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.

So, it is equal to 3 units.

Question: 2

The distance between the points A(0, 6) and B(0, –2)

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

 

  

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

OP 6 0 8 0

 

  

( 6) 8

 36 64

 100

10units

Hence, .

OP 10units

Question: 4

The distance between the points (0, 5) and (–5, 0) is

a. 5

5 2

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

 

 

(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

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

 

  

(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

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

 

 

 16

 4cm

   

   

2 2

QR 3 0 0 0

 



 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

Area of a triangle x y y x y y x y y

   

      

Area of ABC 3 0 4 7(4 0) 8 0 0

 

      

3 4 7 4 8 0

   

12 28 0



 

 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

 

    

PQ 4 4 0 0

 



 64

 8cm

   

   

2 2

QR 0 4 3 0

   

  

2 2

4 3

 16 9

 25

 5cm

 

    

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

a. I quadrant

b. II quadrant

c. III quadrant

d. IV quadrant

Solution:

(d)

Let be the required point.

 

R x,y

Now, let us consider, and



x 7,

 

y 6,



x 3



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

m m







1 2 2 1

1 2

m y m y

m m

So, by the section formula,

  





1 3 2 7

1 2





3 14



 

   





1 4 2 6

1 2





4 12





Here, and







Thus, the point is which lies in the IV



 

 

 

17 8

3 3

quadrant.

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)

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





y 7

Now, find the values of .





x 6

 x 6 6

 x 6 6

x 0

Similarly, find the value of .





y 7

 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)



AP AB

b. (B)

AP PB

c. (C)



PB AB

d. (D)



AP AB

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

m m

So, by the section formula,

  





t 8 1 4

t 1

  2t 2 8t 4

 2 6t





 

Verification:

Now, as so, the point P lies outside the line





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,



AP AB

 

1 1

1 1, whichis true

Hence, proved.

Question: 13

If P is the mid–point of the line segment

 

 

 

joining the points Q and R , then the

 

6,5

 

2,3

value of is

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.

 

 

 

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





 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

   

1 y 25 10y

  

y 10y 26

   

   

2 2

BR 0 4 y 6

 

    

4 y 36 12y

   

16 y 36 12y

  

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

y 10y 26 y 12y 52

    

2 2

y 10y 26 y 12y 52

    10y 26 12y 52

2y 26



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

 

x,y

 

y,x

 

 

 

x y

2 2

 

 

 

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

 

 

 

interior of the circle is

a. (A)



 

 

 

b. (B)

 

 

 

c. (C)



 

 

 

d. (D)

 



 

 

Solution:

(d)

First find the radius of the given circle having one

point as the origin and other point as .

 

 

 

Distance between two points is given by the

formula,

   

   

2 2

1 2 1 2

l x x y y

 

 

   

 

 

Radius 0 0 0

 



 

 



 6.5units

Now, let us consider all the four cases.

a. Distance of the point from



 

 

 

 

0,0

 



 

   

 

 

Distance 0 1 0



 

 

 

 

 

 

25 5

16 4

1.25units

The distance

1.25 6.5

So, the point lies in the interior of the



 

 

 

circle.

b. Distance of the point from

 

 

 

 

0,0

 

 

   

 

 

Distance 2 0 0

 

 

 

 

 



 3.073units

The distance

3.0731 6.5

So, the point lies in the interior of the circle.

 

 

 

c. Distance of the point from



 

 

 

 

0,0

 

 

    

 

 

Distance 5 0 0

 

  

 

 

 



101

 5.0249units

The distance

5.0249 6.5

So, the point lies in the interior of the circle.



 

 

 

d. Distance of the point from .

 



 

 

 

0,0

 

 

    

 

 

Distance 6 0 0

 

 

 

 

 



169



 6.5units

The distance

6.5 6.5

So, the point lies on the circle and not in

 



 

 

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 4

Now, find the value of .

 

 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)

 

 

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

Area of a triangle x y y x y y x y y

 

  

    

   

a c a a b

Area of given triangle b a b (b c

c b c (c a

    



     

2 2 2

ac a a ab ab b

b bc bc c c ac

     

ac ab ab bc bc ac

 

 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

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

 

5 9 p

Take a square of both sides.

 

 

5 9 p

 

25 9 p

 

p 25 9



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 b

a 2b

2a b

 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

Area of a triangle x y y x y y x y y

     

     

0 1 0 b 0 b 2 a 2 0

   

0 b 0 2a

  0 b 2a

2a b

Hence, the answer.is (C).

Exercise 7.2 (12)

State whether the following statements are true or

false. Justify your answer.

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

 

    

AB 2 2 0 0

 

 

4 0

 16

   

   

2 2

BC 0 2 2 0

 

  

2 2

 4 4

 8

 2 2 units

 

    

AC 0 2 2 0

 

 

2 2

 4 4

 8

 2 2 units

In ,

DEF

 

    

DE 4 4 0 0

 

8 0

 64

 8units

   

   

2 2

EF 0 4 4 0

 

  

4 4

 16 16

 32

 4 2 units

 

    

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

Area of a triangle x y y x y y x y y

     

        

Area of ABP 4 6 2 4 2 6 4 6 6

     

     

4 8 4 4 4 12

  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

Area of a triangle x y y x y y x y y

     

 

        

Area of PQR 0 9 6 0 6 5 3 5 9

 

  

0 0 0 3 14

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

   

 

    

BC 0 12 2 2

   

  

2 2

12 4

 144 16

 160

 4 10 units

   

   

2 2

AC 0 3 2 1

 

  

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).

So, radius

 OP

Find the radius 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

 

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

 

  

4 2

 16 4

 20

 2 5 units

   

 

    

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

and







1 2 2 1

1 2

m x m x

m m







1 2 2 1

1 2

m y m y

m m

So, by the section formula,

and

   



 



 



 

t 1 1 7

t 1

   

  

 



 



 

t 5 1 2

t 1

Now, put the values of in the above section

x and y

formula.



 



 



 

t 7

t 1

 

  5 t 1 t 7

  5t 5 t 7

  5t t 7 5

4t 2



 

 

 

 



 

5t 2

and 3

t 1

 

    3 t 1 5t 2

    3t 3 5t 2

  5t 3t 3 2

2t 1



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 .



AB AC

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

Area of a triangle x y y x y y x y y

 

   

         

Area of ABC 6 6 8 4 8 10 3 10 6

     

    

6 14 4 18 3 4

   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

 

  

9 18

 81 324

 405

 9 5 units

 

     

AB 4 6 6 10

 

  

2 4

 4 16

 20

 2 5 units

Now,



AB 2 5

9 5



AB AC

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

But, radius

 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

BD 3 4 0 3

   

   

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

 

      

AB 4 5 2 6

 

  

1 8

 1 64

 65units

So,



AB 65units

 

     

2 2

BC 7 4 5 2

 

 

11 7

 121 49

 170 units

So,



BC 170units

 

    

AC 7 5 5 6

 

  

12 1

 144 1

 145units

So,



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 .

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

   

 

    

AR x 7 0 4

 

   

2 5 x 49 14x 4

Take a square on both sides.

 

   

2 5 x 49 14x 4

    

4 5 x 49 14x 16

  

20 x 14x 65

  

x 14x 45 0

Now, solve the quadratic equation.

   

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

  

 



 

 

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

   

 

     

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

   

 

    

AB 7 2 3 2

 

2 2

5 5

 25 25

 50

 5 2 units

   

    

2 2

BC 11 7 1 3

 

  

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

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

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

 

  

9 a 3 9

Take a square of both sides.

 

   

9 a 9 6a 81

  

81 a 6a 90

   

90 81 a 6a

  

a 6a 9 0

   

a 3a 3a 9 0

   

   a a 3 3 a 3 0

 

 

a 3 0

 a 3 0

 a 3

Hence, the value of is .

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

 

    

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

 

    

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

 

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

   

x 25 10x 4

 

  

x 10x 29 ...... 1

   

 

    

BQ x 4 0 2

   

  

2 2

x 4 2

   

x 16 8x 4

 

  

x 8x 20 ...... 2

Equate equations and and take a square of

 

both sides.

   

    

2 2

x 10x 29 x 8x 20

    

2 2

x 10x 29 x 8x 20

  10x 8x 20 29

 18x 9









Thus, the point Q is .



 

 

 

Now, find the lengths of the sides AQ and BQ by

putting the value of in the equations .

   

1 and 2

 

   

  

   

   

1 1

AQ 10 29

2 2

  

5 29

 



1 20 116



units

 

   

  

   

   

1 1

BQ 8 20

2 2

  

4 20

 



1 16 80



units

Now, find the length of AB using the distance

formula again.

 

      

2 2

AB 4 5 2 2

   

    

2 2

4 5 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

 

 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

of a triangle.

   

 

     

1 2 3 2 3 1 3 1 2

Area of a triangle x y y x y y x y y

     

 

        

Area of PQR 5 3 2m 2 2m 1 8 1 3

     

15 10m 4m 2 32

  

14m 19

Now, Area of .

 PQR 0

  

14m 19 0

  14m 19 0

14m 19



Hence, the value of .

mis

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

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

   

 

    

AP 3 2 8 4

 

2 2

1 12

 1 144

 

 145 ...... 1

   

 

     

AQ 10 2 y 4

   

   

2 2

12 y 4

   

144 y 16 8y

 

  

y 8y 160 ...... 2

Equate equations and and find a square on

 

both sides.

 

  

145 y 8y 160

  

145 y 8y 160

   

y 8y 160 145

  

y 8y 15

  

y 8y 15 0

   

y 3y 5y 15 0

   

   y y 3 5 y 3 0

  

  y 3 y 5 0

 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

 

   

13 11

 169 121

 290 units

Now, find the length of PQ using the coordinates

P(3, 8) and .

 

 10, 5

   

     

2 2

PQ 10 3 5 8

 

   

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

Area of a triangle x y y x y y x y y

     

       

Area of PQR 8 6 9 6 9 4 3 4 6

     

     

8 3 6 5 3 2

  

24 30 6

 

 

30 30

 

 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







1 2 2 1

1 2

m y m 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

formula.

   

 

 



 



 

t 7 1 6

t 1

 0 7t 6

7t 6



Now, and



m 6



m 7

Now, again use the section formula.







1 2 2 1

1 2

m x m x

m m

So, by the section formula,

   

  

 



 



 

6 1 7 4

6 7

 



6 28





Find the value of from equation .

 

   

 

 



 



 

6 7 7 6

6 7



 



 

 

42 42

 0

Hence, the required point of intersection is Z



 

 

 

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







1 2 2 1

1 2

m x m x

m m

So, by the section formula,

 

 



 

 





t 2 1

t 1

Now, put the values of in the above section

formula.







4t 1

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



Now, and



m 1



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

and







1 2 2 1

1 2

m x m x

m m







1 2 2 1

1 2

m y m y

m m

So, by the section formula,

   

 

 



3 8a 1 3a 1

9a 2

3 1

 

 

24a 3a 1

9a 2

  36a 8 27a 1

  36a 27a 1 8

9a 9



a 1

Now, apply the section formula to find the values of

   

 

 



3 5 1 3

3 1



 

15 3

 4b 12

 

 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.

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

 



6 4





 b 1

Now, according to the question,

 

 a 2b 18 ...... 2

Put the value of in the equation .

 

   a 2 1 18

 a 2 18

 a 18 2

a 16

Now, from equation ,

 





10 k

Put the value of in the above equation.





10 k

  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

   

 

    

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

 

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,

10 2

Therefore,

 

AO AB

 

AO 10 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

   

 

     

AO 2a 11 a 7 9

 

    

4a 121 44a a 2

     

2 2

4a 121 44a a 4 4a

…

  

5a 40a 125

 

Equate equations and .

 

  

5a 40a 125 5 2

Take a square of both sides.

 

  

5a 40a 125 5 2

   

2 2

5a 40a 125 5 2

  

5a 40a 50 125

  

5a 40a 75 0

  

a 8a 15 0

   

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.

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

Solution:

Now, let us consider, and



x 3,



y 2,



x 5



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

m m







1 2 2 1

1 2

m y m y

m m

So, by the section formula,

   







1 5 2 3

1 2





5 6



   







1 1 2 2

1 2





1 4



Here, and



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

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.

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

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,

      DBE DEF FEC ADF

Thus, area of the

 

   ABC 4 Area of DEF

Now, find the area of .

DEF

   

 

     

1 2 3 2 3 1 3 1 2

Area of a triangle x y y x y y x y y

     

       

     

     

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

2 2 2 2 2 2

 

     

   

     

     

1 1 1 2 7 1

2 2 2 2 2 2

  

1 1 7

2 4 4

 



1 1 28 7

2 4



1 22

2 4

 

 

 

 

1 11

2 2



Now, area of the

 

   ABC 4 Area of DEF

So, area of the

  

ABC 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 .

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

 

    

a 4 4a 4

   

a 4 4a 16

 

  

a 4a 20 ...... 1

   

   

2 2

BC 5 a 5 5

   

25 a 10a 0

 

  

a 10a 25 ...... 2

   

   

2 2

AC 5 2 5 9

 

  

3 4

 9 16

 25

 

 5units ...... 3

Now, apply the Pythagoras theorem to find the

value of .

According to Pythagoras theorem,

 

 

2 2 2

AB BC AC ...... 4

Put the values of AB, BC and AC in equation .

 

   

     

2 2

2 2 2

a 4a 20 a 10a 25 5

     

2 2

a 4a 20 a 10a 25 25

  

2a 14a 45 25

  

2a 14a 20 0

  

a 7a 10 0

   

a 5a 2a 10 0

   

   a a 5 2 a 5 0

  

  a 5 a 2 0

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

 

and , to find the lengths of sides AB and BC.

 

   

AB 2 4 2 20

  4 8 20

 4units

   

BC 2 10 2 25

  4 20 25

 9

 3units

Now, find the area of .

ABC

  

Area of a triangle base height

   

Area of ABC BC AB

 

= 3 4

 3 2

 6squareunits

Hence, the required value of is 2, and the area of

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 .



PR PQ

Solution:

According to the question,

So,



PR PQ



PQ 5

PR 3





PR RQ 5

PR 3

 

PR RQ 5

PR PR 3

 

RQ 5

PR 3

 

RQ 5

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



m 3



m 2

Now, find the coordinates of R using the section

formula.

and







1 2 2 1

1 2

m x m x

m m







1 2 2 1

1 2

m y m y

m m

So, by the section formula,

   

 





3 2 2 1

3 2





6 2



   







3 5 2 3

3 2





15 6



Hence, the coordinates of R are .

 

 

 

4 21

5 5

Question: 19

Find the values of if the points A , B

 

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

Area of a triangle x y y x y y x y y

Area of is given by,

ABC

        

 

         

k 1 2k 3 5k 3k 5k 2k 5k 1 2k 2k 3 0

    

  

 

 

  

 

       

 

       

 

2 2

k 1 3 3k 3k 3k

5k 1 3 3k 3k 3 3k 9k 0

15k 3 3k 3 9k 15k 3

  

6k 15k 6 0

  

2k 5k 2 0

   

2k 4k k 2 0

   

   2k k 2 1 k 2 0

  

  2k 1 k 2 0



k 2

Hence, the values of are 2 and .

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

m m







1 2 2 1

1 2

m y m y

m m

So, by the section formula,

   







t 2 1 8

t 1







2t 8

t 1

   

 





t 1 1 9

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



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 .

   

   



   

 

   

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

 

    

PQ 4 4 3 3

 

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

 

    

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 .

 

    

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 .

 

    

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 .

 

        

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 .

 

    

  

2 2

0 y 8 0 6y 39

y 6y 39 0

Now, compare the above equation with .



b 4ac

So, and

a 1,

 b 6

 c 39

   

      

b 4ac 6 4 1 39

 36 156

192

Now, find the value of by using the equation

  



b b 4ac

 

  





6 192

2 1





6 8 3

 y 3 4 3

and

 y 3 4 3

3 4 3

So, the third vertex of the triangle is

 



R 0,3 4 3

or .

 



R 0,3 4 3

Now, solve the roots.

 

   

1 1

R 0,3 4 3 R 0,3 4 1.732

 

 

R 0,3 6.9

 



R 0, 9.9

And,

 

   

2 2

R 0,3 4 3 R 0,3 4 1.732

 

 

R 0,3 6.9

 

 

R 0, 3.9

Hence, from the above figure it can be stated, that

the third vertex is so that origin lies

 



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 .

ADE

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



x 7



   



   

   

Similarl

2 y

 

  2 2 y 5 2

 4 2y 10

2y 6



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

 



 

 

Now, find the area of .

ADE

   

 

     

1 2 3 2 3 1 3 1 2

Area of a triangle x y y x y y x y y

 

   

      

   

   

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





squareunits

Hence, area of as area cannot



ADE= squareunits

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

m m







1 2 2 1

1 2

m y m y

m m

So, by the section formula,

 



 



 

 





2 3

x x

2 1 x

2 1

 



2 3 1

x x x

 



 



 

 





2 3

y y

2 1 y

2 1

 



2 3 1

y y y

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

m m







1 2 2 1

1 2

m y m y

m m

So, by the section formula,

 



 



 

 





1 3

x x

2 1 x

2 1

 



1 3 2

x x x

 



 



 

 





1 3

y y

2 1 y

2 1

 



1 3 2

y y y

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

m m







1 2 2 1

1 2

m y m y

m m

So, by the section formula,

 



 



 

 





1 2

x x

2 1 x

2 1

 



1 2 3

x x x

 



 



 

 





1 2

y y

2 1 y

2 1

 



1 2 3

y y y

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

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

Mid–point of

   

 



 

 

4 2 3 3

BD ,

2 2



 



 

 

 

 1,0

Equate the mid–points.

 



 

 

 

 

1 a

,0 1,0



 

1 a

  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

Area of a triangle x y y x y y x y y

 

          

Area of ABD 1 3 3 2 3 2 4 2 3

     

      

3 3 2 3 2 4 5

   

   

6 2 1 20

 

  

6 2 20

 

 

12units ...... 1

Also,

  

Area of a triangle base height

   

Area of a ABD AB h

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

   

 

    

AB 2 1 3 2

 

2 2

1 5

 1 25

 26 units

   

Area of a ABD 26 h

…



26h

units

 

Equate equations and .

 



26h

 26h 12 2



Multiply and divide by .

 

24 26

26 26



24 26



Hence, the height of the parallelogram is

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

 

  

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

AD 7 3 1 5

 

  

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

 

8 0

 64

 8units

   

   

2 2

BD 7 7 1 9

 

  

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

BS 13 5 14 8

 

2 2

8 6

 

2 2

8 6

 100

10km

   

   

2 2

SO 13 13 26 14

 

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