 =1
x3=0
y1=0
y2=2
y3=1
Area= 1/2 | x1(y2 - y3) +x2(y3 - y1) + x3(y1 - y2) |
=> 1/2 | 1(2-1) + 1(1-0) + 0(0-2) |
=>1/2 |2|
=> 1 sq. units.
Finding ratio of two areas of triangles :-
ABC : DEF = 4 :1
*********************************************
(4) Find the area of the quadrilateral whose vertices , taken in order , are ( -4, -2) (-3,-5) (3, -2) and (2,3).
sol) Let A= (-4,--2) B=(-3, -5) C=( 3, -2) D=(2,3)
Two traingles are formed "ABC" & " ACD" by joining "AC
If we find the area of two traingles "ABC and ACD then we can easily get area of quadrilateral "ABCD"
ABCD = /_\ABC + /_\ ACD
Finding area of /_\ ABC :-
A=(-4,-2) B =(-3,-5) C=(3,-2)
x1=-4
x2 =-3
x3=3
y1=-2
y2=-5
y3=-2
Area= 1/2 | x1(y2 - y3) +x2(y3 - y1) + x3(y1 - y2) |
=>1/2 | -4(-5+2) +(-3)(-2+2) +3(-2+5)|
=> 1/2 |12-0+9|
=> 1/2 |21|
=> 21 square units .
2
Area of /_\ ADC :-
A=(-4,-2) D =(2,3) C=(3,-2)
x1=-4
x2 =2
x3=3
y1=-2
y2=3
y3=-2
Area= 1/2 | x1(y2 - y3) +x2(y3 - y1) + x3(y1 - y2) |
=>1/2 | -4(3+2) -3(-2+2) + 3(-2-3)|
=> 1/2 | -20 -0 - 15|
=1/2 | -35|
Since area cannot be negative
=> 35
2
Area of Quadrilateral ABCD = /_\ ABC + /_\ ADC
=> 21 + 35
2 2
=> 56
2
=> 28 sq.units
**********************************************
sol) Let A=(8, -5) B=(-2, -7) C=(5,1)
Using Distance let us find AB=c, BC=a, and CA=b
c = _/(-2-8)^2 + (-7+5)^2
=> _/100 + 4)
=> _/104
a= _/(5+2)^2 + (1+7)^2
=> _/49 + 64
=> _/113
b = _/(5-8)^2 + (1+5)^2
=> _/9 + 36
=> _/ 45
Heron's Formula :-
Using above formula, we can find the area of the triangle, but this is complicated and we can use following easier method.
Area= 1/2 | x1(y2 - y3) +x2(y3 - y1) + x3(y1 - y2) |
A=(8, -5) B=(-2, -7) C=(5,1)
x1=8
x2=-2
x3=5
y1=-7
y2=-7
y3=1
Area = 1/2 | 8(-7-1) -2(1-(-5)) +5(-5-(-7))|
=> 1/2 | 8(-8) -2(6) +5(2) |
=> 1/2 | -66|
Since area cannot be negative
=> 1/2 * 66
=> 33 sq. units.
*****************************************
Straight Lines
ching and ping are discussing to find solutions for a linear equation in two variable.
ching : Can you find solutions for 2x+3y=12
ping : Yes, I have done this, see
2x + 3y = 12
3y = 12 - 2x
y = 12 - 2x
3
ching : can you write these solutions in order pairs.
Ping : Yes (0,4), (3,2), (6,0), (-3,6)
Ching, can you plot these points on the coordinate plane.
Ching : I have done case like this
Ping : What do u observe?
What does this line represent?
Ching : It is a straight Line.
Ping : Can you identify some more points on this line?
Can you help Ching to find some more points on this line?
.........., ..........., ..............., ................,
___
And In this line, what is AB called?
___
AB is a line segment.
*****************************************
Example 22. The end points of a line are (2,3), (4,5). Find the slope of the line.
sol) Points of a line are (2,3) (4,5) then slope of the line
m = y2 - y1
x2 - x1
= 5 - 3 = 2 = 1
4 - 2 2
Slope of the given line is 1.
*****************************************
Example- 23. determine x so that 2 is the slope of the line through P(2,5) and Q(x,3)
sol) Slope of the line passing through P(2,5) and Q(x,3) is 2.
Slope of a PQ = y2 - y1 = 3 - 5 = - 2
x2 - x1 x - 2 x - 2
= - 2 = 2
x - 2
2x - 4 = -2
2x = 2
x = 1
******************************************
Exercise- 7.4
(1) Find the slope of the line joining the two given points.
1) ( 4, -8) and (5, -2)
sol) Slope of line passing through
A=(4, -8) and B=(5, -2)
x1 = 4
x2 = 5
y1 = -8
y2 = -2
=> -2 -(-8)
5 - 4
=> 6 = 6
1
.^. the slope of the line is 6
**************************
2) (0,0) and ( _/3 , 3)
sol) A=(0,0) B=( -/3, 3)
x1 = 0
x2 = _/3
y1= 0
y2 = 3
=> 3 - 0
_/3 - 0
=> 3
_/3
=> _/3 *-/3 (because _/3 * _/3 = 3 )
_/3
_/3 is the slope of the line
*************************
3) (2a, 3b) and ( a , -b)
sol) Let A = (2a, 3b) B =(a, -b)
x1 = 2a
x2 = a
y1 = 3b
y2 = -b
=> -b - 3b
a - 2a
=> -4b
-a
4b is the slope of a line
a
******************************************
4) (a,0) and (0,b)
sol) Let A=(a,0) B=(0,b)
x1 = a
x2 = 0
y1 = 0
y2 =b
=> b - 0
0 - a
=> -b is the slope of a line
a
****************************************
5) A(-1.4, -3.7), B ( -2.4, 1.3)
sol) x1 = -1.4 x2 = -2.4 y1 = -3.7 y2= 1.3
=> 1.3 + 3.7
-2.4 + 1.4
=> 5
-1
=> -5 is the slope of a line.
********************************************
6) A(3, -2), B (-6, -2)
sol) x1 = 3 x2= -6 y1 = -2 y2= -2
=> -2 -(-2)
-6 - 3
=> o is the slope of the line
*********************************************
7) A ( -3 1/2, 3), B( -7, 2 1/2)
sol) x1 = -3 1/2 = -7/2 x2= -7 y1 = 3 y2 = 2*1/2 = 5/2
=> 5/2 - 3
-7 + 7/2
=> 5 -6
2
-14 + 7
2
=> -1/2
-7/2
=> 1/7 is the slope of a line
******************************************
8) A(0,4), B(4,0)
sol) x1 = 0 x2=4 y1=4 y2=0
=> 0 - 4
4 - 0
= -1 is the slope of a line
*****************************************
(*) Optional Exercise
1). Center of the circle Q is on the Y-axis. And the circle passes through the points (0,7) and (0,-1). Circle intersects the positive X-axis at (P,0). What is the value of 'P'.
sol)
Equation of the circle having center at y-axis is of the form
(x-0)^2 + (y-k)^2 = a^2
Since it passes through (0,7)
x = 0, y = 7
(0-0)^2 + (7-k)^2 = a^2
(7 - k)^2 = a^2
49 + k^2 - 14k = a^2 ------- (!)
Also the circle passes through ( 0, -1)
x = 0 , y = -1
(0 - 0)^2 + (-1 - k)^2 = a^2
(-1 - k)^2 = a^2
1 + k^2 +2k = a^2 -------- (!!)
Equating (!) & (!!)
49 + k^2 - 14k = 1 + k^2 + 2k
48 = 16k
48 = k
16
3 = k
substitute k=3 in any one equation
we get
1 + (3)^2 + 2(3) = a^2
1 + 9 + 6 = a^2
16 = a^2 or a = 4
We have
Equation of a circle with center on y-axis is
x^2 + (y-k)^2 = a^2
substitute the value of k=3 and a^2 = 16 we get
x^2 + (y-3)^2 = 16
But the circle intersects x-axis at (p,0),
we get
x = p, y = 0
(p)^2 + (0-3)^2 = 16
p^2 + 9 = 16
p^2 = 7
p = + _/7
****************************************
2). A triangle △ABC is formed by the points A(2,3), B(-2,-3), C(4, -3). What is the point of intersection of side BC and angular bisector of A.
sol) We know that : An angle bisector of a triangle divides the opposite side into two segments that are proportional to the other two sides of the triangle.
BD = AB
DC AC ---------- (!)
Now, using distance formula
Length of A(2,3) B(-2, -3)
x1 = 2 y1 = 3
x2 = -2 y2 = -3
AB = √(-2 -2)^2 + (-3 - 3)^2
AB = √16 + 36
AB = √52
AB = √4 * 13
AB = 2 √13 -------- (!!)
Distance between A(2,3) C(4, -3)
x1 = 2 y1 = 3
x2 = 4 y2 = -3
AC = √(4-2)^2 + (-3-3)^2
AC = √4 + 36
AC = √40
AC = √4 * 10
AC = 2 √10 ------- (!!!)
Substituting (!!) and (!!!) in (!)
we get
BD = AB
DC AC
BD = 2 √13
DC 2 √10
BD = √13
DC √10
BD : DC = √ 13 : √10 -------- (!V)
So, the line segment BC is divided in the ration of √13 : √10
Now, by using section formula
D (x,y) = ( mx2 + nx1 , my2 + ny1 )
m + n m + n
D(x,y) = (√13(4) + √10(-2) , √13(-3) + √10(-3)
√13 + √10 √13 + √10
substituting √13 = 3.606, √10 = 3.162
D(x,y)
= 4(3.606) - 2(3.162), -3(3.606) - 3(3.162)
3.606 + 3.162 3.606 + 3.162
D(x,y) = (14.424 - 6.324, -10.818 - 9.486 )
3.606 + 3.162 3.606 + 3.162
D(x,y) = ( 8.1, -20.304 )
6.768 6.768
D(x,y) = (1.12 , -3)
Therefore the coordinates of point of intersection are (1.12, -3)
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3). The side of BC of an equilateral triangle △ABC is parallel to X-axis. Find the slopes of line along sides BC, CA and AB.
sol) In an equilateral Triangle all angles are equal.
i.e., ∠A=∠B=∠C= 60 deg
Sum of the angles of triangle = 180 deg
Given : BC is parallel to x-axis
We know that the slope of any line parallel to x-axis is 0
.^. Slope of BC = 0 deg
(!!) Slope of CA :-
Tan ∠CEX = Tan 120deg
=> Tan (180 - 60)
=> Tan (-60)
=> - √3
(!!!) Slope of AB :
Tan ∠ADX = Tan 60deg
=> √3
****************************************
4). A right triangle has sides 'a' and 'b' where a>b. If the right angle is bisected then find the distance between orthocenters of the smaller triangles using coordinate geometry.
sol)
*****************************************
5). Find the centroid of the triangle formed by the line 2x + 3y - 6 = 0. With the coordinate axis.
sol) Given Line : 2x + 3y = 6
Divide the Line by constant 6
we get,
2x + 3y = 6
6 6 6
x + y = 1
3 2
(x/3) Since x intercept is at 3. so line will cut the x-axis at (3,0)
(y/2) and y intercept is at 2, so line will cut the y-axis at (0,2)
.^. Coordinates of triangle are
(3,0) (0,2) (0,0)
x1 = 3, x2=0, x3=0
y1 = 0, y2 = 2, y3 = 0
So centroid is
(x,y) = x1 + x2 + x3 y1+y2+y3
3 3
=> 3 + 0 + 0 2+0+0
3 3
Centroid (x,y)= (1, 2/3)
*****************************************
In general for the points A(x1, 0), B(x2, 0) on the X-axis, the distance between A and B is
|x2 - x1|
example : A=(-2, 0) and B=(-6, 0)
|x2 - x1 | = -6 -(-2) | = |-6+2| = |-4| = 4 units
Similarly, if two points lie on Y-axis, then the distance between the points A and B would be the difference between their y-coordinates of the points.
The distance between two points (0, y1) ( 0, y2) would be |y2 - y1|
example: A = (0,2) B=(0,7)
|y2 - y1| = | 7-2| = 5 units.
****************************************
Example-1. What is the distance between A=(4,0) and B=(8,0)
sol) Here the points lie on x-axis
The distance between the points will be The difference in the x-coordinates
|x^2 - x1| = | 8 - 4| = 4 units.
*****************************************
Example-2. A and B are two points given by (8,3) (-4,3). Find the distance between A and B.
sol) Since the y-coordinates(y1 = 3 and y2=3) are equal, points lie on a line, parallel to x-axis.
Here x1 and x2 are lying in two different quadrants and y-coordinates are equal.
Distance AB = |x2 - x1|
= |-4 - 8| = |-12| = 12 units
*****************************************
Example-3. Let's find the distance between two points A(4,2) and B(8,6)
Sol) Compare these points with
x1 = 4 x2 = 8
y1 = 3 y2 = 6
Using distance formula
Distance AB = d
= √(8-4)^2 + (6-3)^2
= √16 + 9
= √25 = 5 units
*****************************************
Example 4. Show that the points
A(4,2),
B(7,5) and
C(9,7) are three points lie on a same line.
Sol) Now, we find the distances AB, BC, AC
AB = (4,2) (7,5)
x1 = 4 x2 = 7
y1 = 2 y2 = 5
= √(7-4)^2 + (5-2)^2
= √9 + 9
=√18
=√9 *2
= 3√2 units
BC = √(9-7)^2 + ( 7-5)^2
= √4 + 4
= √8
= 2√2 units
AC = √(9-4)^2 + (7-2)^2
= √25 + 25
= √50
= 5√2 units
Now AB + BC = 3√2 + 2√2
= 5√2 = AC.
Therefore, that the three points (4,2), (7,5) and (9,7) lie on a straight line.
(Points that lie on the same line are called collinear points).
*****************************************
Example-5. Are the points (3,2), (-2,-3) and (2,3) form a triangle?
sol) Let us apply the distance formula to find the distances PQ, QR and PR, where
P(3,2), Q(-2,-3) and R(2,3) are the given points. We have
PQ = √(-2-3)^2 + (-3-2)^2
= √25 + 25
= √50
= 5√2 = 7.07 units (approx)
QR = √(2-(-2)^2 + (3-(-3)^2
= √(4)^2 + (6)^2
= √52
QR = 7.21 units (approx)
PR =√(2-3)^2 + (3-2)^2
= √(-1)^2 + 1^2
= √2
PR = 1.41 units (approx)
Since the sum of any two of these distances is greater than the third distance, therefore, the points P,Q and R form a triangle and all the sides of triangle is unequal
********************************************
Example-6. Show that the points
(1,7),
(4,2),
(-1,-1) and
(-4,4) are the vertices of a square.
sol) One way of showing that ABCD is a square is to use the property that all sides should be equal and both its diagonals should also be equal.
Now.
AB = √(1-4)^2 + (7-2)^2
= √9+25
= √34 units
BC = √(4+1)^2 + (2+1)^2
= √34 units
CD = √(-1+4)^2 + (-1-4)^2
= √34 units
DA = √(-4-1)^2 + (4+7)^2
= √34 units
Diagonals are
AC = √(1+1)^2 + (7+1)^2
= √68 units
BD = √(4+4)^2 + (2-4)^2
= √68 units
Since AB=BC=CD=DA and AC=BD.
So all the four sides of the quadrilateral ABCD are equal and its diagonals AC and BD are also equal.
Therefore, ABCD is a square.
*****************************************
Example-7. Madhuri, Meena, pallavi are seated at
A(3,1)
B(6,4)
C(8,6) respectively.
Do you think they are seated in a line?
Give reasons to your answer.
sol) Using distance formula, we have
AB = √(6-3)^2 + (4-1)^2
= √18
= √9 * 2
= 3√2 units.
BC = √(18-6)^2 + (6-4)^2
= √4 + 4
=2√2 units.
AC =√(8-3)^2 = (6-1)^2
= √50
=5√2 units
Since, AB + BC = 3√2 + 2√2
= 5√2 = AC,
we can say that the points A,B and C are collinear.
Therefore , they are seated in a line.
*****************************************
Example-8. Find a relation between x and y such that the point (x,y) is equidistant from the points (7,1) and (3,5).
sol) Let P(x,y) be equidistant from the points
A(7,1) and B(3,5).
Given that AP=BP
√(x-7)^2 + (y-1)^2 = √(x-3)^2 + (y-5)^2
√(x^2 - 14x + 49) + (y^2 - 2y +1) = √(x^2 -6x+9) + (y^2 - 10y + 25)
√(x^2 +y^2 -14x-2y+50) - (x^2+y^2-6x-10y+34) = 0
So, AP^2 = BP^2
-8x + 8y = -16
divide by 8
we get,
i.e., x- y = 2 which is the required relation.
******************************************
Example-9. Find the point on the y-axis which is equidistant from both the points A(6,5) and B(-4,3).
sol) We know that a point on the Y-axis is of the form (0,y).
So, let the point P(0,y) be equidistant from A and B. Then
PA = √(6-0)^2 + (5-y)^2
PB = √(-4-0)^2 + (3-4)^2
PA^2 = PB^2
So,
(6-0)^2 + (5-y)^2 = (-4-0)^2 + (3-y)^2
i.e., 36 + 25 + y^2 - 10y = 16 + 9 + y^2 - 6y
i.e., 4y = 36
i.e., y = 9
So, the required point is (0,9)
Let us check our solution :
AP =√(6-0)^2 + (5-9)^2
= √52
BP =√(-4-0)^2 + (3-9)^2
= √52
So, (0,9) is equidistant from (6,5) and (4,3)
*****************************************
Exercise - 7.1
(1) Find the distance between the following pairs of points
1) (2,3) and (4,1)
sol) Let A=(2,3) and B= (4,1)
(x1, y1) = (2,3) and (x2,y2) = (4,1)
x1 = 2
x2 = 4
y1= 3
y2 = 1
Using distance formula :-
AB = _/(4-2)^2 + ( 1-3)^2
AB = _/(2)^2 + (-2)^2
AB = _/ 4 + 4
AB = _/8
AB = _/4*2
AB = 2_/2
*************
2) (-5,7) and ( -1,3)
sol) Let A = (-5,7) and B = (-1, 3)
(x1, y1) = ( -5, 7) and (x2,y2) = ( -1, 3)
x1 = -5
x2 = -1
y1 = 7
y2 = 3
Using Formula :-
AB=_/ (-1-(-5)^2 + ( 3-7)^2
AB = _/ (-1+5)^2 + (-4)^2
AB = _/ (4)^2 + 16
AB = _/ 16 +16
AB = _/32
AB = _/ 16*2
AB = 4_/2 units
*****************
3) (-2, -3) and (3, 2)
sol) Let A = (-2, -3) and B =(3,2)
x1 = -2
x2 = 3
y1= -3
y2 = 2
using Formula :-
AB= _/(3-(-2))^2 + ( 2-(-3))^2
AB = _/ (3+2)^2 + (2+3)^2
AB = _/ (5)^2 + (5)^2
AB = _/ 25 + 25
AB = _/ 50
AB = _/25*2
AB = 5_/2 units.
*********************
4) (a,b) and (-a, -b)
sol) Let A = (a-b) and B =(-a, -b)
(x1, y1) = ( a , b) (x2,y2) = (-a, -b)
x1 = a
x2 = -a
y1 = b
y2 = -b
Using Formula :-
AB = _/(-a -a)^2 + (-b - b)^2
AB = _/ (-2a)^2 + ( -2b)^2
AB = _/ 4a^2 + 4b^2
AB = _/4(a^2 + b^2)
AB = 2_/(a^2 + b^2)
***************************
(2) Find the distance between the points (0,0) and (36,15).
sol) Let A= (0,0) and B =(36,15)
(x1,y1) = (0,0) and (x2, y2) = (36,15).
x1 = 0
x2 = 36
y1 = 0
y2 = 15
Using Formula :-
AB = _/(36-0)^2 + (15-0)^2
AB = _/ 1296 + 225
AB = _/1521
AB = _/(39)^2
AB = 39
***************************
(3) Verify that the point (1,5) (2,3) and (-2, -1) are collinear or not.
sol) Collinear :- Points that lie on the same line are called collinear points.
If AB + BC = AC . We say that points lie on a straight line.
Now,
Let A = (1,5) B = (2,3) C = ( -2, -1)
Now we find the distances AB, BC, AC
Distance between AB:-
(x1,y1) = (1, 5) and ( x2, y2) = ( 2,3)
x1= 1
x2 = 2
y1= 5
y2= 3
Using Formula :-
AB = _/ (2-1)^2 + (3 - 5) ^2
AB = _/ (1)^2 + (-2)^2
AB = _/1 + 4
AB = _/5 -----------(1)
Distance between BC :-
(x1, y1) = ( 2,3) and (x2,y2) = ( -2, -1)
x1 = 2
x2 = -2
y1 = 3
y2 = -1
Using Formula :-
BC = _/ (-2 -2)^2 + ( -1-3)^2
BC = _/ (-4)^2 + (-4)^2
BC = _/ 16 + 16
BC = _/ 32
BC = _/ 16*2
BC = 4_/2 ------------------------(2)
Distance between AC :-
(x1, y1) = (1,5) and (x2, y2) = (-2, -1)
x1 = 1
x2 = -2
y1 = 5
y2= -1
Using Formula :-
AC = _/ (-2-1)^2 + ( -1-5)^2
AC = _/ (-3)^2 + (-6)^2
AC = _/ (9 + 36)
AC = _/ 45
AC = _/ 9 * 5
AC = 3_/5 ----------------(3)
Verifying
AB + BC = AC
_/5 + 4_/2 =/= 3_/5
Here, the sum of any two of these distances is not equal to the third distance.
.^. the three given points do not lie on a straight line
Hence, they are not collinear.
******************************************************************
(4) Check whether (5,-2) , (6,4) and (7, -2) are the vertices of an isosceles triangle.
sol) Let A = (5, -2) B =( 6,4) C= ( 7, -2)
Isosceles triangle :- Any two of the three sides are equal
i.e, AB = BC or BC = CA or CA = AB
Distance between AB :-
(x1, y1) = ( 5, -2) and (x2, y2) = ( 6, 4)
x1 = 5
x2 = 6
y1 = -2
y2 = 4
Using Formula :-
AB = _/ (6 - 5) ^2 + ( 4- (-2)^2
AB = _/ (1)^2 + (6)^2
AB = _/ 1 + 36
AB = _/ 37 --------------------(1)
Distance between BC
(x1, y1) = ( 6 , 4 ) and (x2, y2) = ( 7, -2)
x1 = 6
x2 = 7
y1 = 4
y2 = -2
Using Formula :-
BC = _/ ( 7 - 6)^2 + ( -2 - 4) ^2
BC = _/ (1)^2 + (-6)^2
BC = _/ 1 + 36
BC = _/ 37 --------------------(2)
Distance between AC :-
(x1, y1) = ( 5, -2) and ( x2, y2) = ( 7, -2)
x1 = 5
x2 = 7
y1 = -2
y2 = -2
Using Formula :-
AC = _/ (7 -5)^2 + ( -2 -(-2)^2
AC = _/(2)^2 + ( -2+2)^2
AC = _/4 + 0
AC = 2 ------------------(3)
Since AB = BC
two sides are equal . Hence it is an isosceles triangle.
*********************************************
(5) In a class room, 4 friends are seated at the points A,B,C and D as shown in figure. Jarina and Phani walk into the class and after observing for a few minutes Jarina asks phani " Don't you think ABCD is a square? " Phani disagrees.
using distance formula, find which of them is correct. why?
sol) One way of showing that ABCD is a square is to use the property that all its sides should be equal and both its diagonal should also be equal.
i.e, sides = AB = BC = CD = DA
diagonals = AC = BD
Now from the figure.
A(3, 4), B( 6, 7), C (9, 4) and D( 6, 1)
Distance between AB :-
(x1, y1) = (3, 4) and (x2, y2) = ( 6, 7)
x1 = 3
x2 = 6
y1 = 4
y2 = 7
Using Formula :-
AB=_/(6 -3)^2 + ( 7 - 4)^2
AB = _/ (3)^2 + ( 3)^2
AB = _/ 9 + 9
AB = _/18
AB = _/2*9
AB = 3_/2 ----------------------(1)
Distance between BC :-
(x1, y1) = ( 6, 7) and (x2,y2) = ( 9,4)
x1 = 6
x2 = 9
y1 = 7
y2 = 4
Using Formula :-
BC = _/ ( 9 -6)^2 + ( 4 - 7)^2
BC = _/ (3)^2 + (-3)^2
BC = _/9 + 9
BC = _/18
BC =_/ 2*9
BC = 3_/2 ------------(2)
(x1, y1) = ( 9, 4) and (x2, y2) = ( 6, 1)
x1 = 9
x2 = 6
y1 = 4
y2 = 1
CD = _/(6 - 9)^2 + ( 1-4)^2
CD = _/ (-3)^2 + (-3)^2
CD = _/ (9) + (9)
CD = _/ 18
CD = _/ 2*9
CD =3 _/2---------------(3)
Distance between DA :-
(x1, y1) = ( 6, 1) and ( x2, y2) = ( 3, 4)
x1 = 6
x2 = 3
y1 = 1
y2 = 4
DA = _/ ( (3 -6) ^2 + ( 4 -1)^2
DA = _/ (-3)^2 + (3)^2
DA = _/ 9 + 9
DA = _/18
DA = _/ 2 * 9
DA = 3_/2 --------------(4)
Diagonal AC :-
A =( 3,4) and C = ( 9, 4)
x1 = 3
x2 = 9
y1 = 4
y2 = 4
AC = _/ ( 9-3)^2 + ( 4 -4)^2
AC = _/ (6)^2 + ( 0)^2
AC = _/ 36 + 0
AC = 6
Diagonal BD :-
B( 6,7) and D(6,1)
x1 = 6
x2 = 6
y1 = 7
y2 = 1
BD = _/ (6-6)^2 + ( 1-7)^2
BD = _/ (0)^2 + ( -6)^2
BD = _/ 0+ 36
BD = 6
Here, sides =AB = BC = CD = DA
Diagonals = AC = BD
Hence, ABCD is a square.
*******************************
(6) Show that the following points form a equilateral triangle A(a,0), B(-a,0), C(0, a_/3)
sol) If the distance between all the sides are equal . Then we say that the triangle is equilateral triangle.
Distance between AB :-
(x1,y1) = (a , 0) and (x2, y2) = ( -a, 0)
x1 = a
x2 = -a
y1 = 0
y2 = 0
Using Formula :-
AB = _/(-a -a)^2 + ( 0 -0)^2
AB = _/ (-2a)^2 + 0
AB = _/ 4a^2
AB = 2_/a^2 -----(1)
Distance between BC :-
(x1, y1) = ( -a , 0) and (x2,y2) = ( 0, a_/3)
x1 = -a
x2 = 0
y1 = 0
y2 = a_/3
Using Formula :-
BC = _/(0 + a)^2 + (a_/3 - 0)^2
BC = _/ a^2 + 3a^2
BC = _/4a^2
BC = 2_/a^2 ----------(2)
Distance Between AC :-
(x1, y1) = (a, 0) and (x2, y2) = (0, a_/3)
x1 = a
x2 = 0
y1 = 0
y2 = a_/3
Using Formula :-
AC = _/(0 -a) ^2 + (a_/3 - 0 )^2
AC = _/ a^2 + 3a^2
AC =_/4a^2
Ac =2_/a^2 ----(3)
Here (1), (2), and (3) have same value. Hence they form the equilateral triangle.
********************************************************************
(7) Prove that the points (-7, -3), (5, 10), (15,8) and (3, -5) taken in order are the corners of a parallelogram. And find its area .
( Hint : Area of rhombus = 1/2 * product of its diagonals)
sol) Let A= ( -7, -3), B =( 5, 10), C = ( 15, 8), D =( 3, -5) are the given points.
Proving points taken in order are the corners of a parallelogram :-
In parallellogram opposite sides are equal
Distance between AB :-
(x1, y1) = ( -7, -3) and (x2, y2) = (5, 10)
x1 = -7
x2 = 5
y1 = -3
y2 = 10
Using Formula :-
AB = _/ ( 5 -(-7)^2 + (10-(-3)^2
AB = _/ (5 + 7)^2 + ( 10 + 3)^2
AB = _/ (12)^2 + ( 13)^2
AB = _/ 144 + 169
AB = _/ 313 -----------------(1)
Distance between BC :-
(x1, y1) = (5, 10) and (x2, y2) = ( 15, 8)
x1 = 5
x2 = 15
y1 = 10
y2 = 8
Using Formula :-
BC = _/ (15 - 5)^2 + ( 8 - 10)^2
BC = _/ (10)^2 + ( -2)^2
BC = _/ 100 + 4
BC = _/ 104 --------(2)
Distance between CD :-
(x1, y1) = ( 15, 8) and (x2, y2) = ( 3, -5)
x1 = 15
x2 = 3
y1 = 8
y2 = -5
CD = _/ (3 - 15)^2 + ( -5 -8)^2
CD = _/ 144 + 169
CD = _/ 313 --------------(3)
Distance between DA :-
(x1, y1) = ( 3, -5) and (x2, y2) = ( -7, -3)
x1 = 3
x2 = -7
y1 = -5
y2 = -3
DA = _/ (-7 -3)^2 + ( -3 +5)^2
DA = _/ (-10)^2 + (2)^2
DA = _/ 104 ----------------------(4)
Here AB = CD and BC = DA. Opposite sides are equal. Hence it is a parallelogram.
Finding area :-
Diagonal of a parallelogram divides it into two congruent triangles.
Divide the parallelogram. ABCD into two triangles ABC and ACD
Area (ABCD) = Area ( /_\ ABC) + Area ( /_\ACD)
We have formula for finding area of triangle with three points:-
(x1,y1) = (-7, -3) (x2,y2) = ( 5, 10) (x3,y3) = ( 15, 8)
x1 = -7
x2 = 5
x3 = 15
y1 = -3
y2 = 10
y3 = 8
Area(/_\ ABC) = 1/2 [ x1(y2 -y3) + x2(y3 - y1) + x3(y1-y2)]
=> 1 [ -7 ( 10-8) + 5 ( 8+3) + 15 (-3-10)]
2
=> 1 [ -7(2) + 5(11) + 15(-13)]
2
=> 1 [ -14 + 55 -195]
2
=> 1 [ -154 ]
2
Since the area of two(2) triangles formed inside the parallelogram are equal.
As we got area one triangle we can just simply multiply with "2"
i.e,
*****************************************
(8) Show that the points ( -4,-7), (-1,2), ( 8,5) and (5,-4) taken in order are the vertices of a rhombus.
sol) In rhombus all sides are equal but diagonals are not equal
let A ( -4, -7) , B(-1, 2) , C ( 8, 5) and D ( 5, -4) are the given points,
We have,
AB = _/ (-1 + 4) ^2 + ( 2 +7)^2
AB = _/ 9 + 81
AB = _/ 90
AB = _/9 * 10
AB = 3_/10 ------------------(1)
BC = _/ (8 +1)^2 + ( 5-2)^2 => _/ 81 + 9 => 3_/10
CD = _/(5-8)^2 + ( -4 -5)^2 => _/ (9 + 81) => 3_/10
DA = _/ ( 5+4)^2 + ( -4 +7)^2 => _/( 9 +81) = > 3_/10
Here AB = BC = CD = DA, so all four sides of the quadrilateral ABCD are equal.
Now let us find distances of the diagonals AC and BD
AC = _/(8+4)^2 + (5 +7)^2 => _/ (12)^2 + (12)^2 => 12_/2
BD = _/ ( 5 + 1)^2 + ( -4-2)^2 => _/ (6)^2 + (6)^2 => 6_/2
Here , AC =/= BD
.^. AB, BC, CD, DA are the vertices of a rhombus.
****************************************************
(9) Name the type of quadrilateral formed, if any, by the following points, and give reason for your answer.
(1) (-1, -2) , (1,0),(-1,2), (-3,0)
(2) (-3,5), (3,1), (0,3), (-1, -4)
(3) (4,5), (7,6), (4,3), (1,2)
sol) Let A =(-1, -2) B =(1,0) C =(-1, 2) D =(-3,0)
AB =_/(1+1)^2 + ( 0-2)^2 = _/ (4 + 4) = _/8 = 2_/2
BC = _/(-1-1)^2 + ( 2-0)^2 = _/ (4 + 4) = _/8 = 2_/2
CD = _/(-3+1)^2 + ( 0-2)^2 = _/8 = 2_/2
DA = _/(-3+1)^2 + ( 0 +2)^2 = _/8 = 2_/2
Diagonal :-
AC = _/(-1 +1)^2 + (2 +2)^2 = _/16 = 4
BD = _/(-3-1)^2 + (0+0)^2 = 4
Here, AB =BC = CD = DA and AC =BD
so all the four sides of the quadrilateral ABCD are equal and its diagonal are also equal.
.^. , ABCD is a square.
---------------------------------
Let A = (-3,5) B =(3,1) C = (0,3) D=(-1, -4)
Findng Distance AB :- A=( -3,5) B =(3,1)
x1 = -3
x2 = 3
y1 = 5
y2 = 1
Using Formula :-
AB = _/(3(-3)^2 + (1-5)^2
AB = _/(3+3)^2 + (-4)^2
AB = _/ 36 + 16
AB = _/ 52
AB = _/ 4* 13
AB = 2_/13
Finding Distance BC :- B =( (3, 1) C=(0,3)
x1 = 3
x2 = 0
y1 = 1
y2 = 3
Using Formula :-
BC = _/ (0-3)^2 + (3-1)^2
BC = _/9 + 4
BC = _/ 13
Finding distance CD :- C=( 0, 3) D =(-1, -4)
x1 = 0
x2 = -1
y1= 3
y2 = -4
CD = _/ (-1-0)^2 + ( -4-3)^2
CD = _/ 1 + 49
CD = _/50
Finding Distance DA :- D =( -1, -4) A=(-3, 5)
x1 = -1
x2 = - 3
y1 = -4
y2 = 5
DA = _/(-3-(-1)^2 + ( 5-(-4)^2
DA = _/ (-2)^2 + (9)^2
DA = _/4 + 81
DA = _/85
Since AB=/= BC =/=CD =/= DA
So, it is not a quadrilateral
We can find out what actual figure this points formed by using graph
A=( -3, 5) B=(3,1) C=(0,3) D=(-1, -4)
Here points A,B,C are collinear.
We check if AC + BC = AB
Finding Distance AC :- A =(-3,5) C =( 0, 3)
x1 = -3
x2 = 0
y1 = 5
y2 = 3
AC = _/ (0 -(-3)^2 + ( 3 -5)^2
AC = _/ (9) + ( -2)^2
AC = _/ 9 + 4
AC = _/13
We check if
AC + BC = AB
_/13 + _/13 = 2_/13
Hence, ABCD is a triangle, not a quadrilateral.
------------------------------------------------------------
(3) Let A=( 4,5) B=(7,6) C=(4,3) D=(1,2)
Finding distance AB :- A = ( 4,5) B= ( 7,6)
x1=4
x2=7
y1= 5
y2 =6
AB = _/ ( 7-4)^2 + ( 6 - 5)^2
AB = _/ 9 + 1
AB = _/10
Finding distance BC :- B=(7,6) C=( 4,3)
x1 = 7
x2 = 4
y1= 6
y2 =3
BC = _/((4-7)^2 + ( 3-6)^2
BC = _/ 9 + 9
BC = _/ 18
Finding distance CD :- C= (4,3) D=( 1,2)
x1 = 4
x2 = 1
y1 = 3
y2 = 2
CD = _/(1-4)^2 + ( 2-3)^2
CD = _/9 + 1
CD = _/10
Finding distance DA :- D=(1,2) A=(4,5)
x1 = 1
x2 = 4
y1 = 2
y2 = 5
DA = _/ ( 4-1)^2 + (5-2)^2
DA = _/ (9) + (9)
DA = _/ 18
Here, AB = CD = _/10 and BC=DA = _/18
Since, opposite sides of quadrilateral are equal.
It would be a parallelogram.
(10) Find the point on the x-axis which is equidistant from (2, -5) and (-2,9).
sol) Let A=( 2, -5) B=(-2, 9)
RTP :- Point on x-axis...
As we know on x-axis , y =0
Hence the required point to find on x-axis is C = (x, 0)
Given :- point on x-axis C= (x,0) is equidistant from "A" and "B"
i.e, AC = BC
Now,
Finding distance AC :- A=( 2, -5) C=( x, 0)
x1 = 2
x2 = x
y1 = -5
y2 = 0
AC = _/ ( x -2)^2 + ( 0 - (-5)^2
AC = _/ x^2 + 2^2 - 2(x)(2) + (5)^2
AC = _/ x^2 + 4 - 4x + 25
AC = _/ x^2 - 4x + 29 --------(1)
Finding distance BC :- B = ( -2, 9) C =(x,0)
x1 = -2
x2 = x
y1 = 9
y2 = 0
BC = _/(x -(-2)^2 + ( 0 -9)^2
BC = _/ (x +2)^2 + 81
BC = _/x^2 + 2^2 + 2(2)(x) + 81
BC = _/x^2 + 4 +4x + 81
BC = _/ x^2 + 4x + 85 -------------------(2)
Now AC = BC
_/(x^2 -4x +29) = _/ x^2 + 4x + 85
squaring on both sides
(_/(x^2 -4x +29))^2 = (_/ x^2 + 4x + 85 )^2
-4x - 4x = 85 - 29
-8x = 56
s = - 56
8
x = -7
Hence the required point is C = (-7, 0)
point C = (-7,0) is equidistant from A(2, -5) B(-2. 9)
*********************************************
(11) If the distance between two points (x,7) and (1, 15) is10, find the value of "x"
sol) Let A = (x,7) B=(1,15)
x1 = x
x2 = 1
y1 = 7
y2 = 15
Using Formula :-
AB =_/(1-x)^2 + ( 15-7)^2
10 =_/(1)^2 + (x)^2 - 2(1)(x) + (8)^2
10 = _/ 1 + x^2 - 2x + 64
10 = _/ x^2 -2x + 65
Now, squaring on both sides
(10)^2 = (_/x^2 -2x + 65)^2
100 = x^2 -2x + 65
x^2 - 2x = 100 - 65
x^2 - 2x = 35
x^2 - 2x - 35 = 0
x^2 -7x + 5x - 35 = 0
x(x - 7) + 5(x - 7) = 0
x-7 = 0 or x +5 = 0
x = 7 or x = -5
**************************
(12) Find the values of "y" for which the distance between the points P(2,-3) and Q(10,y) is 10 units.
sol) Given :-P = ( -2, 3) and Q =( 10, y)
Distance between PQ = 10 units
Finding Distance PQ :-
x1 = -2
x2 = 10
y1 = 3
y2 = y
_/( 10-2)^2 + (y-(-3)^2 = 10
_/ 64 + (y +3)^2 = 10
_/ 64 + y^2 + (3)^2 + 2(y)(3) = 10
-/ 64 + y^2 + 9 +6y = 10
_/ y^2 + 6y + 73 = 10
squaring on both sides, we get
y^2 + 6y + 73 = 100
y^2 + 6y + 73 - 100 = 0
y^2 + 6y - 27 = 0
y^2 + 9y - 3y - 27 = 0
y(y + 9) -3 ( y +9) = 0
(y-3) (y+9) = 0
y = 3 or y = -9 is the solution
*************************
(13) Find the radius of the circle whose center is (3,2) and passes through ( -5,6).
sol) Let 'r' be the radius of the given circle.
Radius(r) :- Distance between the center of the circle and any point on the circumference
Given :- Center (p) = ( 3,2) and passing through point Q=(-5,6)
Finding distance PQ :-
x1 = 3
x2 = -5
y1 = 2
y2 = 6
PQ= _/ (-5-3)^2 + ( 6 -2)^2
=> _/ 64 + 16
=> _/ 80
=> _/ 16 * 5
=> 4_/5
r = 4_/5 units
****************************
(14) Can you draw a triangle with vertices (1,5), (5,8) and (13, 14)? Give reason.
sol) Let A = ( 1,5) B = ( 5, 8) C = ( 13, 14)
Finding distance AB :- A=( 1, 5) B =( 5, 8)
x1 = 1
x2 = 5
y1 = 5
y2 = 8
AB = _/ ( 5-1)^2 + (8-5)^2
AB = _/ 16 + 9
AB = _/ 25
AB = 5
finding distance BC :- B=( 5, 8) C = ( 13, 14)
x1 = 5
x2 = 13
y1 = 8
y2 = 14
BC = _/(13 - 5)^2 + ( 14-8)^2
BC = _/ 64 + 36
BC = _/ 100
BC = 10
Finding distance AC :- A = ( 1,5) C = ( 13, 14)
x1 = 1
x2 = 13
y1 = 5
y2 = 14
AC = _/ (13-1)^2 + (14-5)^2
AC = _/ 144 + 81
AC= _/ 225
AC = 15
If A, B, C are collinear triangle is not possible
Lets check
AB = 5
BC = 10
AC = 15
AB + BC = AC
5 + 10 = 15
15 = 15
All points lie on the same line.
Thus, we cannot draw a triangle with the given vertices
*************************************************
(15) Find a relation between "x" and "y" such that the point (x,y) is equidistant from the points (-2,8) and (-3,-5).
sol) Let A = ( -2,8) B=( -3, 5) and P =(x,y)
Given :- point "P" is equidistant from A and B
i.e AP = BP
Distance between AP :- A = ( -2, 8) P = ( x,y)
x1= -2
x2 = x
y1 = 8
y2 = y
AP = _/ (x-(-2)^2 + ( y-8)^2
AP = _/ (x +2)^2 + (y-8)^2
AP = _/(x)^2 + (2)^2 + 2(x)(2) + (y)^2 + (8)^2 - 2(y)(8)
AP = _/ x^2 + 4 + 4x + y^2 + 64 - 16y
AP = _/x^2 + y^2 + 4x - 16y + 68
Distance between BP :- B =( -3, -5) P =(x,y)
x1 = -3
x2 = x
y1 = -5
y2 = y
BP = _/ x-(-3)^2 + ( y - (-5)^2
BP = _/ (x +3)^2 + ( y+5)^2
BP = _/ x^2 + (3)^2 + 2(x)(3) + y^2 + (5)^2 + 2(y)(5)
BP = _/ x^2 + 9 + 6x + y^2 + 25 + 10y
BP = _/ x^2 + y^2 + 6x + 10y + 34
Since AP = BP
4x - 6x -16y - 10y + 68 - 34 = 0
-2x - 26y + 34 = 0
divide by "2"
-x -13y + 17 = 0
x + 13y - 17 = 0
x + 13y = 17 is the required equation
*********************************************************
Section formula
The coordinates of the points P(x,y) which divides the line segment joining the points A(x1,y1) and
B(x2,y2), internally in the ratio m1 : m2 are
(m1x2 + m2x1 , m1y2 + m2y1 )
m1 + m2 m1 + m2
If the ratio in which P divides AB is k:1, then the coordinated of the point P are
(kx2 + x1 , ky2 + y1)
k+1 k+1
*****************************************
Example-10. Find the coordinates of the point which divides the line segment joining the points (4, -3) and (8,5) in the ratio 3:1 internally.
sol) Let P(x,y) be the required point. using the section formula
P(x,y) = (m1x2 + m2x1 , m1y2 + m2y1 )
m1 + m2 m1 + m2
we get,
x = 3(8) + 1(4) = 24 + 4 = 28 = 7
3 + 1 4 4
y = 3(5) + 1(-3) = 15 - 3 = 12 = 3
3+1 4 4
P(x,y) = (7,3) is the required point.
*****************************************
Example-11. Find the mid point of the line segment joining the points (3,0) and (-1,4)
sol) The mid point M(x,y) of the line segment joining the points (x1, y1) and (x2,y2)
M(x,y) = ( x1 + x2, y1 + y2 )
2 2
.^. The mid point of the line segment joining the points (3,0) and (-1,4) is
M(x,y) = (3 +(-1), 0+4) = ( 2, 4 ) = (1,2)
2 2 2 2
*****************************************
(*) Centroid of a Triangle
The coordinates of the centroid are given by
[ x1 + x2 + x3, y1 + y2 + y3 ]
3 3
Example-12. Find the centroid of the traingle whose vertices are
(3,-5)
(-7,4)
(10,-2) respectively.
sol) The coordinated of the centroid are
= [ x1 + x2 + x3, y1+y2+y3]
3 3
= [ 3 + (-7) + 10, (-5) + 4 + (-2) ]
3 3
= (2, -1) is the centroid
*****************************************
Example-13. In what ratio does the point (-4,6) divide the line segment joining the points A(-6,10) and B(3, -8)?
sol) Let (-4,6) divide AB internally in the ratio m1:m2.
Using the section formula, we get
(-4, 6) = (m1x2 + m2x1 , m1y2 + m2y1 )
m1 + m2 m1 + m2
(-4,6) = [3m1 -6m2, -8m1 + 10m2] --(!)
m1+m2 m1+m2
We know that if(x,y)= (a,b) then
x= a and y = b
So,
-4 = 3m1 - 6m2
m1 + m2
and
6 = -8m1 + 10m2
m1+m2
Now,
-4 = 3m1 -6m2 gives us
m1+ m2
-4m1 - 4m2 = 3m1 -6m2
i.e., 7m1 = 2m2
m1 = 2
m2 7
i.e., m1 : m2 = 2:7
We should verify that the ratio satisfies the y-coordinate also.
Now,
-8m1 + 10m2 (Dividing throughout by m2)
m1 + m2
-8 m1 + 10
= m2
m1 + 1
m2
= -8 * (2/7) + 10
(2/7) + 1
= -16/7 + 10
(9/7)
= -16 + 70
9
= 54/9 = 6
Therefore, the point (-4,6) divides the line segment joining the points A(-6,10) B(3,-8) in the ratio 2:7
*****************************************
(*) Trisectional Points of a Line
Example-14. Find the coordinates of the points of trisection(The points which divide a line segment into 3 equal parts are said to be the sectional points) of the line segment joining the points A(2,-2) and B(-7,4).
Sol) Let P and Q be the points of trisection of AB
i.e., AP = PQ = QB
A---------P----------Q---------B
(2,-2) (-7,4)
Therefore, P divides AB internally in the ration 1:2
Therefore, the coordinates of P are ( by applying the section formula)
P(x,y) = (m1x2 + m2x1 , m1y2 + m2y1 )
m1 + m2 m1 + m2
[ 1(-7) + 2(2), 1(4) + 2(-2) ]
1 + 2 1 + 2
i.e., [ -7 + 4 , 4 -4 ]
3 3
= [ -3 , 0 ]
3 3
= ( -1 , 0)
Now, Q also divides AB internally in the ratio 2:1
So, the coordinates of Q are
= [2(-7) + 1(2), 2(4) + 1(-2) ]
2 + 1 2 + 1
i.e., [ -14 + 2 , 8 -2 ]
3 3
= [ -12 , 6 ]
3 3
= [ -4, 2 ]
Therefore, the coordinates of the points of trisection of the line segment are P(-1,0) and Q(-4, 2)
*****************************************
Example-15. Find the ratio in which the y-axis divides the line segment joining the points (5, -6) and (1, -4) . Also find the point of intersection.
sol) Let the ratio be K : 1. Then by the section formula, the coordinates of the point which divides AB in the ratio K : 1 are
[ K(-1) + 1(5), K(-4) + 1(-6) ]
K + 1 K + 1
i.e., [ -K + 5, -4K -6 ]
K + 1 K + 1
This point lies on the y-axis, and we know that on the y-axis the abscissa is 0 .
Therefore,
-K + 5 = 0
K + 1
-K + 5 = 0
K = 5.
So, the ratio is K:1 = 5 : 1
Putting the value of K=5,
we get the point of intersection as
= (-5 + 5, -4(5) -6)
5 + 1 5 + 1
= (0, -20-6)
6
= ( 0, -26 )
6
= (0, -13 )
3
*****************************************
Example -16. Show that the points
A(7,3)
B(6,1)
C(8,2) and
D(9,4)
taken in that order are vertices of parallelogram.
sol) We know that the diagonals of a parallelogram bisect each other.
.^. So the midpoints of the diagonals AC and DB should be equal.
Now, we find the mid points of AC and DB by using
( x1 + x2, y1 + y2 )
2 2
Midpoint of AC = (7 +8, 3+2)
2 2
= ( 15, 5 )
2 2
Midpoint of DB = ( 9 + 6, 4+1 )
2 2
= ( 15, 5 )
2 2
Hence,
midpoint of AC = midpoint of DB
Therefore, the points A,B,C,D are vertices of a parallelogram.
*****************************************
Example-17. If the points
A(6,1)
B(8,2)
C(9,4) and
D(p,3)
are the vertices of a parallelogram, taken in order, find the value of P.
sol) We know that diagonals of a //gram bisect each other.
So,
the coordinates of the midpoint of AC = Coordinates of the midpoint BD
i.e., ( 6 + 9, 1+4) = ( 8 + p , 5 )
2 2 2 2
( 15, 5 ) = ( 8 + p, 5 )
2 2 2 2
15 = 8 +P
2 2
15 = 8 + p
p = 7.
*****************************************
Exercise - 7.2
(1) Find the co-ordinates of the point which divides the join of (-1, 7) and (4, -3) in the ratio 2:3.
sol) Let A= ( -1, 7) B=(4, -3) and m1:m2 = 2; 3
x1 = -1
x2 = 4
y1 = 7
y2 = -3
Let P(x,y) be the required point. Using the section formula.
Finding "x" :-
x = m1x2 + m2x1
m1 + m2
x1 = -1
x2 = 4
m1 :m2 = 2 : 3
x = 2(4) + 3(-1)
2 + 3
x = 8 - 3
5
x =
5
x = 1
Finding "y" :-
y = m1y2 + m2y1
m1 + m2
y1 = 7
y2 = -3
m1 : m2 = 2 : 3
y = 2(-3) + 3(7)
2 + 3
y = -6 + 21
5
y = 15
5
y = 3
Hence the required p(x,y) = p( 1, 3)
*********************************
(2) Find the coordinates of the points of trisection of the line segment joining (4, -1) and (-2, -3).
sol) Trisection :- It means dividing a line segment in three equal parts (OR) dividing a line segment in the ratio 1:2 and 2 :1 internally
A-------------P--------Q---------B
(4, -1) ( -2, -3)
P & Q are two points on AB such that
AP = PQ = QB (equal parts)
Finding "P":-
A----------P--------Q--------B
----1--- --------2--------
Case (1) :- When m1 : m2 = 1 :2
A=( 4,-1) and B=( -2, -3)
x1= 4
x2 = -2
y1 = -1
y2 = -3
Let P(x,y) be the required point, using section formula.
x = m1x2 + m2x1
m1 + m2
x = 1(-2) + 2(4)
1 + 2
x = -2 + 8
3
x =
3
x = 2
Finding "y":-
y = m1y2 + m2y1
m1 + m2
y = 1(-3) + 2(-1)
1 + 2
y = -3 -2
3
y = -5
3
Hence, point P(x,y) = P ( 2, -5/3)
Finding Q :-
A------P----Q--------B
------2---- ---1---
case(2) :- When m1 : m2 = 2 ;1
A=( 4,-1) and B=( -2, -3)
x1= 4
x2 = -2
y1 = -1
y2 = -3
Let Q(x,y) be the required point . Using section formula
x = m1x2 + m2x1
m1 + m2
x = 2(-2) + 1(4)
2 + 1
x = -4 + 4
3
x = 0
Finding "y":-
y = m1y2 + m2y1
m1 + m2
y = 2(-3) + 1(-1)
2 + 1
y = -6 -1
3
y = -7
3
Hence , Point Q(x,y) = Q( 0, -7/3)
A --------P -------------Q-------------B
(4, -1) (2, -5/3) (0, -7/3) (-2, -3)
*************************************
(3) Find the ratio in which the line segment joining the points ( -3, 10) and (6, -8) is divided by (-1, 6).
sol) Given :- Point P(-1,6) divide the line segment joining the points (-3,10) and (6,3)
A------------------P--------------------B
(-3,10) (-1,6) (6, -8)
RTP :- m1:m2=??
Using section formula, we get
x1 = -3
x2 = 6
y1 = 10
y2 = -8
P(-1, 6) = m1(6) + m2(-3) , m1(-8) + m2(10)
m1 + m2 m1 + m2
Now,
-1 = 6m1 - 3m2
m1 + m2
-1(m1 + m2) = 6m1 - 3m2
-m1 - m2 = 6m1 - 3m2
-m1 - 6m1 = -3m2 + m2
-7m1 = - 2m2
m1 = 2
m2 7
m1:m2 = 2:7
.^. , the point (-1,6) divides the line segment joining the points A(-3, 10) and B(6, -8) in the ratio 2 :7
*********************************************************************************
(4) If (1,2) , (4,y) , (x ,6) and (3,5) are the vertices of a parallelogram taken in order, find "x" and "y".
sol) Let A=(1,2) B=(4,y) C=(x,6) D=(3,5) are the vertices of the parallelogram
In parallelogram diagonals bisect each other.
.^. , the mid-points of the diagonals AC and BD should be equal.
Let us now find the mid-points of diagonal(AC) and diagonal(BD)
mid-point of AC :- A(1,2) C(x,6)
x1 = 1
x2 = x
y1 = 2
y2 = 6
x1 + x2 = 1 +x
2 2
y1 + y2 = 2 + 6 = 4
2 2
AC = ( 1 + x , 4) -------------(1)
2
Mid-point BD :- B( 4,y) D(3,5)
x1 = 4
x2 = 3
y1 = y
y2 = 5
x1 + x2 = 4 + 3 = 7
2 2 2
y1 + y2 = y + 5
2 2
BD = ( 7 , y +5 ) -------(2)
2 2
Since diagonals in parallelogram is equal we can equate AC and BD
equating (1) and (2)
(1+x , 4) = ( 7 , y +5 )
2 2 2
Finding "x"
1 + x = 7
1+x = 7
x = 6
Finding "y" :-
4 = y + 5
2
8 = y + 5
y = 3
******************************************************
(5) Find the coordinates of a point A, where AB is the diameter of a circle whose centee is (2, -3) and B is ( 1,4).
sol) Let point A=(x,y)
Given :- center(C) = (2, -3) and B =(1, 4)
We know that the mid-point M(x,y) of the line segment joining the points ( x1, y1) and (x2, y2) is
M(x,y) = (x1 + x2 , y1 + y2 )
2 2
Here
M(x,y) = C(2,-3)
A(x,y) B( 1, 4)
x1 = x
x2 = 1
y1 = y
y2 = 4
Now,
C(2,-3) = (x + 1 , y + 4 )
2 2
2 = x +1
2
4 = x +1
3 = x
- 3 = y + 4
2
-6 = y +4
y = -10
Hence, A( 3, -10)
**********************
(6) If A and B are ( -2, -2) and (2, -4) respectively. Find the coordinates of "P" such that AP = 3/7 AB and "P" lies on the segment AB.
sol) Let P=(x,y)
Given :- AP = 3 AB
7
A--------P-------------------------------B
(-2, -2) (x,y) (2, -4)
=> 7AP = 3AB
-----------------------
But AB = AP +PB
-----------------------
=> 7AP = 3(AP + PB)
=> 7AP = 3AP + 3PB
=> 7AP - 3AP = 3BP
=> 4AP = 3PB
AP = 3
PB 4
Here the point "P" divides "AB" in the ratio of 3:4
Finding Point P(x,y) :-
We have, m1 :m2 = 3 :4
A(-2, -2) B(2, -4)
x1= -2
x2 = 2
y1 = -2
y2 = -4
using section formula :-
Finding "x" :-
x = m1x2 + m2 x1
m1 + m2
x = 3(2) + 4(-2)
3 + 4
x = 6 - 8
7
x = -2
7
Finding "y" :-
y = m1y2 + m2y1
m1 + m2
y = 3(-4) + 4(-2)
3 + 4
y = -12 - 8
7
y = -20
7
Thus, the co-ordinates of "P" are P ( -2/7, -20/7)
A ------------------P---------------------------------B
(-2, -2) (-2/7, -20/7) (2, -4)
*******************************************
(7) Find the coordinates of points which divide the line segment joining A(-4,0) and B (0,6) into four equal parts.
sol) Given :- A(-4, 0) B(0,6)
Given that lines segment divided into 4 equal parts which have A(-4,0) starting point and B(0,6) ending point
A---------P-------Q-------R--------B
(-4,0) (x1,y1) (x2,y2) (x3,y3) (0,6)
Given all 4 parts are equal
i.e, AP = PQ= QR= RB
Finding co-ordinates of "P" :- A (-4, 0) B(0,6)
A-----------P-------Q------R------B
|------1----| |-------------3----------}
Here m1 :m2 = 1 ; 3
x1 = -4
x2 = 0
y1 = 0
y2 = 6
Using section formula :-
p(x,y) = ( 1(0) + 3(-4) , 1(6) + 3(0) }
1 + 3 1 + 3
p(x,y) = -12 , 6
4 4
p(x,y) = (-3, 3/2)
Finding co-ordinates of "Q" :- A( -4, 0) B(0,6)
A------P-----Q------R------B
------2------ ------2------
m1 :m2 = 2: 2
x1 = -4
x2 = 0
y1 = 0
y2 = 6
Using section formula:-
Q(x,y) = ( 2(0) + 2(-4) , 1(6) + 3(0) )
4 4
Q(x,y) = (-2, 3)
Finding co-ordinates of "R" :- A( -4, 0) B(0,6)
A------P------Q--------R-----B
|---------------3-------| |----1--|
m1:m2 = 3 : 1
x1 = -4
x2 = 0
y1 = 0
y2 = 6
Using section Formula :-
R(x,y) = (3(0) + 1(-4) , 3(6) + 1(0)
4 4
R(x,y) = (-1, 9/2)
A---------------P----------Q------------R--------B
(-4,0) (-3, 3/2) (-2,3) (-1, 9/2) (0,6)
******************************************
(8) Find the coordinates of the points which divides the line segment joining A(-2, 2) and B (2,8) into four equal parts.
sol) Given :- A (-2, 2) B(2,8)
Given line segment is divided into 4 equal parts
A----------P-------Q-----R-----B
AP = PQ = QR = RB
Finding co-ordinates of "P" :- A (-2, 2) B(2,8)
A-----------P-------Q------R------B
|------1----| |-------------3----------}
Here m1 :m2 = 1 ; 3
x1 = -2
x2 = 2
y1 = 2
y2 = 8
Using section formula :-
P(x,y) = {(1(2) + 3(-2) , 1(8) + 3(2)}
4 4
P(x,y) = ( -1, 7/2)
Finding co-ordinates of "Q" :- A (-2, 2) B(2,8)
A-----------P-------Q------R------B
|------2 ------------| |-----2-------}
Here m1 :m2 = 2 :2
x1 = -2
x2 = 2
y1 = 2
y2 = 8
Using section formula :-
Q(x,y) = ( 2(2) + 2(-2) , 2(8) + 2(2) }
4 4
Q(x,y) = (0,5)
Finding co-ordinates of "R" :- A (-2, 2) B(2,8)
A-----------P-------Q------R------B
|------3--------------------| |---1--|
Here m1 :m2 = 3 : 1
x1 = -2
x2 = 2
y1 = 2
y2 = 8
Using section formula :-
R(x,y) = ( 3(2) + 1(-2) , 3(8) + 1(2) )
4 4
R(x,y) = (1, 13/2 )
A---------------P----------Q------------R--------B
(-2,2) (-1, 7/2) (0,5) (1, 13/2) (2,8)
*******************************************
(9) Find the coordinates of the point which divides the line segment joining the points ( a+b, a-b) and (a-b, a+b) in the ratio 3 :2 internally.
sol) Let P(x,y) be the required point.
Given :- m1 m2 = 3 :2
A(a+b, a-h) B(a-b, a+b)
x1 = a+b
x2 = a-b
y1 = a-h
y2 = a +b
using Section Formula :-
x = 3(a-b) + 2(a+b)
3 + 2
x = 3a - 3b + 2a + 2b
5
x = 5a - b
5
y = 3(a+b) + 2(a-b)
3 + 2
y = 3a + 3b + 2a - 2b
5
y = 5a + b
5
P(x,y) = ( 5a - b, 5a +b ) is the required point
5 5
******************************************
(10) Find the coordinates of centriod of the following
1) (-1, 3), (6, -3) and (-3, 6)
2) (6,2), (0,0) and (4, -7)
3) (1,-1), (0,6) and (-3,0)
sol) Let A ( -1, 3) B(6,3) C( -3,6)
x1 = -1
x2 = 6
x3 = -3
y1 = 3
y2 = -3
y3 = 6
The co-ordinates of the centroid are
G(x,y) = ( x1 + x2 + x3, y1+y2+y3)
3 3
G(x,y) = ( -1 + 6 +(-3). 3 + (-3) +6)
3 3
G(x,y) = (2, 2 )
3
2) (6,2), (0,0) and (4, -7)
sol) Let A ( 6, 2) B(0,0) C( 4, -7)
x1 = 6
x2 = 0
x3 = 4
y1 = 2
y2 = 0
y3 = -7
The co-ordinates of the centroid are
G(x,y) = ( x1 + x2 + x3, y1+y2+y3)
3 3
G(x,y) = ( 6+0+4). 2+0-7)
3 3
G(x,y) = (10, -5 )
3 3
3) (1,-1), (0,6) and (-3,0)
sol) Let A (1, -1) B(0,6) C( -3,0)
x1 = 1
x2 = 0
x3 = -3
y1 = -1
y2 = 6
y3 = 0
The co-ordinates of the centroid are
G(x,y) = ( x1 + x2 + x3, y1+y2+y3)
3 3
G(x,y) = ( 1 +0-3). -1+6+0)
3 3
G(x,y) = (-2, 5 )
3 3
*****************************************
Example 18. Find the area of a triangle whose vertices are (1, -1), (-4, 6) and (-3, -5)
sol) The area of the triangle formed by the vertices
A(1,-1) B(-4, 6) C(-3,-5)
Using Formula
=1 |x1(y2 - y3) + x2(y3 - y1) + x3 (y1-y2) |
2
= 1 |1(6+5) + (-4)(-5+1) + (-3)(-1-6) |
2
= 1 | 11 + 16 + 21 |
2
= 24
So the area of the triangle is 24 square units.
*****************************************
Example-19. Find the area of a traingle formed by the points
A(5,2)
B(4,7) and
C(7,-4)
sol) The area of the triangle formed by the vertices is given by
1 |5(7+4) + 4(-4-2) + 7(2-7)|
2
= 1 | 55 -24 -35|
2
= |-4| = |-2|
2
Since area is a measure, which cannot be negative, we will take the numberical value 2 or absolute value
i.e., |-2| = 2
Therefore,
the area of the triangle = 2 square units.
*****************************************
Example-20. If A(-5,7), B(-4,-5), C(-1,-6) and D(4,5) are the vertices of a quadrilateral. Then, find the area of the quadrilateral ABCD.
sol) By joining B to D, you will get two triangles ABD and BCD.
The area of △ABD
= 1 |-5(-5-5) + (-4)(5-7) + 4(7+5)|
2
= 1 |50 + 8 + 48|
2
= 106 = 53 square units
2
Also,
The area of △BCD
= 1 |-4(-6-5) - 1(5+5) + 4(-5+6)|
2
= 1 | 44 - 10 + 4|
2
= 19 square units.
Area of △ABD + Area of △BCD
So, the area of quadrilateral
ABCD = 53 + 19= 72 square units.
*****************************************
(*) Collinearity
Example-21. The points (3,-2)(-2,8) and (0,4) are three points in a plane.Show that these points are collinear.
sol) By using area of the triangle formula
△ = 1 | 3(8-4) + (-2)(4-(-2) + 0(-2-8)|
2
= 1 | 12-12| = 0
2
The area of the triangle is 0.
Hence the three points are collinear or lie on the same line.
****************************************
(*) Area of Triangle-'Heron's Formula'
A triangle whose lengths of sides are a,b, and c.
______________
A = √S(S-a)(S-b)(S-c)
where
S = a + b + c
2
Example-22. Find the value of 'b' for which the points are collineary.
sol). Let given points
A(1,2), B(-1,b), C(-3, -4)
Then
x1=1 y1=2
x2 = -1 y2 = b
x3 = -3 y3 = -4
We know, area of
△ABC = 1 |x1(y2-y3) + x2(y3-y1) +x3(y1-y2)|
2
.^. 1 |1(b +4) + (-1)(-4-2) + (-3)(2-b)| = 0
2
(= 0 because The given points are collinear)
|b + 4 + 6 -6 + 36 | = 0
|4b + 4| = 0
4b + 4 = 0
.^. b = -1
*****************************************
Exercise- 7.3
(1) Find the area of the triangle whose vertices are
1) (2,3)(-1, 0), (2,-4)
2) (-5, -1), (3,-5), (5,2)
3) (0,0), (3,0) and (0,2)
sol) Area of triangle formed by the vertices
A( 2,3) B(-1,0) C(2, -4)
x1 = 2
x2 = -1
x3 = 2
y1 = 3
y2 =0
y3 = -4
Formula :-
Area= 1/2 | x1(y2 - y3) +x2(y3 - y1) + x3(y1 - y2) |
=> 1/2 |2(0 +4) + (-1)( -4-3) + 2(3-0) |
=> 1/2 |8 + 7 +6 |
=> 21 sq.units
2
2) (-5, -1), (3,-5), (5,2)
sol) Area of triangle formed by the vertices
A( -5, -1) B(3, -5) C(5, 2)
x1 = -5
x2 = 3
x3 = 5
y1 = -1
y2 = -5
y3 = 2
Formula :-
Area= 1/2 | x1(y2 - y3) +x2(y3 - y1) + x3(y1 - y2) |
=> 1/2 |-5(-5-2) + 3( 2+1) + 5(-1+5) |
=> 1/2 |35+9+20 |
=> 64 sq.units
2
=> 32 sq.units
(3) (0,0) (3,0) and ( 0,2)
sol) Area of triangle formed by the vertices
A( 0,0) B(3, 0) C(0, 2)
x1 = 0
x2 = 3
x3 = 0
y1 = 0
y2 = 0
y3 = 2
Formula :-
Area= 1/2 | x1(y2 - y3) +x2(y3 - y1) + x3(y1 - y2) |
=> 1/2 |0(0-2) + 3( 2-0) + 0(0-0) |
=> 1/2 |6 |
=> 3 sq.units.
********************************************
(2) Find the value of 'K' for which the points are collinear.
1) (7,-2) (5,1) (3,k)
2) (8,1) (k, -4) (2, -5)
3) K,K), (2,3) and (4, -1).
Sol) When points are collinear they lie on same line. It means it cannot form Triangle and the area /_\ = 0
1) (7,-2) (5,1) (3,k)
x1 = 7
x2 = 5
x3 = 3
y1 = -2
y2 = 1
y3 = k
Area= 1/2 | x1(y2 - y3) +x2(y3 - y1) + x3(y1 - y2) |
0 = 1/2 | 7(1-k) + 5 (k-(-2) + 3(-2-1) |
0 = 1/2 |7-7k + 5k + 10 -9 |
0 = 1/2 | 8 - 2k |
0 = 1/
4-k = 0
4=k
--------------------------
2) (8,1) (k, -4) (2, -5)
----------------------------
sol) x1 = 8
x2 = k
x3 = 2
y1 = 1
y2 = -4
y3 = -5
Area= 1/2 | x1(y2 - y3) +x2(y3 - y1) + x3(y1 - y2) |
0 = 1/2 |8(-4+5) + k(-5-1) + 2 (1+4)|
0 = 1/2 |8 -6k +10|
0= 1/2 |18-6k|
0 = 1/2 ||6(3-k)|
0 = 3(3-k)
k = 3
----------------------------------
3) K,K), (2,3) and (4, -1).
---------------------------------
sol) x1 = k
x2 = 2
x3 = 4
y1 = k
y2 = 3
y3 = -1
Area= 1/2 | x1(y2 - y3) +x2(y3 - y1) + x3(y1 - y2) |
0 = 1/2 | k(3+1) + 2(-1-k) +4(k-3) |
0 = 1/2 |4k -2 -2k + 4k -12|
0 = 1/2 |6k - 14|
0 = 3k - 7
3k = 7
k = 7/3
(3) Find the area of the triangle formed by joining the mid-points of the sides of the triangle w hose vertices are (0, -1), (2,1) and (0,3) . find the ratio of this area to the area of the given triangle.
sol) let A=(0,-1) B =(2,1) C=(0,3) are the vertices of the triangle
RTP :- triangle formed by joining the mid-points of the sides of the traingle.
The mid-point of AB be D :-
A=(0,-1) B =(2,1)
D(x,y)
x1= 0
x2 =2
y1= -1
y2= 1
D(x,y) =( x1 +x2, y1+y2)
2 2
=> (0+2, -1+1) => (1,0)
2 2
The mid-point of BC be E :-
B =(2,1) C=(0,3)
E(x,y)
x1= 2
x2 =0
y1= 1
y2= 3
E(x,y) =( x1 +x2, y1+y2)
2 2
E(x,y) = (2+0, 1+3 ) => (1,2)
2 2
The mid-point of AC be F :-
A=(0,-1) C =(0,3)
F(x,y)
x1= 0
x2 =0
y1= -1
y2= 3
F(x,y) =( x1 +x2, y1+y2)
2 2
F(x,y) =( 0+0, -1+3)
2 2
F(x,y) = (0,1)
Finding area of triangle ABC
A=(0,-1) B =(2,1) C=(0,3)
x1=0
x2 =2
x3=0
y1=-1
y2=1
y3=3
Area= 1/2 | x1(y2 - y3) +x2(y3 - y1) + x3(y1 - y2) |
=> 1/2 |0(1-3) + 2(3+1) + 0(-1-1)|
=> 1/2 |8|
=> 4 sq.units
Finding area of triangle DEF
D=(1,0) E =(1,2) F=(0,1)
x1=1
x2 =1
x3=0
y1=0
y2=2
y3=1
Area= 1/2 | x1(y2 - y3) +x2(y3 - y1) + x3(y1 - y2) |
=> 1/2 | 1(2-1) + 1(1-0) + 0(0-2) |
=>1/2 |2|
=> 1 sq. units.
Finding ratio of two areas of triangles :-
ABC : DEF = 4 :1
*********************************************
(4) Find the area of the quadrilateral whose vertices , taken in order , are ( -4, -2) (-3,-5) (3, -2) and (2,3).
sol) Let A= (-4,--2) B=(-3, -5) C=( 3, -2) D=(2,3)
Two traingles are formed "ABC" & " ACD" by joining "AC
If we find the area of two traingles "ABC and ACD then we can easily get area of quadrilateral "ABCD"
ABCD = /_\ABC + /_\ ACD
Finding area of /_\ ABC :-
A=(-4,-2) B =(-3,-5) C=(3,-2)
x1=-4
x2 =-3
x3=3
y1=-2
y2=-5
y3=-2
Area= 1/2 | x1(y2 - y3) +x2(y3 - y1) + x3(y1 - y2) |
=>1/2 | -4(-5+2) +(-3)(-2+2) +3(-2+5)|
=> 1/2 |12-0+9|
=> 1/2 |21|
=> 21 square units .
2
Area of /_\ ADC :-
A=(-4,-2) D =(2,3) C=(3,-2)
x1=-4
x2 =2
x3=3
y1=-2
y2=3
y3=-2
Area= 1/2 | x1(y2 - y3) +x2(y3 - y1) + x3(y1 - y2) |
=>1/2 | -4(3+2) -3(-2+2) + 3(-2-3)|
=> 1/2 | -20 -0 - 15|
=1/2 | -35|
Since area cannot be negative
=> 35
2
Area of Quadrilateral ABCD = /_\ ABC + /_\ ADC
=> 21 + 35
2 2
=>
2
=> 28 sq.units
**********************************************
(5) Find the area of the triangle formed by the points ( 8, -5) ( -2, -7) and (5,1) by using Heron's formula.
sol) Let A=(8, -5) B=(-2, -7) C=(5,1)
Using Distance let us find AB=c, BC=a, and CA=b
c = _/(-2-8)^2 + (-7+5)^2
=> _/100 + 4)
=> _/104
a= _/(5+2)^2 + (1+7)^2
=> _/49 + 64
=> _/113
b = _/(5-8)^2 + (1+5)^2
=> _/9 + 36
=> _/ 45
Heron's Formula :-
Using above formula, we can find the area of the triangle, but this is complicated and we can use following easier method.
Area= 1/2 | x1(y2 - y3) +x2(y3 - y1) + x3(y1 - y2) |
A=(8, -5) B=(-2, -7) C=(5,1)
x1=8
x2=-2
x3=5
y1=-7
y2=-7
y3=1
Area = 1/2 | 8(-7-1) -2(1-(-5)) +5(-5-(-7))|
=> 1/2 | 8(-8) -2(6) +5(2) |
=> 1/2 | -66|
Since area cannot be negative
=> 1/2 * 66
=> 33 sq. units.
*****************************************
Straight Lines
ching and ping are discussing to find solutions for a linear equation in two variable.
ching : Can you find solutions for 2x+3y=12
ping : Yes, I have done this, see
2x + 3y = 12
3y = 12 - 2x
y = 12 - 2x
3
ching : can you write these solutions in order pairs.
Ping : Yes (0,4), (3,2), (6,0), (-3,6)
Ching, can you plot these points on the coordinate plane.
Ching : I have done case like this
Ping : What do u observe?
What does this line represent?
Ching : It is a straight Line.
Ping : Can you identify some more points on this line?
Can you help Ching to find some more points on this line?
.........., ..........., ..............., ................,
___
And In this line, what is AB called?
___
AB is a line segment.
*****************************************
Example 22. The end points of a line are (2,3), (4,5). Find the slope of the line.
sol) Points of a line are (2,3) (4,5) then slope of the line
m = y2 - y1
x2 - x1
= 5 - 3 = 2 = 1
4 - 2 2
Slope of the given line is 1.
*****************************************
Example- 23. determine x so that 2 is the slope of the line through P(2,5) and Q(x,3)
sol) Slope of the line passing through P(2,5) and Q(x,3) is 2.
Slope of a PQ = y2 - y1 = 3 - 5 = - 2
x2 - x1 x - 2 x - 2
= - 2 = 2
x - 2
2x - 4 = -2
2x = 2
x = 1
******************************************
Exercise- 7.4
(1) Find the slope of the line joining the two given points.
1) ( 4, -8) and (5, -2)
sol) Slope of line passing through
A=(4, -8) and B=(5, -2)
x1 = 4
x2 = 5
y1 = -8
y2 = -2
=> -2 -(-8)
5 - 4
=> 6 = 6
1
.^. the slope of the line is 6
**************************
2) (0,0) and ( _/3 , 3)
sol) A=(0,0) B=( -/3, 3)
x1 = 0
x2 = _/3
y1= 0
y2 = 3
=> 3 - 0
_/3 - 0
=> 3
_/3
=> _/3 *
_/3 is the slope of the line
*************************
3) (2a, 3b) and ( a , -b)
sol) Let A = (2a, 3b) B =(a, -b)
x1 = 2a
x2 = a
y1 = 3b
y2 = -b
=> -b - 3b
a - 2a
=> -4b
-a
4b is the slope of a line
a
******************************************
4) (a,0) and (0,b)
sol) Let A=(a,0) B=(0,b)
x1 = a
x2 = 0
y1 = 0
y2 =b
=> b - 0
0 - a
=> -b is the slope of a line
a
****************************************
5) A(-1.4, -3.7), B ( -2.4, 1.3)
sol) x1 = -1.4 x2 = -2.4 y1 = -3.7 y2= 1.3
=> 1.3 + 3.7
-2.4 + 1.4
=> 5
-1
=> -5 is the slope of a line.
********************************************
6) A(3, -2), B (-6, -2)
sol) x1 = 3 x2= -6 y1 = -2 y2= -2
=> -2 -(-2)
-6 - 3
=> o is the slope of the line
*********************************************
7) A ( -3 1/2, 3), B( -7, 2 1/2)
sol) x1 = -3 1/2 = -7/2 x2= -7 y1 = 3 y2 = 2*1/2 = 5/2
=> 5/2 - 3
-7 + 7/2
=> 5 -6
2
-14 + 7
2
=> -1/2
-7/2
=> 1/7 is the slope of a line
******************************************
8) A(0,4), B(4,0)
sol) x1 = 0 x2=4 y1=4 y2=0
=> 0 - 4
4 - 0
= -1 is the slope of a line
*****************************************
(*) Optional Exercise
1). Center of the circle Q is on the Y-axis. And the circle passes through the points (0,7) and (0,-1). Circle intersects the positive X-axis at (P,0). What is the value of 'P'.
sol)
Equation of the circle having center at y-axis is of the form
(x-0)^2 + (y-k)^2 = a^2
Since it passes through (0,7)
x = 0, y = 7
(0-0)^2 + (7-k)^2 = a^2
(7 - k)^2 = a^2
49 + k^2 - 14k = a^2 ------- (!)
Also the circle passes through ( 0, -1)
x = 0 , y = -1
(0 - 0)^2 + (-1 - k)^2 = a^2
(-1 - k)^2 = a^2
1 + k^2 +2k = a^2 -------- (!!)
Equating (!) & (!!)
49 + k^2 - 14k = 1 + k^2 + 2k
48 = 16k
48 = k
16
3 = k
substitute k=3 in any one equation
we get
1 + (3)^2 + 2(3) = a^2
1 + 9 + 6 = a^2
16 = a^2 or a = 4
We have
Equation of a circle with center on y-axis is
x^2 + (y-k)^2 = a^2
substitute the value of k=3 and a^2 = 16 we get
x^2 + (y-3)^2 = 16
But the circle intersects x-axis at (p,0),
we get
x = p, y = 0
(p)^2 + (0-3)^2 = 16
p^2 + 9 = 16
p^2 = 7
p = + _/7
****************************************
2). A triangle △ABC is formed by the points A(2,3), B(-2,-3), C(4, -3). What is the point of intersection of side BC and angular bisector of A.
sol) We know that : An angle bisector of a triangle divides the opposite side into two segments that are proportional to the other two sides of the triangle.
BD = AB
DC AC ---------- (!)
Now, using distance formula
Length of A(2,3) B(-2, -3)
x1 = 2 y1 = 3
x2 = -2 y2 = -3
AB = √(-2 -2)^2 + (-3 - 3)^2
AB = √16 + 36
AB = √52
AB = √4 * 13
AB = 2 √13 -------- (!!)
Distance between A(2,3) C(4, -3)
x1 = 2 y1 = 3
x2 = 4 y2 = -3
AC = √(4-2)^2 + (-3-3)^2
AC = √4 + 36
AC = √40
AC = √4 * 10
AC = 2 √10 ------- (!!!)
Substituting (!!) and (!!!) in (!)
we get
BD = AB
DC AC
BD = 2 √13
DC 2 √10
BD = √13
DC √10
BD : DC = √ 13 : √10 -------- (!V)
So, the line segment BC is divided in the ration of √13 : √10
Now, by using section formula
D (x,y) = ( mx2 + nx1 , my2 + ny1 )
m + n m + n
D(x,y) = (√13(4) + √10(-2) , √13(-3) + √10(-3)
√13 + √10 √13 + √10
substituting √13 = 3.606, √10 = 3.162
D(x,y)
= 4(3.606) - 2(3.162), -3(3.606) - 3(3.162)
3.606 + 3.162 3.606 + 3.162
D(x,y) = (14.424 - 6.324, -10.818 - 9.486 )
3.606 + 3.162 3.606 + 3.162
D(x,y) = ( 8.1, -20.304 )
6.768 6.768
D(x,y) = (1.12 , -3)
Therefore the coordinates of point of intersection are (1.12, -3)
*****************************************
3). The side of BC of an equilateral triangle △ABC is parallel to X-axis. Find the slopes of line along sides BC, CA and AB.
sol) In an equilateral Triangle all angles are equal.
i.e., ∠A=∠B=∠C= 60 deg
Sum of the angles of triangle = 180 deg
Given : BC is parallel to x-axis
We know that the slope of any line parallel to x-axis is 0
.^. Slope of BC = 0 deg
(!!) Slope of CA :-
Tan ∠CEX = Tan 120deg
=> Tan (180 - 60)
=> Tan (-60)
=> - √3
(!!!) Slope of AB :
Tan ∠ADX = Tan 60deg
=> √3
****************************************
4). A right triangle has sides 'a' and 'b' where a>b. If the right angle is bisected then find the distance between orthocenters of the smaller triangles using coordinate geometry.
sol)
*****************************************
5). Find the centroid of the triangle formed by the line 2x + 3y - 6 = 0. With the coordinate axis.
sol) Given Line : 2x + 3y = 6
Divide the Line by constant 6
we get,
2x + 3y = 6
6 6 6
x + y = 1
3 2
(x/3) Since x intercept is at 3. so line will cut the x-axis at (3,0)
(y/2) and y intercept is at 2, so line will cut the y-axis at (0,2)
.^. Coordinates of triangle are
(3,0) (0,2) (0,0)
x1 = 3, x2=0, x3=0
y1 = 0, y2 = 2, y3 = 0
So centroid is
(x,y) = x1 + x2 + x3 y1+y2+y3
3 3
=> 3 + 0 + 0 2+0+0
3 3
Centroid (x,y)= (1, 2/3)
*****************************************
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