Lesson: Coordinate Geometry

Exercise 7.1 (10)

Question: 1

Find the distance between the following pairs of

points:

(i) (2, 3), (4, 1)

(ii) (-5, 7), (-1, 3)

(iii)

 

a,b ,

 

 a, b

Solution:

(i) Let the distance between two points be .

Now, distance between two points is given by,

   

   

2 2

1 2 1 2

l x x y y ......(1)

Here, and

 

1 1

x 2,y 3

 

2 2

x 4, y 1

Put these values in equation .

 

   

   

2 2

l 2 4 3 1

   

  

2 2

4 4 

8

2 2

Hence, distance between the points and

 

2, 3

is given by .

 

4,1

2 2

(ii) Let the distance between two points be .

Now, distance between two points is given by,

   

   

2 2

1 2 1 2

l x x y y ......(1)

Here, and .

 

x 5,



y 7

 

x 1,



y 3

Put these values in equation .

 

   

  

2 2

4 4

 16 16

 32

 4 2

Hence, distance between the points and

 

5,7

is given by .

 

1,3

4 2

(iii) Let the distance between two points be .

Now, distance between two points is given by,

   

   

2 2

1 2 1 2

l x x y y ......(1)

Here, and



x a,



y b

 

x a,

 

y b

Put these values in equation .

 

     

2 2

l a a b b

   

 

2 2

2a 2b

 

2 2

4a 4b

 

2 2

2 a b

Hence, distance between the points and

 

a,b

is given by .

 

 a, b



2 2

2 a b

Question: 2

Find the distance between the points (0, 0) and (36,

15). Can you now find the distance between the two

towns A and B discussed in Section 7.2.

Solution:

Let the distance between two points be .

Now, distance between two points is given by,

   

   

2 2

1 2 1 2

l x x y y ......(1)

Here, and



x 0,



y 0



x 36,



y 15

Put these values in equation .

 

   

   

2 2

l 0 36 0 15

   

   

2 2

36 15

 1296 225

 1521

 39

Thus, distance between the points and

 

0,0

 

36,15

is given by .

Yes, we can find the distance between the given

towns A and B.

Let town A be at origin point (0, 0).

Then, town B will be at point (36, 15) with respect

to town A.

Hence, distance between the towns A and B will be

39 Km.

Question: 3

Determine if the points (1, 5), (2, 3) and (– 2, – 11)

are collinear.

Solution:

Let us consider the points (1, 5), (2, 3), and (−2, −11)

be representing the vertices A, B, and C any triangle

respectively.

Then, and

 

A 1,5 ,

 

B 2,3

 

  C 2, 11

Now, find the distance between the points AB, BC

and AC.

So, by the formula,

   

   

2 2

1 2 1 2

l x x y y

   

   

2 2

AB 1 2 5 3

   

  

2 2

1 2

 1 4

 5

 

     

2 2

BC 2 2 3 11

   

 

2 2

4 14

 16 196

 212

Now,

 

     

2 2

AC 1 2 5 11

   

 

2 2

3 16

 9 256

 265

 AB BC AC

Hence, the points (1, 5), (2, 3), and (−2, −11) are not

collinear.

Question: 4

Check whether (5, – 2), (6, 4) and (7, – 2) are the

vertices of an isosceles triangle.

Solution:

Let us consider the points (5, – 2), (6, 4) and (7, – 2)

be representing the vertices A, B, and C any triangle

respectively.

Then, and

 

 A 5, 2 ,

 

B 6,4

 

 C 7, 2

Now, find the distance between the points AB, BC

and AC.

So, by the formula,

   

   

2 2

1 2 1 2

l x x y y

   

    

2 2

AB 5 6 2 4

   

   

2 2

1 6

 1 36

 37

   

 

    

BC 6 7 4 2

   

  

2 2

1 6

 1 36

 37

Now,

   

 

     

AC 5 7 2 2

   

  

2 2

2 0

 2

AB BC

Hence, the points (1, 5), (2, 3), and (−2, −11) are the

vertices of an isosceles triangle, as two of the sides

of the triangle are equal in length.

Question: 5

In a classroom, 4 friends are seated at the points A,

B, C and D as shown in given figure. Champa and

Chameli walk into the class and after observing for

a few minutes Champa asks Chameli, “Don’t you

think ABCD is a square?” Chameli disagrees. Using

distance formula, find which of them is correct.

Fig. Exc_7.1_5 (Ques)

Solution:

According to the question,

The positions of these 4 friends is given by, A(3, 4),

B(6, 7), C(9, 4), and D(6, 1).

Now, find the distance between the points AB, BC,

CD and AD.

Now, distance between two points is given by,

   

   

2 2

1 2 1 2

l x x y y

So, by using the above formula,

   

   

2 2

AB 3 6 4 7

   

   

2 2

3 3

 9 9

 18

 3 2

   

   

2 2

BC 6 9 7 4

   

  

2 2

3 3

 9 9

 18

 3 2

   

   

2 2

CD 9 6 4 1

   

 

2 2

3 3

 9 9

 18

 3 2

Now,

   

   

2 2

AD 3 6 4 1

   

  

2 2

3 3

 9 9

 18

 3 2

Also, find the lengths of the diagonals AC and BD.

   

   

2 2

AC 3 9 4 4

 

  

6 0

 6

   

   

2 2

BD 6 6 7 1

 

 

0 6

 6

Fig. Exc_7.1_5 (Sol)

  So, AB BC CD AD

And diagonals,

AC BD

So, ABCD is a square.

Hence, Champa was correct.

Question: 6

Name the type of quadrilateral formed, if any, by

the following points, and give reasons for your

answer:

(i) (−1, −2), (1, 0), (−1, 2), (−3, 0)

(ii) (− 3, 5), (3, 1), (0, 3), (−1, −4)

(iii) (4, 5), (7, 6), (4, 3), (1, 2)

Solution:

(i) Let us consider the points (−1, −2), (1, 0), (−1,

2), and (−3, 0) be the vertices A, B, C, and D of

the quadrilateral respectively.

Then, and

 

  A 1, 2 ,

 

B 1,0 ,

 

 C 1, 2

 

 D 3,0

Now, find the distance between the points AB,

BC, CD and AD.

So, by the formula,

   

   

2 2

1 2 1 2

l x x y y

   

     

2 2

AB 1 1 2 0

   

   

2 2

 4 4

 8

 2 2

 

    

BC 1 1 0 2

   

  

2 2

 4 4

 8

 2 2

 

     

CD 1 3 2 0

   

 

2 2

 4 4

 8

 2 2

 

      

AD 1 3 2 0

   

  

2 2

 4 4

 8

 2 2

Also, find the lengths of the diagonals AC and

BD.

 

      

AC 1 1 2 2

 

  

0 4

 16

 4

 

    

BD 1 3 0 0

 

2 2

4 0

 16

Now, all sides of this quadrilateral are of the

same length and the diagonals are of the same

length too.

Hence, the points are the vertices of a square.

(ii) Let us consider the points (− 3, 5), (3, 1), (0, 3),

and (−1, −4) be representing the vertices A, B,

C, and D of any quadrilateral respectively.

Then, and

 

 A 3,5 ,

 

B 3,1 ,

 

C 0,3

 

  D 1, 4

Now, find the distance between the points AB,

BC, CD and AD.

So, by the formula,

   

   

2 2

1 2 1 2

l x x y y

   

    

2 2

AB 3 3 5 1

   

  

2 2

6 4

 36 16

 52

 2 13

   

   

2 2

BC 3 0 1 3

   

  

2 2

3 2

 9 4

 13

 

     

2 2

CD 0 1 3 4

   

 

2 2

1 7

 1 49

 50

 5 2

 

      

2 2

AD 3 1 5 4

   

  

2 2

2 9

 4 81

 85

Now, all sides of this quadrilateral are of the

different lengths.

Hence, the given points are the vertices of any

general quadrilateral, and not a specific

quadrilateral, such as square, rectangle, etc.

(iii) Let us consider the points (4, 5), (7, 6), (4, 3),

and (1, 2) be representing the vertices A, B, C,

and D of any quadrilateral respectively.

Then, and

 

A 4,5 ,

 

B 7,6 ,

 

C 4,3

 

D 1, 2

Now, find the distance between the points AB,

BC, CD and AD.

So, by the formula,

   

   

2 2

1 2 1 2

l x x y y

   

   

2 2

AB 4 7 5 6

   

   

2 2

3 1

 9 1

 10

   

   

2 2

BC 7 4 6 3

   

 

2 2

3 3

 9 9

 18

 3 2

   

   

2 2

CD 4 1 3 2

   

 

2 2

3 1

 9 1

 10

   

   

2 2

AD 4 1 5 2

   

 

2 2

3 3

 9 9

 18

 3 2

Also, find the length of the diagonals AC and

BD.

   

   

2 2

AC 4 4 5 3

 

 

0 2

 0 4

 2

   

   

2 2

BD 7 1 6 2

 

2 2

6 4

 36 16

 52

 2 13

Now, the opposite sides of this quadrilateral

are of same length.

But, the diagonals are of different lengths.

Hence, the given points are the vertices of a

parallelogram.

Question: 7

Find the point on the axis which is equidistant

x

from (2, -5) and (-2, 9).

Solution:

According to the question,

The point should be on the axis.

x

Thus, its coordinate will be 0.

y

Let the point on axis be .

x

 

a,0

Now, find the distance of point from the

 

a,0

points (2, -5) and (-2, 9).

So, by the formula,

   

   

2 2

1 2 1 2

l x x y y

Distance between the points and

 

a,0

 

2, 5

   

 

    

a 2 0 5

   

  

2 2

a 2 5 ......(1)

Now, distance between the points

   

a,0 and 2, 9

 

    

a 2 0 9

   

  

2 2

a 2 9 ......(2)

By the given condition in the question, these

distances are equal in measure, so equations (1) and

(2) are equal.

       

    

2 2 2 2

a 2 5 a 2 9

   

    

2 2

a 2 25 a 2 81

 8a 25 81

 8a 56

 a 7

Hence, the point is (−7, 0).

Question: 8

Find the values of for which the distance between

the points P(2, –3) and Q is 10 units.

 

10, y

Solution:

According to the question,

Distance between the points (2, –3) and is 10.

 

10, y

So, by the formula,

   

   

2 2

1 2 1 2

l x x y y

   

    

2 2

2 10 3 y 10

   

   

2 2

8 3 y 10

Now, take squares on both the sides.

 

  

64 y 3 100

 

 

y 3 36

  y 3 6

 y 3 6

  y 3 6

Hence, or .

y 3

 9

Question: 9

If Q(0, 1) is equidistant from P(5, –3) and R ,

 

x,6

find the values of . Also, find the distances QR and

PR.

Solution:

According to the question,

PQ QR

So, by the formula,

   

   

2 2

1 2 1 2

l x x y y

Now, find the distances PQ and QR and equate

them.

       

       

2 2 2 2

5 0 3 1 0 x 1 6

       

     

2 2 2 2

5 4 x 5

  

25 16 x 25

 

41 x 25



16 x

 x 4

So, the value of or .

 x 4

4

Thus, point R will be or .

 

4,6

 

4,6

Take, point R as and find the distances PR

 

4,6

and QR.

So, by the formula,

   

   

2 2

1 2 1 2

l x x y y

   

    

2 2

PR 5 4 3 6

 

  

1 9

 1 81

 82

   

   

2 2

QR 0 4 1 6

   

   

2 2

4 5

 16 25

 41

Thus, distances PR and QR are and

respectively.

Now, take, point R as and find the distances

 

4,6

PR and QR.

So, by the formula,

   

   

2 2

1 2 1 2

l x x y y

 

     

PR 5 4 3 6

   

  

2 2

9 9

 81 81

 9 2

 

    

QR 0 4 1 6

   

  

2 2

4 5

 16 25

 41

Hence, distances PR and QR are and

9 2

respectively.

Question: 10

Find a relation between and such that the point

is equidistant from the point (3, 6) and (–3, 4).

 

x, y

Solution:

According to the question,

Point is equidistant from the points (3, 6) and

 

x, y

(−3, 4).

So, the distance between the point and (3, 6)

 

x, y

is equal to the distance between the point

 

x, y

and (−3, 4).

So, by the formula,

   

   

2 2

1 2 1 2

l x x y y

     

 

       

2 2 2

x 3 y 6 x 3 y 4

       

      

2 2 2 2

x 3 y 6 x 3 y 4

Take square on both sides.

       

      

2 2 2 2

x 3 y 6 x 3 y 4

          

2 2 2 2

x 9 6x y 36 12y x 9 6x y 16 8y

    36 16 6x 6x 12y 8y

 20 12x 4y

 3x y 5

  3x y 5 0

Hence, the relation between and is given by

  3x y 5 0

Exercise 7.2 (10)

Question: 1

Find the coordinates of the point which divides the

join of (–1, 7) and (4, –3) in the ratio 2 : 3.

Solution:

Let be the required point.

 

R x,y

Now, let us consider, and

 

x 1,



y 7,



x 4

 

y 3

And the ratio,



1 2

2:3 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,

 

   





2 4 3 1

2 3





8 3



1

 

   





2 3 3 7

2 3

 



6 21



 3

Here, and

x 1

y 3

Hence, the point is .

 

1,3

Question: 2

Find the coordinates of the points of trisection of

the line segment joining (4, –1) and (–2, –3).

Solution:

Fig. Exc_7.2_2

Let R and S be the points of trisection

 

1 1

x ,y

 

2 2

x , y

of the line segment joining the given points i.e.,

 PR RS SQ

So, point R divides PQ internally in the ratio .

1:2

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 2 2 4

1 2

 



2 8



 2

   

    





1 3 2 1

1 2

 



3 2





Thus, point R

 



 



 

 

1 1

x ,y 2,

Point S divides PQ internally in the ratio .

2:1

Again, apply the section formula.

So, by the section formula,

 

   





2 2 1 4

2 1

 



4 4



 0

   

    





2 3 1 1

2 1

 



6 1





Thus, point S

 



 



 

 

2 2

x , y 0,

Hence, the coordinates are and .



 

 

 



 

 

 

Question: 3

To conduct Sports Day activities, in your

rectangular shaped school ground ABCD, lines have

been drawn with chalk powder at a distance of 1 m

each. 100 flower pots have been placed at a distance

of 1 m from each other along AD, as shown in the

given figure. Niharika runs the distance AD on

the 2

line and posts a green flag. Preet runs

the distance AD on the eighth line and posts a red

flag. What is the distance between both the flags? If

Rashmi has to post a blue flag exactly halfway

between the line segment joining the two flags,

where should she post her flag?

Fig. Exc_7.2_3 (Ques)

Solution:

According to the question,

Niharika posted the green flag at of the

distance AD.

So, it is equal to metres from the

 

 

 

 

100 m 25

starting point of 2

line.

Let us consider this point as R.

Therefore, the coordinates of this point R are (2, 25).

Similarly, Preet posted red flag at of the

distance AD.

So, it is equal to metres from the

 

 

 

 

100 m 20

starting point of 8

line.

Let us consider this point as T.

Therefore, the coordinates of this point T are (8, 20).

Fig. Exc_7.2_3 (Sol) (i)

Now, distance between two flags is given by,

   

   

2 2

1 2 1 2

l x x y y

So, by using the above formula,

   

   

2 2

RT 8 2 25 20

 36 25

 61metres

The point at which Rashmi should post her blue

flag is the mid–point of the line joining these two

points R and T.

Let this point be S .

 

x, y

Fig. Exc_7.2_3 (Sol) (ii)

Now, let us consider, and



x 2,



y 25,



x 8



y 20

And the ratio,



1 2

1:1 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 2 1 8

1 1





2 8



  





1 25 1 20

1 1





25 20



Here, and

x 5

y 22.5

Thus, the point is .

 

5, 22.5

Hence, Rashmi should post her blue flag at 22.5 m

on the 5

line.

Question: 4

Find the ratio in which the line segment joining the

points (–3, 10) and (6, –8) is divided by (–1, 6).

Solution:

Let us consider, the line segment joining (−3, 10)

and (6, −8) is divided by point (−1, 6) in the ratio

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 6 1 3

t 1

   t 1 6t 3

2 7t

Hence, the required ratio is .

2:7

Question: 5

Find the ratio in which the line segment joining A

(1, –5) and B (–4, 5) is divided by the axis. Also,

x

find the coordinates of the point of division.

Solution:

Let the line segment joining A (1, −5) and B (−4, 5)

is divided by axis in the ratio .

x

t :1

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,

 

   





t 4 1 1

t 1

 





4t 1

t 1

 

   





t 5 1 5

t 1







5t 5

t 1

Thus, the coordinates of the point of division are

  

 

 

 

 

4t 1 5t 5

t 1 t 1

Now, coordinate of any point on axis is 0.

y

x







5t 5

So, 0

t 1

 5t 5 0

5t 5



t 1

Therefore, axis divides AB in the ratio .

x

1:1

Now, find the coordinates of the division point

 

x, y

 

   

  

 



 

 

 

4 1 1 5 1 5

x, y ,

1 1 1 1

  

 



 

 

4 1 5 5

2 2



 



 

 

Hence the coordinates of the point of division are



 

 

 

Question: 6

If and are the vertices of a

 

1,2 ,

 

4, y ,

 

x,6

 

3,5

parallelogram taken in order, find .

x and y

Solution:

Fig. Exc_7.2_6

Let P(1, 2), Q , R and S(3, 5) be the

 

4, y)

 

x,6

coordinates of vertices of a parallelogram PQRS.

The intersection point of the diagonals, PR and QS

also divides these diagonals in two equal halves.

Thus, O is the mid–point of PR and QS.

Now, O is the mid–point of PR and the coordinates

of O are

  

   



   

   

1 x 2 6 1 x

, ,4

2 2 2

O is also the mid–point of QS, so the coordinates of

O are

  

   



   

   

4 3 5 y 7 5 y

, ,

2 2 2 2

Now, since both the coordinates are of the same

point O,

Equate both the coordinates of point O.





x 1 7

2 2

 2x 2 14

2x 12



x 6

Similarly,





5 y

 8 5 y

y 3

Hence, and

x 6

y 3

Question: 7

Find the coordinates of a point A, where AB is the

diameter of a circle whose centre is (2, –3) and B is

(1, 4).

Solution:

Fig. Exc_7.2_7

Let the coordinates of point A be and the

 

x, y

mid–point of the circle be O.

According to the question,

Coordinates of O are (2, −3).

Now, the coordinates of O can also be written as,

 

 

 

 

x 1 y 4

2 2

So,

 

 

 

 

 

 

x 1 y 4

2, 3 ,

2 2





x 1

 4 x 1

x 3

And,



 

y 4

  6 y 4

 y 10

Hence, the coordinates of A are .

 

3, 10

Question: 8

If A and B are (–2, –2) and (2, –4), respectively, find

the coordinates of P, such that and P lies



AP AB

on the line segment AB.

Solution:

Let the coordinates of P be .

 

x, y

Fig. Exc_7.2_8

According to the question,

The coordinates of A and B are (–2, –2) and (2, –4)

and .



AP AB

So,

AP:PB 3:4

Now, let us consider, and

 

x 2,

 

y 2,



x 2

 

y 4

And the ratio,



1 2

3:4 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,

 

   





3 2 4 2

3 4





6 8





   

    





3 4 4 2

3 4

 



12 8





Here, and









Hence, the coordinates of point P are .

 

 

 

 

2 20

7 7

Question: 9

Find the coordinates of the points which divide the

line segment joining A (–2, 2) and B (2, 8) into four

equal parts.

Solution:

Fig. Exc_7.2_9

Let the points P, Q and R divide the line segment

AB in the ratios and respectively.

1:3,

1:1

3:1

Now, apply the section formula to find the

coordinates of points P, Q and 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 for coordinates of point

 

   





1 2 3 2

1 3





2 6





 1

  





1 8 3 2

1 3





8 6



Thus, the coordinates of point P, are .

 



 

 

Apply the section formula for coordinates of point

 

   





1 2 1 2

1 1

 

 



2 2

 0

  





1 8 1 2

1 1





8 2



 5

Thus, the coordinates of point Q are .

 

0,5

Now, apply the section formula for coordinates of

point R,

 

   





3 2 1 2

3 1





6 2



1

  





3 8 1 2

3 1





24 2



Thus, the coordinates of point R are .

 

 

 

Question: 10

Find the area of a rhombus if its vertices are (3, 0),

(4, 5), (–1, 4) and (–2, –1) taken in order. [Hint: Area

of a rhombus (product of its diagonals)]



Solution:

Fig. Exc_7.2_10

Let (3, 0), (4, 5), (−1, 4) and (−2, −1) be the

coordinates of the vertices P, Q, R, S of a rhombus

PQRS.

Apply the distance formula.

Now, distance between two points is given by,

   

   

2 2

1 2 1 2

l x x y y

Find the length of diagonal PR by the distance

formula.

So, length of PR

 

    

3 1 0 4

 

  

4 4

 16 16

 32

 4 2

Now, find the length of diagonal QS by the distance

formula.

So, length of QS

 

     

2 2

4 2 5 1

 

2 2

6 6

 36 36

 72

 6 2

Area of rhombus where and are

 

  

1 2

d d

the lengths of the diagonal.

Hence, area of rhombus

  

4 2 6 2

 24 squareunits

Exercise 7.3 (5)

Question: 1

Find the area of the triangle whose vertices are:

(i) (2, 3), (–1, 0), (2, –4)

(ii) (–5, –1), (3, –5), (5, 2)

Solution:

(i) Three vertices of a triangle are (2, 3), (–1, 0)

and (2, –4).

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

   

 

   

 

 



     

2 0 4

Area of given triangle

1 4 3 2 3 0

  

8 7 6



 

 



Hence, area of the triangle is square units.

(ii) Three vertices of a triangle are (–5, –1), (3, –5),

(5, 2).

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

   

 

   

 

  



      

5 5 2

Area of given triangle

3 2 1 5 1 5

  

35 9 20



 

 

 32

Hence, area of the triangle is 32 square units.

Question: 2

In each of the following find the value of ‘k’, for

which the points are collinear.

(i) (7, –2), (5, 1),

 

3,k

(ii) (8, 1), , (2, –5)

 

k, 4

Solution:

(i) Area of a triangle formed by three collinear

points is 0.

Now, area of the triangle is given by,

   

 

     

1 2 3 2 3 1 3 1 2

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

Area of the given triangle

 0

   

 

       

7 1 k 5 k 2 3 2 1 0

    7 7k 5k 10 9 0

  2k 8 0

2k 8



k 4

Hence the value of .

k is4

(ii) Area of a triangle formed by three collinear

points is 0.

Now, area of the triangle is given by,

   

 

     

1 2 3 2 3 1 3 1 2

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

Area of the given triangle

 0

   

 

         

8 4 5 k 5 1 2 1 4 0

  8 6k 10 0

  6k 18 0

6k 18



k 3

Hence the value of .

k is3

Question: 3

Find the area of the triangle formed by joining the

mid–points of the sides of the triangle whose

vertices are (0, –1), (2, 1) and (0, 3). Find the ratio of

this area to the area of the given triangle.

Solution:

Let the vertices of the triangle be P(0, −1), Q(2, 1),

R(0, 3).

Let D, E, F be the mid–points of the sides PQ, QR

and PR of this triangle.

Now, find the coordinates of D, E, and F.

 

  

 

 

 

 

0 2 1 1

D , 1,0

2 2

 

 

 

 

 

 

2 0 1 3

E , 1,2

2 2

 

 

 

 

 

 

0 0 3 1

F , 0,1

2 2

Fig. Exc_7.3_3

Now, find the area of the triangle formed by joining

the mid–points of triangle 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 DEF 1 2 1 1 1 0 0 0 2

  

1 1 0



 

 

1

Thus, area of the triangle DEF is 1 square units.

Now, find the area of .

PQR

   

 

        

Area of PQR 0 1 3 2 3 1 0 1 1

  

0 8 0



 

 

 4

Thus, area of the triangle PQR is 4 square units.

Hence, required ratio is .

1:4

Question: 4

Find the area of the quadrilateral whose vertices,

taken in order, are (–4, –2), (–3, –5), (3, –2) and (2,

3).

Solution:

Let the vertices of the quadrilateral be P(−4, −2),

Q(−3, −5), R(3, −2), and S(2, 3).

Now, join RP to form two triangles and

PQR

PRS

Fig. Exc_7.3_4

   

 

     

1 2 3 2 3 1 3 1 2

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

     

 

     

 

   

 

   

      

   

4 5 2

Area of PQR 3 2 2

3 2 5

       

     

4 3 3 0 3 3

  

12 0 9



 

 



Thus, area of the triangle DEF is square units.

Now, find the area of .

PRS

   

 

   

 

     

 

   

4 2 3 3 3 2

Area of PRS

2 2 2

      

    

4 5 3 5 2 0

  

20 15 0



 

 



Thus, area of the triangle PRS is square units.

   Area of quadrilateral PQRS Area of PQR Area of PRS

 Area of quadrilat

era

1 35

l PQRS





21 35



 28

Hence, area of the given triangle is 28 square units.

Question: 5

You have studied in Class IX, (Chapter 9, Example

3), that a median of a triangle divides it into two

triangles of equal areas. Verify this result for

ABC

whose vertices are A(4, –6), B(3, –2) and C(5, 2).

Solution:

According to the question,

The vertices of the triangle are A(4, −6), B(3, −2),

and C(5, 2).

Let D be the mid–point of side BC of .

ABC

So, AD is the median in .

ABC

Now, find the coordinates of point D.

  

 



 

 

3 5 2 2

D ,

2 2

 

D 4,0

Fig. Exc_7.3_5

Now, find the area of triangles ABD and ADC.

   

 

     

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 4 2 0 3 0 6 4 6 2

     

    

4 2 3 6 4 4

   

8 18 16

 

. .

 

  

 3

However, area cannot be

negative.

Thus, area of is 3 square units.

ABD

   

 

        

Area of ADC 4 0 2 4 2 6 5 6 0

     

    

4 2 4 8 5 6

   

8 32 30

 

 

  

 3

However, area cannot be negative.

Thus, area of is 3 square units.

ADC

Hence, the median AD divides in two

ABC

triangles of equal areas.

Exercise 7.4 (8)

Question: 1

Determine the ratio in which the line

  2x y 4 0

divides the line segment joining the points A(2, –2)

and B(3, 7).

Solution:

Let us consider, the ratio in which the line,

divides the line segment joining the

  2x y 4 0

points A (2, – 2) and B (3, 7) be .

t :1

Find the coordinates of the point of division by

applying 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,

Coordinates

 

 



 

 

 

3t 2 7t 2

t 1 t 1

The above point also lies on the line .

  2x y 4 0

Put the coordinates in the equation of the line

  2x y 4 0

 

   

  

   

 

   

3t 2 7t 2

2 4 0

t 1 t 1

Now, solve the above equation.

    





6t 4 7t 2 4t 4

t 1







9t 2

t 1

 9t 2 0

9t 2



Hence, the required ratio is .

2:9

Question: 2

Find a relation between if the points

x and y

 

x, y

(1, 2) and (7, 0) are collinear.

Solution:

For any three points to be collinear, the area of the

triangle formed by these points will be 0.

   

 

     

1 2 3 2 3 1 3 1 2

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

     

     

Area x 2 0 1 0 y 7 y 2

   

0 2x y 7y 14

  

0 2x 6y 14

  2x 6y 14 0

  x 3y 7 0

Hence, is the required relation.

  x 3y 7 0

Question: 3

Find the centre of a circle passing through the

points (6, –6), (3, –7) and (3, 3).

Solution:

Let O be the centre of the circle.

 

x, y

And let the coordinates (6, −6), (3, −7), and (3, 3) be

P, Q, and R on the circumference of the circle.

Now, find the distance of the points P, Q and R

from the point O.

Distance between two points is given by,

   

   

2 2

1 2 1 2

l x x y y

   

   

2 2

OP x 6 y 6 ......(1)

   

   

2 2

OQ x 3 y 7 ......(2)

   

   

2 2

OR x 3 y 3 ......(3)

Now, (Radii of the same circle)

OP OQ

Equate equations and .

 

       

      

2 2 2 2

x 6 y 6 x 3 y 7

Take square of both sides.

       

      

2 2 2 2

x 6 y 6 x 3 y 7

          

2 2 2 2

x 36 12x y 36 12y x 9 6x y 49 14y

   6x 2y 14 0

 3x y 7 ......(4)

Similarly, (Radii of the same circle)

OP OR

Equate equations and

 

       

      

2 2 2 2

x 6 y 6 x 3 y 3

Take square of both sides.

       

      

2 2 2 2

x 6 y 6 x 3 y 3

          

2 2 2 2

x 36 12x y 36 12y x 9 6x y 9 6y

   6x 18y 54 0

   3x 9y 27 ......(5)

Add equations .

   

4 and 5

    3x y 3x 9y 7 27

 10y 20





 y 2

Substitute the value of in equation .

 

 3x 2 7

 3x 7 2

3x 9



x 3

Hence, the centre of the circle is given by .

 

3, 2

Question: 4

The two opposite vertices of a square are (–1, 2) and

(3, 2). Find the coordinates of the other two vertices.

Solution:

Let PQRS be a square having (−1, 2) and (3, 2) as

vertices P and R respectively.

Let and be the coordinates of vertices

 

x, y

 

1 1

x ,y

Q and S respectively.

Fig. Exc_7.4_4

Now, find the distance between the points PQ, QR

and PR.

Distance between two points is given by,

   

   

2 2

1 2 1 2

l x x y y

   

   

2 2

PQ x 1 y 2 ......(1)

   

   

2 2

QR x 3 y 2 ......(2)

   

   

2 2

PR 3 1 2 2

 

PR 4 0

PR 16

PR 4 ......(3)

The sides of a square are equal.

So,

PQ QR

Equate equations .

   

1 and 2

       

      

2 2 2 2

x 1 y 2 x 3 y 2

Take square of both sides.

       

      

2 2 2 2

x 1 y 2 x 3 y 2

          

2 2 2 2

x 1 2x y 4 4y x 9 6x y 4 4y

8x 8



Now, in a square, all interior angles are of .

90

In ,

PQR

(By Pythagoras theorem)

 

2 2 2

PQ QR PR

Substitute the values of PQ, QR and PR in the

above equation.

   

 

   

 

       

2 2

2 2 2 2 2

x 1 y 2 x 3 y 2 4

Substitute the value of in the above equation.

         

   

       

   

2 2 2 2

1 1 y 2 1 3 y 2 4

     

   

      

   

2 2 2

2 y 2 2 y 2 16

       

2 2

4 y 4 4y 4 y 4 4y 16

  

2y 16 8y 16

 

2y 8y 0

 

y 4y 0

 

 y y 4 0

y 0or 4

In a square, the diagonals bisect each other at

90

and are of equal length.Let O be the mid–point of

PR.

So, it is also the mid–point of QS.

Now, find the coordinates of point O.

Coordinates

  

 



 

 

1 3 2 2

2 2

 



 

 

2 4

2 2

 

 1, 2

The coordinates of point O can also be derived from

the side QS.

Coordinates

 

 



 

 

1 1

x x y y

2 2

Equate the coordinates.

 

 

 



 

 

1 1

x x y y

, 1, 2 ......(4)

2 2

Put the value of .

 

xin equation 4

 

 

 



 

 

1 1

1 x y y

, 1, 2

2 2



 



 

 

1 x

 

x 1 2

 

x 2 1



x 1

And,



 



 

 

y y

 

y y 4

Now, if

y 0,



y 4

y 4,



y 0

Hence, the required coordinates are (1, 0) and (1, 4).

Question: 5

The Class X students of a secondary school in

Krishinagar have been allotted a rectangular plot of

land for their gardening activity. Sapling of

Gulmohar are planted on the boundary at a

distance of 1 m from each other. There is a

triangular grassy lawn in the plot as shown in the

given figure. The students are to sow seeds of

flowering plants on the remaining area of the plot.

(i) Taking A as origin, find the coordinates of the

vertices of the triangle.

(ii) What will be the coordinates of the vertices of

if C is the origin? Also calculate the

PQR

areas of the triangles in these cases. What do

you observe?

Fig. Exc_7.4_5

Solution:

(i) Take A as the origin, AD as the axis and AB

x

as the axis.

y

So, the coordinates of point P, Q, and R are (4,

6), (3, 2), and (6, 5) respectively.

Now, find area of the triangle 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 4 2 5 3 5 6 6 6 2

     

    

4 3 3 1 6 4

   

12 3 24



squareunits

(ii) Let us take C as origin, CB as the axis, and

x

CD as the axis.

y

So, the coordinates of vertices P, Q, and R are

(12, 2), (13, 6), and (10, 3) respectively.

Now, find area of the triangle 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 12 6 3 13 3 2 10 2 6

     

   

12 3 13 1 10 4

  

36 13 40



squareunits

Hence, the area of the triangle is same in both

the cases.

Question: 6

The vertices of a are A(4, 6), B(1, 5) and C(7,

ABC

2). A line is drawn to intersect sides AB and AC at D

and E respectively, such that

 

AD AE 1

AB AC 4

.Calculate the area of the and compare it

ADE

with the area of .

ABC

Solution:

Fig. Exc_7.4_6

According to the question,

 

AD AE 1

AB AC 4

Now, put AB as and AC as .

AD DB

AE EC

 

 

AD AE 1

AD DB AE EC 4

 

AD AE 1

DB EC 3

So, D and E are two points on sides AB and AC

respectively, such that they divide sides AB and AC

in the ratio .

1:3

Now, find the coordinates of points D and E by

applying 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,

Coordinates of D

     

 



 

 

 

1 1 3 4 1 5 3 6

1 3 1 3

 



 

 

13 23

4 4

Coordinates of E

     

 



 

 

 

1 7 3 4 1 2 3 6

1 3 1 3

 



 

 

19 20

4 4

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 23 20 13 20 19 23

Area of ADE 4 6 6

2 4 4 4 4 4 4

 

 

   

   

   

   

1 13 20 24 19 24 23

23 20

2 4 4 4 4



   

  

   

   

1 13 4 19 1

2 4 4 4 4

  

1 13 19

2 4 16

 



1 48 52 19

2 16



squareunits

Now, find the area of .

ABC

     

      

Area of ABC 4 5 2 1 2 6 7 6 5

     

   

4 3 1 4 7 1

  

12 4 7



 



squareunits

Hence, the ratio between the areas of

is .

 ADEand ABC

1:16

Alternatively,

A line segment is parallel to the third side of the

triangle If it divides its two sides in the same ratio.

Thus, the two triangles so formed i.e. and

ADE

will be similar to each other.

ABC

Therefore, the ratio between the areas of these two

triangles will be the square of the ratio between the

sides of these two triangles.

Hence, ratio between the areas of and

ADE

 

  

 

 

1 1

ABC

4 16

Question: 7

Let A(4, 2), B(6, 5) and C(1, 4) be the 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 do yo observe?

(v) If A , B and C are the

 

1 1

x ,y

 

2 2

x , y

 

3 3

x ,y

vertices of , find the coordinates of the

ABC

centroid of the triangle.

Solution:

Fig. Exc_7.4_7

(i) Median AD of the triangle will divide the side

BC in two equal parts.

So, D is the mid–point of side BC.

Now, find the coordinates of point D.

Coordinates

 

   

 

   

   

6 1 5 4 7 9

, ,

2 2 2 2

(ii) According to the question,

Point P divides the side AD in a ratio .

2:1

Now, find the coordinates of point P by

applying 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,

Coordinates

 

     

 

 

 

   

 

 

 

 

7 9

2 1 4 2 1 2

11 11

2 2

, ,

2 1 2 1 3 3

(iv) According to the question,

Median BE divides the side AC in two equal

parts.

So, E is the mid–point of the side AC.

Now, find the coordinates of point E.

Coordinates

 

   

 

   

   

4 1 2 4 5

, ,3

2 2 2

Now, point Q divides the side BE in a ratio

2:1

Find the coordinates of point P by applying

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,

Coordinates

 

  

 

  

 

 

   

 

 

 

 

2 1 6

2 3 1 5 11 11

, ,

2 1 2 1 3 3

The median CF will divide the AB in two equal

parts.

So, F is the mid–point of side AB.

Find the coordinates of point F.

Coordinates

 

   

 

   

   

4 6 2 5 7

, 5,

2 2 2

The point R divides the side CF in a ratio .

2:1

So, find the coordinates of point R by applying

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,

Coordinates

 

  

 

  

 

 

   

 

 

 

 

2 1 4

2 5 1 1 11 11

, ,

2 1 2 1 3 3

(v) The coordinates of point P, Q, R are the same.

So, all these points are representing the same

point on the plane, which is the centroid of the

triangle.

(v) According to the question,

Let us consider, , have its vertices as A

ABC

, B , and C .

 

1 1

x ,y

 

2 2

x , y

 

3 3

x ,y

Median AD will divide the side BC in two

equal parts.

So, D is the mid–point of side BC.

Now, find the coordinates of point D.

Coordinates

 

 



 

 

2 3 2 3

x x y y

2 2

Let the centroid of this triangle be O. Point O

divides the side AD in a ratio .

2:1

So, find the coordinates of point O by applying

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,

Coordinates

 

 

     

 



 

 

 

 

2 3 2 3

1 1

x x y y

2 1 x 2 1 y

2 2

2 1 2 1

   

 



 

 

1 2 3 1 2 3

x x x y y y

3 3

Hence, the coordinates of the centroid O are

   

 

 

 

1 2 3 1 2 3

x x x y y y

3 3

Question: 8

ABCD is a rectangle formed by the points A(–1, –1),

B(–1, 4), C(5, 4) and D(5, – 1). P, Q, R and S are the

mid–points of AB, BC, CD and DA respectively. Is

the quadrilateral PQRS a square? a rectangle? or a

rhombus? Justify your answer.

Solution:

According to the question,

P, Q, R and S are the mid–points of sides AB, BC,

CD and AD.

So, find the coordinates of points P, Q, R and S.

Coordinates of P

   

   

  

   

   

1 1 1 4 3

, 1,

2 2 2

Coordinates of Q

 

  

 

 

 

 

1 5 4 4

, 2,4

2 2

Coordinates of R

  

   

 

   

   

5 5 1 4 3

, 5,

2 2 2

Coordinates of S

 

   

 

  

 

 

1 5 1 1

, 2, 1

2 2

Fig. Exc_7.4_8

Now, find the lengths of the sides PQ, QR, RS and

SP by the distance formula.

Distance between two points is given by,

   

   

2 2

1 2 1 2

l x x y y

 

 

    

 

 

PQ 1 2 4

 



 

 

   

 

 

QR 2 5 4

 



 

 

   

 

 

RS 5 2 1

 



 

 

    

 

 

SP 2 1 1

 



Also, find the lengths of the diagonals PR and QS

by the distance formula.

 

 

    

 

 

3 3

PR 1 5

2 2

 

 

 6

   

   

2 2

QS 2 2 4 1

 



 5

Now, all sides of the given quadrilateral are of the

same length but, the diagonals have different

measurements.

Hence, PQRS is a rhombus.