Lesson: Coordinate Geometry
Exercise 7.1
Question: 1
The distance of the point P(2, 3) from the x-axis is
a. 2
b. 3
c. 1
d. 5
Solution:
(b)
The perpendicular distance of the point P(2, 3) from
x-axis is equal to the coordinate.
y
So, it is equal to 3 units.
Question: 2
The distance between the points A(0, 6) and B(0, –2)
is
a. 6
b. 8
c. 4
d. 2
Solution:
(b)
Distance between two points is given by the
formula,
2 2
1 2 1 2
l x x y y
2 2
AB 0 0 2 6
2
0 8
64
8units
Hence, .
AB 8units
Question: 3
The distance of the point P(–6, 8) from the origin is
a. 8
b. (B)
2 7
c. 10
d. 6
Solution:
(c)
Coordinate of origin is O (0, 0).
Now, find the distance of the point P(–6, 8) from
the point O.
Distance between two points is given by the
formula,
2 2
1 2 1 2
l x x y y
2
2
( 6) 8
36 64
100
10units
Hence, .
OP 10units
Question: 4
The distance between the points (0, 5) and (–5, 0) is
a. 5
b.
5 2
c.
2 5
d. 10
Solution:
(b)
Let the two points be A(0, 5) and B .
5,0
Distance between two points is given by the
formula,
2 2
1 2 1 2
l x x y y
2 2
AB 5 0 0 5
2
2
(5) 5
25 25
50
5 2 units
Hence, .
AB 5 2 units
Question: 5
AOBC is a rectangle whose three vertices are
vertices A(0, 3), O(0, 0) and B(5, 0). The length of its
diagonal is
a. 5
b. 3
c.
34
d. 4
Solution:
(c)
The coordinates of A and B are and .
0,3
5,0
Fig. Exm_7.1_5
The length of diagonal
AB
Distance between two points is given by the
formula,
2 2
1 2 1 2
l x x y y
2 2
AB 5 0 0 3
2
2
(5) 3
25 9
34
Hence,
AB 34
Question: 6
The perimeter of a triangle with vertices (0, 4), (0, 0)
and (3, 0) is
a. 5
b. 12
c. 11
d.
7 5
Solution:
(b)
Perimeter of any
PQR PQ QR PR
Let P(0, 4), Q(0, 0) and R(3, 0) be the three vertices
of .
PQR
Now, find the lengths of the sides PQ, QR and PR.
Distance between two points is given by the
formula,
2 2
1 2 1 2
l x x y y
2 2
PQ 0 0 0 4
2
4
16
4cm
2 2
QR 3 0 0 0
2
3
9
3cm
2 2
PR 3 0 0 4
2 2
3 4
9 16
25
5cm
Now perimeter of
PQR PQ QR PR
Perimeter
4 3 5
12cm
Hence, the perimeter is 12 cm.
Question: 7
The area of a triangle with vertices A(3, 0), B(7, 0)
and C(8, 4) is
a. 14
b. 28
c. 8
d. 6
Solution:
(c)
According to the question,
The vertices of the triangle are A(3, 0), B(7, 0) and
C(8, 4).
Now, find the area of the .
ABC
1 2 3 2 3 1 3 1 2
1
Area of a triangle x y y x y y x y y
2
1
Area of ABC 3 0 4 7(4 0) 8 0 0
2
1
3 4 7 4 8 0
2
1
12 28 0
2
1
16
2
1
16
2
8squareunits
Hence, the area of is 8 square units.
ABC
Question: 8
The points (–4, 0), (4, 0), (0, 3) are the vertices of a
a. right triangle
b. isosceles triangle
c. equilateral triangle
d. scalene triangle
Solution:
(b)
Let us consider a having vertices P ,
PQR
4,0
Q(4, 0) and R(0, 3).
Now, find the lengths of the sides PQ, QR and PR.
Distance between two points is given by the
formula,
2 2
1 2 1 2
l x x y y
2
2
PQ 4 4 0 0
2
8
64
8cm
2 2
QR 0 4 3 0
2 2
4 3
16 9
25
5cm
2
2
PR 0 4 3 0
2 2
4 3
16 9
25
5cm
Now, and
QR PR 5cm
PQ 8cm
Thus, the triangle is an isosceles triangle.
Question: 9
The point which divides the line segment joining
the points (7, –6) and (3, 4) in ratio internally
1:2
lies in the
Solution:
(d)
Let be the required point.
R x,y
Now, let us consider, and
1
x 7,
1
y 6,
2
x 3
2
y 4
And the ratio,
1 2
1:2 m :m
Now, apply the section formula.
and
1 2 2 1
1 2
m x m x
x
m m
1 2 2 1
1 2
m y m y
y
m m
So, by the section formula,
1 3 2 7
x
1 2
3 14
3
17
3
1 4 2 6
y
1 2
4 12
3
8
3
Here, and
17
x
3
8
y
3
Thus, the point is which lies in the IV
17 8
,
3 3
Question: 10
The point which lies on the perpendicular bisector
of the line segment joining the points A(–2, –5) and
B(2, 5) is
a. (0, 0)
b. (0, 2)
c. (2, 0)
d. (–2, 0)
Solution:
(a)
The perpendicular bisector of the line segment AB,
passes through the mid–point of AB.
Let the mid–point be O .
x,y
Now, find the coordinates of O.
Coordinates
2 2 5 5
,
2 2
0 0
,
2 2
0,0
Thus, the perpendicular bisector passes through the
point .
0,0
Question: 11
The fourth vertex D of a parallelogram ABCD
whose three vertices are A(–2, 3), B(6, 7) and C(8, 3)
is
a. (0, 1)
b. (0, –1)
c. (–1, 0)
d. (1, 0)
Solution:
(b)
Let the fourth vertex of the parallelogram be D
.
x,y
The diagonals AC and BD of parallelogram ABCD
bisect each other.
So,
mid point of BD mid point of AC
Now, find the mid–points of BD and AC.
Mid–point of BD
x 6 y 7
,
2 2
Mid–point of AC
2 8 3 3
,
2 2
6 6
,
2 2
3,3
Equate mid–points of BD and AC.
x 6 y 7
, 3,3
2 2
and
x 6
3
2
y 7
3
2
Now, find the values of .
x
x 6
3
2
x 6 6
x 6 6
x 0
Similarly, find the value of .
y
y 7
3
2
y 7 6
y 6 7
y 1
Thus, the fourth vertex of parallelogram is .
0, 1
Question: 12
If the point P(2, 1) lies on the line segment joining
points A(4, 2) and B(8, 4), then
a. (A)
1
AP AB
3
b. (B)
AP PB
c. (C)
1
PB AB
3
d. (D)
1
AP AB
2
Solution:
(d)
Let us consider, the ratio in which the line segment
joining A(4, 2) and B(8, 4), is divided by point P(2, 1)
be .
t :1
Now, apply the section formula.
1 2 2 1
1 2
m x m x
x
m m
So, by the section formula,
t 8 1 4
2
t 1
2t 2 8t 4
2 6t
2
t
6
1
t
3
Verification:
Now, as so, the point P lies outside the line
1
t
3
segment AB.
So,
AP 1
PB 3
AP i.e., 1 part outside AB and AB
1
2
Thus, AP and AB units
1x unit
3x 1x 2x
So,
1
AP AB
2
1
1 1
2
1 1, whichis true
Hence, proved.
Question: 13
If P is the mid–point of the line segment
a
,4
3
joining the points Q and R , then the
6,5
2,3
value of is
a
a. (A)
4
b. (B)
12
c. (C) 12
d. (D)
6
Solution:
(b)
According to the question,
P is the mid–point of the line segment QR.
a
,4
3
Find the coordinates of the mid–point of the line
segment QR.
Coordinates
6 2 5 3
,
2 2
8 8
,
2 2
Now, equate the coordinates of P with the
coordinates of mid–point of the line segment QR.
a 8 8
,4 ,
3 2 2
a 8
3 2
2a 24
24
a
2
a 12
Hence,
a 12
Question: 14
The perpendicular bisector of the line segment
joining the points A(1, 5) and B(4, 6) cuts the
y
axis at
a. (0, 13)
b. (B)
0, 13
c. (0, 12)
d. (13, 0)
Solution:
(a)
Let the perpendicular bisector cut the axis at R
y
.
0,y
Find the distances of points A and B from R.
Distance between two points is given by the
formula,
2 2
1 2 1 2
l x x y y
2 2
AR 0 1 y 5
2
1 y 25 10y
2
y 10y 26
2 2
BR 0 4 y 6
2
2
4 y 36 12y
2
16 y 36 12y
2
y 12y 52
Now, (R is mid–point of AB)
AR BR
2 2
y 10y 26 y 12y 52
Taking square on both sides.
2 2
2 2
y 10y 26 y 12y 52
2 2
y 10y 26 y 12y 52
10y 26 12y 52
2y 26
26
y
2
y 13
Thus, the coordinates of R are (0, 13).
Question: 15
The coordinates of the point which is equidistant
from the three vertices of the as shown in
AOB
the given figure is
a.
x,y
b.
y,x
c.
x y
,
2 2
d.
y x
,
2 2
Solution:
(a)
Let the mid–point of the hypotenuse AB be M.
In a right-angled triangle, the mid–point of the
hypotenuse is equidistant from the three vertices of
triangle.
So, find the coordinates of M.
Coordinates
0 2x 2y 0
,
2 2
2x 2y
,
2 2
x,y
Hence, the coordinates are .
x,y
Question: 16
A circle drawn with origin as the centre passes
through . The point which does not lie in the
13
,0
2
interior of the circle is
a. (A)
3
,1
4
b. (B)
7
2,
3
c. (C)
1
5,
2
d. (D)
5
6,
2
Solution:
(d)
First find the radius of the given circle having one
point as the origin and other point as .
13
,0
2
Distance between two points is given by the
formula,
2 2
1 2 1 2
l x x y y
2
2
13
2
2
13
2
13
2
6.5units
Now, let us consider all the four cases.
a. Distance of the point from
3
,1
4
0,0
2
2
3
Distance 0 1 0
4
2
3
1
4
9
1
16
25 5
16 4
1.25units
The distance
1.25 6.5
So, the point lies in the interior of the
3
,1
4
circle.
b. Distance of the point from
7
2,
3
0,0
2
2
7
Distance 2 0 0
3
2
2
7
2
3
49
4
9
85
9
3.073units
The distance
3.0731 6.5
So, the point lies in the interior of the circle.
7
2,
3
c. Distance of the point from
1
5,
2
0,0
2
2
1
Distance 5 0 0
2
2
2
1
5
2
1
25
4
101
4
5.0249units
The distance
5.0249 6.5
So, the point lies in the interior of the circle.
1
5,
2
d. Distance of the point from .
5
6,
2
0,0
2
2
5
Distance 6 0 0
2
2
2
5
6
2
25
36
4
169
4
13
2
6.5units
The distance
6.5 6.5
So, the point lies on the circle and not in
5
6,
2
the interior of the circle.
Question: 17
A line intersects the axis and axis at the
y
x
points P and Q, respectively. If (2, –5) is the mid–
point of PQ, then the coordinates of P and Q are,
respectively
a. (A) and (2, 0)
0, 5
b. (B) (0, 10) and
4,0
c. (C) (0, 4) and
10,0
d. (D) and (4, 0)
0, 10
Solution:
(d)
According to the question,
P lies on the axis so, coordinates of P are .
y
0,y
And, Q lies on the axis so, coordinates of Q are
x
.
x,0
Now, find the coordinates of the mid–point of PQ.
Coordinates
0 x y 0
,
2 2
x y
,
2 2
Also, it is given than (2, –5) is the mid–point of PQ.
So, equate the coordinates.
x y
, 2, 5
2 2
Find the value of .
x
x
2
2
x 4
Now, find the value of .
y
y
5
2
y 10
So, the coordinates of P and Q are and
0, 10
respectively.
4,0
Question: 18
The area of a triangle with vertices ,
a,b c
and is
b,c a
c,a b
a. (A)
2
a b c
b. (B) 0
c. (C)
a b c
d. (D)
abc
Solution:
(b)
Let us consider a whose vertices are P
PQR
, Q and R .
a,b c
b,c a
c,a b
Now, find the area of the triangle.
1 2 3 2 3 1 3 1 2
1
Area of a triangle x y y x y y x y y
2
a c a a b
1
Area of given triangle b a b (b c
2
c b c (c a
2 2 2
2 2 2
ac a a ab ab b
1
2
b bc bc c c ac
1
ac ab ab bc bc ac
2
1
0
2
0
Hence, the area of given triangle is 0.
Question: 19
If the distance between the points and (1, 0) is
4,p
5, then the value of is
p
a. 4 only
b. (B)
4
c. (C) only
4
d. (D) 0
Solution:
(b)
Let the two points be A and B .
4,p
1,0
Distance between the points A and B is 5 units.
So,
AB 5units
Now, distance between two points is given by the
formula,
2 2
1 2 1 2
l x x y y
2 2
AB 4 1 p 0
2 2
5 3 p
2
5 9 p
Take a square of both sides.
2
2
2
5 9 p
2
25 9 p
2
p 25 9
2
p 16
p 16
p 4
Hence,
p 4
Question: 20
If the points A(1, 2), O(0, 0) and C are
a,b
collinear, then
a.
a b
b.
a 2b
c.
2a b
d.
a b
Solution:
(c)
Points A(1, 2), O(0, 0) and C will be collinear, if
a,b
the area of is zero.
AOC
Now, take the formula for area of a triangle.
1 2 3 2 3 1 3 1 2
1
Area of a triangle x y y x y y x y y
2
1
0 1 0 b 0 b 2 a 2 0
2
1
0 b 0 2a
2
0 b 2a
2a b
Exercise 7.2 (12)
State whether the following statements are true or
Question: 1
with vertices A , B(2, 0) and C(0, 2) is
ABC
2,0
similar to with vertices D , E(4, 0) and
DEF
4,0
F(0, 4).
Solution:
The statement is true.
Now, if,
ABC DEF:
AB AC BC
DE DF EF
By using the distance formula, find the sides AB,
BC and AC of , and sides DE, EF and DF of
ABC
.
DEF
2
2
AB 2 2 0 0
2
4 0
16
2 2
BC 0 2 2 0
2
2
2 2
4 4
8
2 2 units
2
2
AC 0 2 2 0
2
2
2 2
4 4
8
2 2 units
In ,
DEF
2
2
DE 4 4 0 0
2
8 0
64
8units
2 2
EF 0 4 4 0
2
2
4 4
16 16
32
4 2 units
2
2
DF 0 4 4 0
2 2
4 4
16 16
32
4 2 units
Now,
AB 4 1
DE 8 2
AC 2 2 1
DF 2
4 2
BC 2 2 1
EF 2
4 2
Thus,
AB AC BC 1
DE DF EF 2
Hence,
ABC DEF:
Question: 2
Point P lies on the line segment joining the
4, 2
points A and B .
4,6
4, 6
Solution:
The statement is true.
The point P will lie on the line joining the
4, 2
points A and B if, area of is
4,6
4, 6
ABP
zero.
Now, find the area of the .
ABP
1 2 3 2 3 1 3 1 2
1
Area of a triangle x y y x y y x y y
2
1
Area of ABP 4 6 2 4 2 6 4 6 6
2
 
1
4 8 4 4 4 12
2
32 16 48
0
Hence, point P lies on the line joining A and B.
Question: 3
The points (0, 5), and (3, 6) are collinear.
0, 9
Solution:
The statement is false.
Let the three points be P(0, 5), Q and R(3, 6).
0, 9
These points will be collinear if the area
PQR 0
Now, find the area of the .
PQR
1 2 3 2 3 1 3 1 2
1
Area of a triangle x y y x y y x y y
2
1
Area of PQR 0 9 6 0 6 5 3 5 9
2
 
1
0 0 0 3 14
2
0 42
42 0
Hence, the three given points are not collinear.
Question: 4
Point P(0, 2) is the point of intersection of axis
y
and perpendicular bisector of line segment joining
the points A and B(3, 3).
1,1
Solution:
The statement is false.
As the point, P(0, 2) is the perpendicular bisector of
the line joining the points A and B(3, 3), then
1,1
point P must be equidistant from A and B.
So, PA should be equal to PB.
Now, find the distances PA and PB.
Distance between two points is given by the
formula,
2 2
1 2 1 2
l x x y y
2 2
PA 1 0 1 2
2 2
1 1
1 1
2 units
2 2
PB 3 0 3 2
2 2
3 1
9 1
10 units
Thus,
PA PB
Hence, the given statement is not correct.
Question: 5
Points A(3, 1), B and C(0, 2) cannot be the
12, 2
vertices of a triangle.
Solution:
The statement is true.
The points A, B and C can from a triangle, if the
sum of any two sides is greater than the third side.
Now, find the sides AB, BC and AC of .
ABC
Distance between two points is given by the
formula,
2 2
1 2 1 2
l x x y y
2 2
AB 12 3 2 1
2 2
9 3
81 9
90
3 10 units
2
2
BC 0 12 2 2
2 2
12 4
144 16
160
4 10 units
2 2
AC 0 3 2 1
2
2
3 1
9 1
10 units
So, and
AB 3 10,
BC 4 10
AC 10 units
Now,
AB AC 3 10 10 4 10
This is equal to the side BC.
Hence, the points A, B, C cannot form a triangle.
Question: 6
Points A(4, 3), B(6, 4), C and D are the
5, 6
3,5
vertices of a parallelogram.
Solution:
The statement is false.
ABCD will be a parallelogram if,
mid–point of diagonal mid–point of diagonal
AC
BD.
This is because, diagonals of a parallelogram bisect
each other.
Now, find the mid–points of diagonals AC and BD.
Mid–point of AC
3 6
4 5
,
2 2
9 3
,
2 2
Mid–point of BD
6 3
4 5
,
2 2
3 9
,
2 2
Now,
9 3 3 9
, ,
2 2 2 2
Thus, mid–point of diagonal mid–point of
AC
diagonal BD
Hence, ABCD is not a parallelogram.
Question: 7
A circle has its centre at the origin and a point P(5,
0) lies on it. The point Q(6, 8) lies outside the circle.
Solution:
The statement is true.
The point Q lies in the exterior of the circle, if the
distance of point Q from the centre O(0, 0) is
greater than the radius of the circle.
The point, P(5, 0) lies on the circle and the
coordinates of centre are (0, 0).
OP
Distance between two points is given by the
formula,
2 2
1 2 1 2
l x x y y
2 2
OP 5 0 0 0
2
5 0
25
5units
Now, find the distance between the point Q and the
centre.
2 2
OQ 6 0 8 0
2 2
6 8
36 64
100
10units
Thus,
OQ OP
So, the point Q lies in the exterior of the given
circle.
Question: 8
The point A(2, 7) lies on the perpendicular bisector
of line segment joining the points P(6, 5) and Q
.
0, 4
Solution:
The statement is false.
The point A(2, 7) on perpendicular bisector should
be equidistant from the points P(6, 5) and Q .
0, 4
So,
AP AQ
Now, find the distances AP and AQ.
Distance between two points is given by the
formula,
2 2
1 2 1 2
l x x y y
2 2
AP 2 6 7 5
2
2
4 2
16 4
20
2 5 units
2
2
AQ 2 0 7 4
2 2
2 11
4 121
125
5 5 units
Now,
2 5 5 5
So,
AP AQ
So, the point A does not lie on the perpendicular
bisector of PQ.
Question: 9
Point P is one of the two points of trisection
5, 3
of the line segment joining the points A and
7, 2
B .
1, 5
Solution:
The statement is true.
Let the point P divides the line AB in ratio .
t :1
Now, apply the section formula to find the value of
.
t
and
1 2 2 1
1 2
m x m x
x
m m
1 2 2 1
1 2
m y m y
y
m m
So, by the section formula,
and
t 1 1 7
x
t 1
t 5 1 2
y
t 1
Now, put the values of in the above section
x and y
formula.
t 7
5
t 1
5 t 1 t 7
5t 5 t 7
5t t 7 5
4t 2
2
t
4
1
t
2
5t 2
and 3
t 1
3 t 1 5t 2
  3t 3 5t 2
5t 3t 3 2
2t 1
1
t
2
So, P divides AB in the ratio .
1:2
Hence, P is one point of the trisection point of AB.
Question: 10
Points A , B and C are
6,10
4,6
3, 8
collinear such that .
2
AB AC
9
Solution:
The statement is true.
The points A, B and C will be collinear if, area of
ABC 0
Now, find the area of the .
ABC
1 2 3 2 3 1 3 1 2
1
Area of a triangle x y y x y y x y y
2
1
Area of ABC 6 6 8 4 8 10 3 10 6
2
 
1
6 14 4 18 3 4
2
84 72 12
0
So, the points A, B and C are collinear.
Now, find the lengths of AB and AC by using the
distance formula.
Distance between two points is given by the
formula,
2 2
1 2 1 2
l x x y y
2 2
AC 3 6 8 10
2
2
9 18
81 324
405
9 5 units
2
2
AB 4 6 6 10
2
2
2 4
4 16
20
2 5 units
Now,
AB 2 5
AC
9 5
2
AB AC
9
Hence, the statement is true.
Question: 11
The point P lies on a circle of radius 6 and
2,4
centre C(3, 5).
Solution:
The statement is false.
The point P lies on a circle if, distance between P
and the centre is equal to the radius.
So, find the distance of P from the centre.
Distance between two points is given by the
formula,
2 2
1 2 1 2
l x x y y
2 2
PC 2 3 4 5
2 2
5 1
25 1
26
6units
So,
26 6
Thus, the point P does not lie on the circle.
Question: 12
The points A , B(4, 3), C(2, 5) and D
1, 2
3,0
in that order form a rectangle.
Solution:
The statement is true.
ABCD will form a rectangle if it is a parallelogram
and its diagonals are equal.
Now, if ABCD is a parallelogram, its diagonals
bisect each other.
So, mid–point of mid–point of BD
AC
Now, find the mid–points.
Mid–point of
1 2 2 5
AC ,
2 2
1 3
,
2 2
Mid–point of
4 3 3 0
BD ,
2 2
1 3
,
2 2
Now,
1 3 1 3
, ,
2 2 2 2
So, ABCD is a parallelogram.
Now, find the lengths of diagonals AC and BD by
the distance formula.
Distance between two points is given by the
formula,
2 2
1 2 1 2
l x x y y
2 2
AC 2 1 5 2
2 2
3 7
9 49
2 2
7 3
49 9
Now, length of diagonal length of diagonal
AC
BD.
Hence, ABCD is a rectangle.
Exercise 7.3 (20)
Question: 1
Name the type of triangle formed by the points A
, B and C(7, 5).
5,6
4, 2
Solution:
Given points are A , B and C(7, 5).
5,6
4, 2
Now, find the lengths of the sides AB, BC and AC
of .
ABC
Distance between two points is given by the
formula,
2 2
1 2 1 2
l x x y y
2
2
AB 4 5 2 6
2
2
1 8
1 64
65units
So,
2
AB 65units
2 2
BC 7 4 5 2
2
2
11 7
121 49
170 units
So,
2
BC 170units
2
2
AC 7 5 5 6
2
2
12 1
144 1
145units
So,
2
AC 145units
As, .
AB BC AC
So, the is a scalene triangle.
ABC
Now,
2 2
AC AB 145 65 210
Which is not equal to .
2
BC
So, is not a right-angled triangle.
ABC
Hence, the given triangle is a scalene triangle.
Question: 2
Find the points on the axis which are at a
x
distance of from the point . How many
2 5
7, 4
such points are there?
Solution:
Let a point R be a point on axis which is at
x,0
x
a distance of from the point A .
2 5
7, 4
So,
AR 2 5
Now, find the distance between the points A and R.
Distance between two points is given by the
formula,
2 2
1 2 1 2
l x x y y
2
2
AR x 7 0 4
2
2
2 5 x 49 14x 4
Take a square on both sides.
2
2
2
2
2 5 x 49 14x 4
2
4 5 x 49 14x 16
2
20 x 14x 65
2
x 14x 45 0
2
x 9x 5x 45 0
x x 9 5 x 9 0
x 5 x 9 0
So, .
x 9or x 5
Thus, the total number of points on the axis,
x
whose distance from point A is .
2 5
Hence, the required points are (9, 0), (5, 0).
Question: 3
What type of a quadrilateral do the points A ,
2, 2
B(7, 3), C and D taken in that order,
11, 1
6, 6
form?
Solution:
Let us consider the following cases:
(i) If the mid points of diagonals AC and BD are
same, then the quadrilateral is a parallelogram.
For parallelogram with vertices A , B(7,
2, 2
3), C , D , find whether mid-point
11, 1
6, 6
of mid-point of BD
AC
Mid-point of
AC
11 2 2 1
,
2 2
13 3
,
2 2
Mid-point of BD
7 6 3 6
,
2 2
13 3
,
2 2
So, mid-point of mid-point of BD
AC
13 3
,
2 2
Thus, ABCD is a parallelogram.
(ii) A parallelogram is not a rectangle, if diagonals
.
AC BD
Let us find the lengths of the diagonals, AC
and BD.
Distance between two points is given by the
formula,
2 2
1 2 1 2
l x x y y
2
2
AC 11 2 1 2
2 2
9 1
81 1
82 units
2 2
BD 6 7 6 3
2 2
1 9
1 81
82 units
So,
AC BD 82
Thus, ABCD is a rectangle or a square.
(iii) A parallelogram may be a square or a rhombus
if .
AB BC
Let us find the lengths of the sides, AB and BC.
Distance between two points is given by the
formula,
2 2
1 2 1 2
l x x y y
2
2
AB 7 2 3 2
2 2
5 5
25 25
50
5 2 units
2 2
BC 11 7 1 3
2
2
4 4
16 16
32
4 2 units
Now,
AB BC
Thus, ABCD is not a square or a rhombus.
Hence, ABCD is a rectangle.
Question: 4
Find the value of , if the distance between the
a
points A and B is 9 units.
3, 14
a, 5
Solution:
According to the question,
Distance between the points A and B is 9 units.
So,
AB 9
Now, find the value of , using the distance
a
formula.
Distance between two points is given by the
formula,
2 2
1 2 1 2
l x x y y
2 2
AB a 3 5 14
2
2
9 a 3 9
Take a square of both sides.
2
2
2
9 a 9 6a 81
2
81 a 6a 90
2
90 81 a 6a
2
a 6a 9 0
2
a 3a 3a 9 0
a a 3 3 a 3 0
2
a 3 0
a 3 0
a 3
Hence, the value of is .
a
3
Question: 5
Find a point which is equidistant from the points A
and B ? How many such points are
5,4
1,6
there?
Solution:
Let us consider a point T equidistant from the
x,y
points A and B .
5,4
1,6
So,
AT BT
Find the lengths of AT and BT, using the distance
formula.
Distance between two points is given by the
formula,
2 2
1 2 1 2
l x x y y
2
2
AT x 5 y 4
2 2
x 5 y 4
2 2
x 25 10x y 16 8y
2 2
x y 10x 8y 41 ...... 1
2
2
BT x 1 y 6
2 2
x 1 y 6
2 2
x 1 2x y 36 12y
2 2
x y 2x 12y 37 ...... 2
Equate equations and and take a square of
1
2
both sides.
2 2
2 2 2 2
x y 10x 8y 41 x y 2x 12y 37
2 2 2 2
x y 10x 8y 41 x y 2x 12y 37
10x 2x 8y 12y 37 41
8x 4y 4
2x y 1
The above equation shows that there are infinite
number of points which are equidistant from AB.
This is because all the points on perpendicular
bisector of AB will be at an equal distance from AB.
Now, find one point, which is equidistant from AB.
The mid–point of AB, is equidistant from the points
A and B.
The coordinates of mid–point of AB are given by,
5 1
4 6
,
2 2
Coordinates
6 10
,
2 2
3,5
Hence, the point is equidistant from the
3,5
points A and B.
Question: 6
Find the coordinates of the point Q on the axis
x
which lies on the perpendicular bisector of the line
segment joining the points A and B .
5, 2
4, 2
Name the type of triangle formed by the points Q,
A and B.
Solution:
Let Q be a point on axis which lies on the
x,0
x
perpendicular bisector of AB.
Now,
AQ BQ
Find the lengths of AQ and BQ, using the distance
formula.
Distance between two points is given by the
formula,
2 2
1 2 1 2
l x x y y
2 2
AQ x 5 0 2
2 2
x 5 2
2
x 25 10x 4
2
x 10x 29 ...... 1
2
2
BQ x 4 0 2
2 2
x 4 2
2
x 16 8x 4
2
x 8x 20 ...... 2
Equate equations and and take a square of
1
2
both sides.
2 2
2 2
x 10x 29 x 8x 20
2 2
x 10x 29 x 8x 20
10x 8x 20 29
18x 9
9
x
18
1
x
2
Thus, the point Q is .
1
,0
2
Now, find the lengths of the sides AQ and BQ by
putting the value of in the equations .
x
1 and 2
2
1 1
AQ 10 29
2 2
1
5 29
4
1 20 116
4
97
4
97
units
2
2
1 1
BQ 8 20
2 2
1
4 20
4
1 16 80
4
97
4
97
units
2
Now, find the length of AB using the distance
formula again.
2 2
AB 4 5 2 2
2 2
4 5 2 2
2
9 0
81
9units
Thus,
AQ BQ AB
Hence, is an isosceles triangle.
QAB
Question: 7
Find the value of if the points (5, 1), and
m
2, 3
are collinear.
8, 2m
Solution:
Let us consider the points be P(5, 1), Q and
2, 3
R .
8, 2m
The points P, Q and R will be collinear, if the area of
is 0.
PQR
Now, find the value of using the formula for area
m
of a triangle.
1 2 3 2 3 1 3 1 2
1
Area of a triangle x y y x y y x y y
2
1
Area of PQR 5 3 2m 2 2m 1 8 1 3
2
1
15 10m 4m 2 32
2
1
14m 19
2
Now, Area of .
PQR 0
1
14m 19 0
2
14m 19 0
14m 19
19
m
14
Hence, the value of .
19
mis
14
Question: 8
If the point A is equidistant from P(3, 8) and
2, 4
Q , find the values of . Also find distance
10,y
y
PQ.
Solution:
According to the question,
AP AQ
Now, find the lengths of AP and AQ, using the
distance formula.
Distance between two points is given by the
formula,
2 2
1 2 1 2
l x x y y
2
2
AP 3 2 8 4
2 2
1 12
1 144
145 ...... 1
2
2
AQ 10 2 y 4
2 2
12 y 4
2
144 y 16 8y
2
y 8y 160 ...... 2
Equate equations and and find a square on
1
2
both sides.
2
2
2
145 y 8y 160
2
145 y 8y 160
2
y 8y 160 145
2
y 8y 15
2
y 8y 15 0
2
y 3y 5y 15 0
y y 3 5 y 3 0
y 3 y 5 0
or
y 3
y 5
Now, find the length of PQ using the coordinates
P(3, 8) and Q .
10, 3
2 2
PQ 10 3 3 8
2
2
13 11
169 121
290 units
Now, find the length of PQ using the coordinates
P(3, 8) and .
Q
10, 5
2 2
PQ 10 3 5 8
2
2
13 13
169 169
338
13 2 units
Hence, and
y 3, 5
units and units.
PQ 290
13 2
Question: 9
Find the area of the triangle whose vertices are
and .
8, 4 ,
6,6
3, 9
Solution:
Let the vertices of the triangle be
P 8,4 ,
Q 6,6
and .
R 3,9
Now, find the area of .
PQR
1 2 3 2 3 1 3 1 2
1
Area of a triangle x y y x y y x y y
2
1
Area of PQR 8 6 9 6 9 4 3 4 6
2
1
8 3 6 5 3 2
2
1
24 30 6
2
1
30 30
2
1
0
2
0
Hence, area of the given triangle is 0.
Question: 10
In what ratio does the axis divide the line
x
segment joining the points ?
4, 6 and 1,7
Find the coordinates of the point of division.
Solution:
Let the point Z , on the axis, intersect the
x,0
x
line joining the points P and Q in the
4,6
1,7
ratio .
t :1
Now, apply the section formula to find the value of
.
t
1 2 2 1
1 2
m y m y
y
m m
So, by the section formula,
t 7 1 6
y ...... 1
t 1
Now, put the values of in the above section
y
formula.
t 7 1 6
0
t 1
0 7t 6
7t 6
6
t
7
Now, and
1
m 6
2
m 7
Now, again use the section formula.
1 2 2 1
1 2
m x m x
x
m m
So, by the section formula,
6 1 7 4
x
6 7
6 28
13
34
13
Find the value of from equation .
y
1
6 7 7 6
y
6 7
42 42
13
0
Hence, the required point of intersection is Z
.
34
,0
13
Question: 11
Find the ratio in which the point P divides
3 5
,
4 12
the line segment joining the points A and B
1 3
,
2 2
.
2, 5
Solution:
Let the point P divides the line segment AB in the
ratio .
t :1
Now, apply the section formula to find the value of
.
t
1 2 2 1
1 2
m x m x
x
m m
So, by the section formula,
1
t 2 1
2
x
t 1
Now, put the values of in the above section
x
formula.
4t 1
3
2
4 t 1
3 4t 1
4 2 t 1
6 t 1 16t 4
6t 6 16t 4
16t 6t 6 4
10t 2
2
t
10
1
t
5
Now, and
1
m 1
2
m 5
Hence, P divides AB in the ratio .
1:5
Question: 12
If P divides line segment joining A
9a 2, b
and B in the ratio , find the
3a 1, 3
8a,5
3:1
values of .
a andb
Solution:
According to the question,
Point P divides the line segment joining
9a 2, b
the points A and B in the ratio
3a 1, 3
8a,5
.
3:1
Now, apply the section formula to find the values of
.
a
and
1 2 2 1
1 2
m x m x
x
m m
1 2 2 1
1 2
m y m y
y
m m
So, by the section formula,
3 8a 1 3a 1
9a 2
3 1
24a 3a 1
9a 2
4
36a 8 27a 1
36a 27a 1 8
9a 9
9
a
9
a 1
Now, apply the section formula to find the values of
.
b
3 5 1 3
b
3 1
15 3
b
4
4b 12
12
b
4
b 3
Hence, the values of
a andb are1and 3
respectively.
Question: 13
If is the mid–point of the line segment joining
a,b
the points A and B and ,
10, 6
k,4
a 2b 18
find the value of and the distance AB.
k
Solution:
Let us consider, Z be the mid–point of the line
a,b
segment joining the points A and B .
10, 6
k,4
So, find the coordinates of Z.
10 k
a ...... 1
2
6 4
b
2
2
b
2
b 1
Now, according to the question,
a 2b 18 ...... 2
Put the value of in the equation .
b
2
a 2 1 18
a 2 18
a 18 2
a 16
Now, from equation ,
1
10 k
a
2
Put the value of in the above equation.
a
10 k
16
2
16 2 10 k
10 k 32
k 32 10
k 22
Now, the coordinates of A and B are and
10, 6
respectively.
22, 4
Find the length of AB using the distance formula.
Distance between two points is given by the
formula,
2 2
1 2 1 2
l x x y y
2
2
AB 22 10 4 6
2 2
12 10
144 100
244
2 61units
Hence, the required values of
k 22,
a 16,
b 1
and .
AB 2 61units
Question: 14
The centre of a circle is . Find the values
2a,a 7
of if the circle passes through the point
a
11, 9
and has diameter units.
10 2
Solution:
Let O be the centre of the circle and it
2a,a 7
passes through the point A .
11, 9
According to the question,
AB
10 2
Therefore,
1
AO AB
2
1
AO 10 2
2
AO 5 2 ...... 1
Now, find the length of AO using the distance
formula.
Distance between two points is given by the
formula,
2 2
1 2 1 2
l x x y y
2
2
AO 2a 11 a 7 9
2
2
4a 121 44a a 2
2 2
4a 121 44a a 4 4a
2
5a 40a 125
2
Equate equations and .
1
2
2
5a 40a 125 5 2
Take a square of both sides.
2
2
2
5a 40a 125 5 2
2 2
5a 40a 125 5 2
2
5a 40a 50 125
2
5a 40a 75 0
2
a 8a 15 0
2
a 3a 5a 15 0
a a 3 5 a 3 0
a 3 a 5 0
a 5
a 3
Hence, the required values of are 5 and 3.
a
Question: 15
The line segment joining the points A(3, 2) and B(5,
1) is divided at the point P in the ratio and it
1:2
lies on the line . Find the value of .
3x 18y k 0
k
Solution:
Now, let us consider, and
1
x 3,
1
y 2,
2
x 5
2
y 1
And the ratio,
1 2
1:2 m :m
Now, apply the section formula.
and
1 2 2 1
1 2
m x m x
x
m m
1 2 2 1
1 2
m y m y
y
m m
So, by the section formula,
1 5 2 3
x
1 2
5 6
3
11
3
1 1 2 2
y
1 2
1 4
3
5
3
Here, and
11
x
3
5
y
3
So, the coordinates of point P are .
11 5
,
3 3
Now, P lies on the line .
11 5
,
3 3
3x 18y k 0
So, find the value of , by putting the coordinates
k
of P in the equation of the given line.
11 5
3 18 k 0
3 3
11 6 5 k 0
11 30 k 0
k 19 0
k 19
Hence, the required value of is 19 units.
k
Question: 16
If D , E(7, 3) and F , are the mid–points
1 5
,
2 2
7 7
,
2 2
of sides of , find the area of the .
ABC
ABC
Solution:
According to the question,
The points D , E(7, 3) and F are the
1 5
,
2 2
7 7
,
2 2
mid–points of sides AB, BC and AC of .
ABC
So,
Thus, area of the
ABC 4 Area of DEF
Now, find the area of .
DEF
1 2 3 2 3 1 3 1 2
1
Area of a triangle x y y x y y x y y
2
1 1 7 7 5 7 5
Area of DEF 3 7 3
2 2 2 2 2 2 2
1 1 6 7 7 5 7 5 6
7
2 2 2 2 2 2
1 1 1 2 7 1
7
2 2 2 2 2 2
1 1 7
7
2 4 4
1 1 28 7
2 4
1 22
2 4
1 11
2 2
11
4
Now, area of the
ABC 4 Area of DEF
So, area of the
11
ABC 4
4
11
Hence, the area of triangle is 11 square units.
Question: 17
The points A(2, 9), B and C(5, 5) are the
a,5
vertices of a triangle ABC right angled at B. Find
the values of and hence the area of .
a
ABC
Solution:
First, find the lengths of the sides AB, BC and AC
using the distance formula.
Distance between two points is given by the
formula,
2 2
1 2 1 2
l x x y y
2 2
AB a 2 5 9
2
2
a 4 4a 4
2
a 4 4a 16
2
a 4a 20 ...... 1
2 2
BC 5 a 5 5
2
25 a 10a 0
2
a 10a 25 ...... 2
2 2
AC 5 2 5 9
2
2
3 4
9 16
25
5units ...... 3
Now, apply the Pythagoras theorem to find the
value of .
a
According to Pythagoras theorem,
2 2 2
AB BC AC ...... 4
Put the values of AB, BC and AC in equation .
4
2 2
2 2 2
a 4a 20 a 10a 25 5
2 2
a 4a 20 a 10a 25 25
2
2a 14a 45 25
2
2a 14a 20 0
2
a 7a 10 0
2
a 5a 2a 10 0
a a 5 2 a 5 0
a 5 a 2 0
or
a 5
a 2
Now, if , then the coordinates of B will be (5,
a 5
5).
Then, the length of side BC will become 0.
But, this is not possible so,
a 2
Now, substitute the value of in the equations
a
1
and , to find the lengths of sides AB and BC.
2
2
AB 2 4 2 20
4 8 20
4 8 20
4units
2
BC 2 10 2 25
4 20 25
9
3units
Now, find the area of .
ABC
1
Area of a triangle base height
2
1
Area of ABC BC AB
2
1
= 3 4
2
3 2
6squareunits
Hence, the required value of is 2, and the area of
a
is 6 square units.
ABC
Question: 18
Find the coordinates of the point R on the line
segment joining the points P and Q(2, 5) such
1,3
that .
3
PR PQ
5
Solution:
According to the question,
So,
3
PR PQ
5
PQ 5
PR 3
PR RQ 5
PR 3
PR RQ 5
PR PR 3
RQ 5
1
PR 3
RQ 5
1
PR 3
RQ 5 3
PR 3
RQ 2
PR 3
It can also be written as
PR 3
RQ 2
So,
PR:RQ 3: 2
Thus, and
1
m 3
2
m 2
Now, find the coordinates of R using the section
formula.
and
1 2 2 1
1 2
m x m x
x
m m
1 2 2 1
1 2
m y m y
y
m m
So, by the section formula,
3 2 2 1
x
3 2
6 2
5
4
5
3 5 2 3
y
3 2
15 6
5
21
5
Hence, the coordinates of R are .
4 21
,
5 5
Question: 19
Find the values of if the points A , B
k
k 1,2k
and C are collinear.
3k,2k 3
5k 1,5k
Solution:
The given points A , B and C
k 1,2k
3k,2k 3
are collinear so, the area of is
5k 1,5k
ABC
equal to 0.
1 2 3 2 3 1 3 1 2
1
Area of a triangle x y y x y y x y y
2
Area of is given by,
ABC
1
k 1 2k 3 5k 3k 5k 2k 5k 1 2k 2k 3 0
2
2 2
2 2
k 1 3 3k 3k 3k
5k 1 3 3k 3k 3 3k 9k 0
15k 3 3k 3 9k 15k 3
2
6k 15k 6 0
2
2k 5k 2 0
2
2k 4k k 2 0
2k k 2 1 k 2 0
2k 1 k 2 0
or
1
k
2
k 2
Hence, the values of are 2 and .
k
1
2
Question: 20
Find the ratio in which the line
2x 3y 5 0
divides the line segment joining the points
8, 9
and (2, 1). Also, find the coordinates of the point of
division.
Solution:
Let the given line divide the line segment joining
the points P and Q(2, 1) at the point O
8, 9
x,y
in the ratio .
t :1
Now, find the coordinates of O using the section
formula.
and
1 2 2 1
1 2
m x m x
x
m m
1 2 2 1
1 2
m y m y
y
m m
So, by the section formula,
t 2 1 8
x
t 1
2t 8
t 1
t 1 1 9
y
t 1
t 9
t 1
So, the coordinates of O are .
2t 8 t 9
,
t 1 t 1
Now, according to the question.
The point O lies on the line .
2x 3y 5 0
So, substitute the values of in the equation
x and y
of given line.
2t 8 t 9
2 3 5 0
t 1 t 1
Multiply by on both sides.
t 1
2 2t 8 3 t 9 5 t 1 0
4t 16 3t 27 5t 5 0
2t 16 0
2t 16
16
t
2
t 8
So, the ratio is .
8:1
Now, the coordinates of O are .
2t 8 t 9
,
t 1 t 1
Substitute the value of .
t
2 8 8 8 9 24 1
, ,
8 1 8 1 9 9
8 1
,
3 9
Hence, the line divides AB in ratio
2x 3y 5 0
at point O .
8:1
8 1
,
3 9
Exercise 7.4 (6)
Question: 1
If and (4, 3) are two vertices of an equilateral
4,3
triangle, find the coordinates of the third vertex,
given that the origin lies in the interior of the
triangle.
Solution:
Let the three vertices of the triangle be P ,
4,3
Q(4, 3) and R .
x,y
Now, is equilateral.
PQR
So, all the three sides are equal.
Thus,
PQ QR PR
Find the lengths of the sides PQ, QR and PR.
Distance between two points is given by the
formula,
2 2
1 2 1 2
l x x y y
2
2
PQ 4 4 3 3
2
8 0
64
8units ...... 1
2 2
QR x 4 y 3
2 2
x 16 8x y 9 6y
2 2
x y 8x 6y 25 ...... 2
2
2
PR x 4 y 3
2 2
x 16 8x y 9 6y
2 2
x y 8x 6y 25 ...... 3
Now,
PQ PR
Equate equations and .
1
3
2 2
8 x y 8x 6y 25
Take a square of both sides.
2 2
x y 8x 6y 25 64
2 2
x y 8x 6y 39 ...... 4
Also,
PQ QR
Equate equations and .
1
2
2 2
x y 8x 6y 25 8
Take a square of both sides.
2 2
x y 8x 6y 25 64
2 2
x y 8x 6y 39 ...... 5
Subtract equation from .
4
5
2 2 2 2
x y 8x 6y x y 8x 6y 39 39
2 2 2 2
x y 8x 6y x y 8x 6y 0
16x 0
x 0
Substitute the value of in equation .
x
4
2 2
2
0 y 8 0 6y 39
y 6y 39 0
Now, compare the above equation with .
2
b 4ac
So, and
a 1,
b 6
c 39
2
2
b 4ac 6 4 1 39
36 156
192
Now, find the value of by using the equation
y
2
b b 4ac
y
2a
6 192
y
2 1
6 8 3
y
2
y 3 4 3
and
y 3 4 3
3 4 3
So, the third vertex of the triangle is
1
R 0,3 4 3
or .
2
R 0,3 4 3
Now, solve the roots.
1 1
R 0,3 4 3 R 0,3 4 1.732
1
R 0,3 6.9
1
R 0, 9.9
And,
2 2
R 0,3 4 3 R 0,3 4 1.732
2
R 0,3 6.9
2
R 0, 3.9
Hence, from the above figure it can be stated, that
the third vertex is so that origin lies
2
R 0,3 4 3
inside the triangle.
Question: 2
A(6, 1), B(8, 2) and C(9, 4) are three vertices of a
parallelogram ABCD. If E is the mid–point of DC,
find the area of .
Solution:
According to the question,
ABCD is a parallelogram.
So, mid–point of diagonal BD mid–point of
diagonal AC
Mid–point of BD
8 x 2 y
,
2 2
Mid–point of AC
6 9 1 4
,
2 2
15 5
,
2 2
Equate both the mid–points.
8 x 2 y 15 5
, ,
2 2 2 2
8 x 15
2 2
2 8 x 15 2
16 2x 30
2x 14
14
x
2
x 7
Similarl
5
2
y,
2 y
2
2 2 y 5 2
4 2y 10
2y 6
6
y
2
y 3
So, the coordinates of the fourth vertex are D .
7,3
Now, E is the mid–point of DC.
So, find the coordinates of E.
Coordinates
7 9 3 4
,
2 2
16 7
,
2 2
7
8,
2
Now, find the area of .
1 2 3 2 3 1 3 1 2
1
Area of a triangle x y y x y y x y y
2
1 7 7
Area of ADE 6 3 7 1 8 1 3
2 2 2
1 6 7 7 2
6 7 8 2
2 2 2
1 1 5
6 7 16
2 2 2
1 35
3 16
2 2
1 6 35 32
2 2
1 3
2 2
3
squareunits
4
Hence, area of as area cannot
3
4
be negative.
Question: 3
The points A , B and C are the
1 1
x ,y
2 2
x , y
3 3
x , y
vertices of .
ABC
(i) The median from A meets BC at D. Find the
coordinates of the point D.
(ii) Find the coordinates of the point P on AD such
that
AP:PD 2:1
(iii) Find the coordinates of points Q and R on
medians BE and CF, respectively such that
and
BQ :QE 2:1
CR:RF 2:1
(iv) What are the coordinates of the centroid of the
triangle ABC?
Solution:
(i) Median from A meets BC at D.
So, D is the mid–point of BC.
Find the coordinates of D.
Coordinates
2 3 2 3
x x y y
,
2 2
(ii)
Now, apply the section formula.
and
1 2 2 1
1 2
m x m x
x
m m
1 2 2 1
1 2
m y m y
y
m m
So, by the section formula,
2 3
1
x x
2 1 x
2
x
2 1
2 3 1
x x x
3
2 3
1
y y
2 1 y
2
y
2 1
2 3 1
y y y
3
Hence, the required point is
.
2 3 1 2 3 1
x x x y y y
P ,
3 3
(iii) The median BE meets the side AC at its mid–
point E.
So, find the coordinates of E.
Coordinates
1 3 1 3
x x y y
,
2 2
Now, apply the section formula to find the
coordinates of Q.
and
1 2 2 1
1 2
m x m x
x
m m
1 2 2 1
1 2
m y m y
y
m m
So, by the section formula,
1 3
2
x x
2 1 x
2
x
2 1
1 3 2
x x x
3
1 3
2
y y
2 1 y
2
y
2 1
1 3 2
y y y
3
Hence, the coordinates of point Q are
2 3 1 2 3 1
x x x y y y
,
3 3
Now, median CF meets the side AB at its mid–
point F.
So, find the coordinates of F.
Coordinates
1 2 1 2
x x y y
,
2 2
Now, apply the section formula to find the
coordinates of R.
and
1 2 2 1
1 2
m x m x
x
m m
1 2 2 1
1 2
m y m y
y
m m
So, by the section formula,
1 2
3
x x
2 1 x
2
x
2 1
1 2 3
x x x
3
1 2
3
y y
2 1 y
2
y
2 1
1 2 3
y y y
3
Hence, the coordinates of point R are
2 3 1 2 3 1
x x x y y y
,
3 3
(iv) Coordinates of the centroid G of are
ABC
.
2 3 1 2 3 1
x x x y y y
,
3 3
Now, the coordinates of P, Q, R and G are
same.
Hence, the medians intersect at the same point
i.e., at the centroid G, which divides the
medians in the ratio .
2:1
Question: 4
If the points A , B(2, 3) C and D
1, 2
a,2
4, 3
form a parallelogram, find the value of and height
a
of the parallelogram taking AB as base.
Solution:
According to the question,
ABCD is a parallelogram with points A , B(2,
1, 2
3), C and D .
a,2
4, 3
Now, diagonals of parallelogram bisect each other.
So, mid–point of diagonal mid–point of
AC
diagonal BD
Mid–point of
1 a 2 2
AC ,
2 2
1 a
,0
2
Mid–point of
4 2 3 3
BD ,
2 2
2
,0
2
1,0
Equate the mid–points.
1 a
,0 1,0
2
1 a
1
2
1 a 2
a 2 1
a 3
Now, find the height of the parallelogram.
For this, let us take the formula for area of triangle.
1 2 3 2 3 1 3 1 2
1
Area of a triangle x y y x y y x y y
2
1
Area of ABD 1 3 3 2 3 2 4 2 3
2
1
3 3 2 3 2 4 5
2
1
6 2 1 20
2
1
6 2 20
2
1
24
2
12units ...... 1
Also,
1
Area of a triangle base height
2
1
Area of a ABD AB h
2
Now, find the length AB of the parallelogram, using
the distance formula.
Distance between two points is given by the
formula,
2 2
1 2 1 2
l x x y y
2
2
AB 2 1 3 2
2 2
1 5
1 25
26 units
1
Area of a ABD 26 h
2
26h
units
2
2
Equate equations and .
1
2
26h
12
2
26h 12 2
24
h
26
Multiply and divide by .
26
24 26
h
26 26
24 26
26
12
26
13
Hence, the height of the parallelogram is
12
26
13
units.
Question: 5
Students of a school are standing in rows and
columns in their playground for a drill practice. A,
B, C and D are the positions of four students as
shown in given figure. Is it possible to place Jaspal
in the drill in such a way that he is equidistant from
each of the four students A, B, C and D? If so, what
should be his position?
Solution:
Coordinates of A, B, C and D from the given figure
are A(3, 5), B(7, 9), C(11, 5), and D(7, 1).
Now, let us find the shape of the quadrilateral
ABCD.
For this, find the lengths AB, BC, CD and AD.
Distance between two points is given by the
formula,
2 2
1 2 1 2
l x x y y
2 2
AB 7 3 9 5
2 2
4 4
16 16
32
4 2 units
2 2
BC 11 7 5 9
2
2
4 4
16 16
32
4 2 units
2 2
CD 7 11 1 5
2 2
4 4
16 16
32
4 2 units
2 2
2
2
4 4
16 16
32
4 2 units
Now,
AB BC CD AD 4 2 units
So, ABCD is either a square or a rhombus.
Let us find the length of the diagonals AC and BD.
2 2
AC 11 3 5 5
2
8 0
64
8units
2 2
BD 7 7 1 9
2
0 8
64
8units
Now,
AC BD 8units
So, the quadrilateral ABCD is a square.
Now, find the point, which is equidistant from the
points A, B, C and D of the square ABCD.
This point will be at the intersecting point of
diagonals as in a square, diagonals bisect each
other.
So, find the coordinates of the required point.
Coordinates
7 7 9 1
,
2 2
14 10
,
2 2
7,5
Hence, the required point is (7, 5).
Question: 6
Ayush starts walking from his house to office.
Instead of going to the office directly, he goes to a
bank first, from there to his daughter’s school and
then reaches the office. What is the extra distance
travelled by Ayush in reaching his office? (Assume
that all distances covered are in straight lines). If
the house is situated at (2, 4), bank at (5, 8), school
at (13, 14) and office at (13, 26) and coordinates are
in km.
Solution:
Let us consider the coordinates of house, bank,
school and office be H(2, 4), B(5, 8), S(13, 14) and
O(13, 26) respectively.
Now, find the distances from house to bank (HB),
bank to school (BS), and school to office (SO).
Distance between two points is given by the
formula,
2 2
1 2 1 2
l x x y y
2 2
HB 5 2 8 4
2 2
3 4
9 16
25
5km
2 2
8 6
2 2
8 6
100
10km
2 2
SO 13 13 26 14
2
0 12
144
12km
So, the total distance travelled by Ayush from
house to bank to school and then to office
HB BS SO 5 10 12 27km
Now, direct distance from house to office
HO
Find the distance HO.
2 2
HO 13 2 26 4
2 2
11 22
121 484
605
24.6km
Thus, the extra distance travelled by Ayush
27 24.6
2.4km
Hence, extra distance travelled by Ayush is .
2.4km