CO-ORDINATE GEOMETRY - Testlabz.com

Math Class IX 1 Question Bank

CO-ORDINATE

GEOMETRY Q.1. For each equation given below, name the dependent and independent

variables.

(i) y x −4

= 73

(ii) x y= 9 + 4

(iii) y

x5 + 3

=2

(iv) y x1

= (6 + 5)7

Ans. (i) 4

7,3

y x= − Here, x is independent and y is dependent.

(ii) 9 4,x y= + Here, y is independent and x is dependent.

(iii) 5 3

,2

yx

+= Here, y is independent and x is dependent.

(iv) 1

(6 5),7

y x= + Here, x is independent and y is dependent.

Q.2. Plot the following points on the same graph paper :

(i) (8, 7) (ii) (3, 6)

(iii) (0, 4) (iv) (0, –4)

(v) (3, –2) (vi) (–2, 5)

(vii) (–3, 0) (viii) (5, 0)

(ix) (–4, –3)

Ans. The above points are shown in the adjoining graph.


Q.3. Find the values of x and y if:

(i) x y−( 1, + 3) = (4, 4) (ii) x y − −(3 + 1, 2 7) = (9, 9)

(iii) x y y x− − −(5 3 , 3 ) = (4, 4)

Ans. (i) ( 1, 3) (4, 4)x y− + = (Given)

If ( 1) 4x − = then 4 1 5x = + =

If ( 3) 4y + = then 4 3 1y = − =

∴ 5, 1x y= =

(ii) (3 1, 2 7) (9, 9)x y+ − = − (Given)

If (3 1) 9x + = then 3 9 1x = − ⇒ 3 8x = ⇒8 2

23 3

x = =

If (2 7) 9y − = − then 2 9 7y = − +

⇒ 2 2y = − ⇒2

12

y = − = −

∴ 2

2 , 13

x y= = −

(iii) (5 3 , 3 ) (4, 4)x y y x− − = − (Given)

If (5 3 ) 4x y− = then 5 4 3x y= +

4 3

5

yx

+= ...(i)

If ( 3 ) 4y x− = −

4 3y x= − + ...(ii)

Putting the value of x in (ii), we get

4 3

4 35

yy

+ = − + ×

⇒ 5 20 12 9y y= − + +

9 5 20 12y y− = − ⇒ 4 8y =

8

24

y = =

Putting the value of 2y = in (i), we get

4 3 2 10

25 5

x+ ×

= = =

Hence, 2, 2.x y= =


Q.4. Use the graph given below, to find the co-ordinate of the point(s) satisfying

the given condition :

(i) the abscissa is 2. (ii) the ordinate is 0.

(iii) the ordinate is 3. (iv) the ordinate is −4.

(v) the abscissa is 5. (vi) the abscissa is equal to the ordinate.

(vii) the ordinate is half of the abscissa.

Ans. (i) The required point, A (2, 2)=

(ii) The required point is, B (5, 0)=

(iii) The required points are, C ( 4, 3)= − and E (6, 3)=

(iv) The required point is, D (2, 4)= −

(v) The required points are, B (5, 0) and H (5, 5)= =

(vi) The required points are, A (2, 2) and H (5, 5)= =

(vii) The required point is, E (6, 3)=

Q.5. In each of the followings; find the co-ordinates of the point whose abscissa is

the solution of the first equation and ordinate is the solution of the second

equation :

(i) x y y− −1

3 2 = 7; 2 + 1 = 10 22

(ii) a a b b− −

−2 15 4 2 1

1 = ; =3 2 7 3

(iii) y

x x x y− − − −1 4 +

5 (5 ) = (3 ); 4 3 =2 3


Ans. (i) 1

3 2 7; 2 1 10 22

x y y− = + = − (Given)

3 2 7x− =

⇒ 2 3 7 4x = − = −

⇒ 4

22

x−

= = − ∴ 2x = −

1

2 1 10 22

y y+ = − ⇒5

2 1 102

y y+ = −

⇒ 5

2 10 12

y y+ = − ⇒4 5

10 12

y y+= −

⇒ 9

92

y= ⇒ 9 9 2 18y = × = ⇒

182

9y = =

∴ The co-ordinates of the point are ( 2, 2)−

(ii) 2 15 4 2 1

1 ;3 2 7 3

a a b b− −− = = (Given)

⇒ 2

13 2

a a− = ⇒

21

3 2

a a− =

⇒ 4 3

16

a a−=

⇒ 16

a= ∴ 6a =

Again, 15 4 2 1

7 3

b b− −=

⇒ 45 12 14 7b b− = −

⇒ 12 14 7 45b b− − = − − ⇒ 26 52b− = −

⇒ 52

226

b−

= =−

∴ 2b =

∴ The co-ordinates of the point are (6, 2)

(iii) 1 4

5 (5 ) (3 ); 4 32 3

yx x x y

+− − = − − =

⇒ 1

5 (5 ) (3 )2

x x x− − = − ⇒1

5 5 (3 )2

x x x− + = −

⇒ 1

6 5 (3 )2

x x− = − ⇒ 2(6 5) 3x x− = −

⇒ 12 10 3x x− = − ⇒ 12 10 3x x+ = +


⇒ 13 13x = ⇒13

113

x = =

Again, 4

4 33

yy

+− =

⇒ 3(4 3 ) 4y y− = + ⇒ 12 9 4y y− = +

⇒ 9 12 4y y− − = − + ⇒ 10 8y− = − ⇒8 4

10 5y = =

∴ The co-ordinates of the point are 4

1,5

.

Q.6. Write down the co-ordinates of each of the following points A, B, C, D, E

shown below on the graph paper.

Ans. Co-ordinates of A are ( 3, 2)− , co-ordinates of B are (2, 1)

co-ordinates of C are (4, 2)− , co-ordinates of D are ( 1, 3)− −

and co-ordinates of E are (0, 1).−

Q.7. Plot each of the following points on a graph paper.

(i) A (6, 3) (ii) B (–4, 1)

(iii) C (–2, –5) (iv) D (2, –5)

(v) P (4, 0) (vi) Q (0, 3)

(vii) R (–3, –3) (viii) S (0, –3)

Ans. We plot the given points on the graph after drawing the co-ordinate axes XOX′

and YOY′ as shown.


Q.8. On which axis does the following points lie ?

(i) (5, 0) (ii) (0, –2)

(iii) (0, 3) (iv) (–3, 0)

Ans. (i) In point (5, 0), y co-ordinate is zero. Therefore, this point lies on y-axis.

(ii) In point, x co-ordinate is zero. Therefore, this point lies on y-axis.

(iii) In the point (0, 3), x-co-ordinate is zero. Therefore the point lies on y-axis.

(iii) In point ( 3, 0),− y co-ordinate is zero. Therefore this point lies on x-axis.

Q.9. In which quadrant, does the given point lie ?

(i) A (–3, 2) (ii) B (–5, –3)

(iii) C (2, –7) (iv) D (–2, –2)

Ans. (i) In point, A ( 3, 2)− x co-ordinate is negative and y co-ordinate is positive.

∴ This point lies in 2nd quadrant.

(ii) In point B ( 5, 3)− − Both co-ordinates are negative.

∴ This point lies in third quadrant.

(iii) In point C (2, 7)− x co-ordinate is positive but y co-ordinate is negative.

∴ This point lies in the fourth quadrant.

(iv) In point D ( 2, 2)− − Both co-ordinates are negative.

∴ This point lies in third quadrant.

Q.10. The points A (2, –2) B (8, 4) and C (5, 7) are three vertices of a rectangle

ABCD. Plot these points on a graph paper and hence find the co-ordinates of

its fourth vertex D.

Ans. We plot the points A (2, 2),− B (8, 4) and C (5, 7) on the graph. Join AB, BC

and through C draw a line parallel to AB and through A, draw a line parallel to

BC which intersect each other at D.

∴ D is the fourth vertex of the rectangle ABCD.


Hence, the co-ordinates of D are ( 1,1).−

Q.11. The points A (3, 2) B (0, 5) and D (0, –1) are the three vertices of a square

ABCD. Plot these points on a graph paper and hence find the co-ordinates of

the vertex C.

Ans. We plot the points A (3, 2), B (0, 5) and D ( 0,1)− on the graph. Join AB and

AD.

∴ Through D and B, draw perpendiculars meeting each other at C.

Hence, the co-ordinates of C are ( 3, 2).−


Q.12. In each of the followings; the co-ordinates of the three vertices of a

rectangle ABCD are given. By plotting the given points; find, in each case,

the co-ordinates of the fourth vertex:

(i) A (2, 0) B (8, 0) and C (8, 4).

(ii) A (4, 2), B (–2, 2) and D (4, –2).

(iii) A (–4, –6), C (6, 0) and D (–4, 0).

(iv) B (10, 4), C (0, 4) and D (0, –2).

Ans. (i) A (2, 0), B (8, 0) and C (8, 4)

Plot points A, B and C as shown in the above figure. Join AB and BC and

form the rectangle ABCD. It is clear from the graph the co-ordinates of the

fourth vertex D are (2, 4).

(ii) Plot points A (4, 2) B ( 2, 2)− and D (4, 2)− on a graph paper as shown in

the adjoining figure. Join AB and BD and complete the rectangle ABCD.

Then from the graph, we find that, the co-ordinates of the fourth vertex C are

( 2, 2).− −


(iii)

Plot points A, C and D on a graph paper as shown in the above figure, join

AD and CD. Complete the rectangle ABCD. Then, from the graph, we find

that, the co-ordinates of the fourth vertex B are (6, 6).−

(iv) B (10, 4), C (0, 4) and D (0, 2).−


Plot points B, C and D as shown in the graph paper in the above figure. Join

BC, and CD and complete the rectangle ABCD. Then from the graph, we

find that the co-ordinates of the fourth vertex A are (10, 2).−

Q.13. A (–2, 2), B (8, 2) and C (4, –4) are the vertices of a parallelogram ABCD.

By plotting the given points on a graph paper; find the co-ordinates of the

fourth vertex D. Also from the same graph, state the co-ordinates of the mid-

points of the sides AB and CD.

Ans. A ( 2, 2),− B (8, 2) and C (4, 4)− .

Plot points A, B and C on a graph paper as shown in the adjoining figure.

Join AB, AD, DC and BC and complete the parallelogram ABCD. Then, the

co-ordinates of the fourth vertex D are ( 6, 4).− −

The co-ordinates of the mid-point P of AB are (3, 2) and the co-ordinates of

the mid-points Q of CD are ( 1, 4).− −


Q.14. In rectangle OABC; point O is the origin, OA=10 units along x-axis and

BC = 8 units. Find the co-ordinates of vertices A, B and C.

Ans. In rectangle OABC, point O is the origin i.e. O (0, 0).

OA 10= units along x-axis.

AB 8= units along y-axis.

∴ Co-ordinates of rectangle OABC will be A (10, 0), B (10, 8) and of C (0, 8).

(∵ OC is along y-axis).

Q.15. Draw the graph for each linear equation, given below :

(i) x + 3 = 0 (ii) x − 5 = 0

(iii) y = 4 (iv) x −2 7 = 0

(v) y + 6 = 0 (vi) y −2 7 = 0

(vii) y x−= (viii) y3 + 5 = 0

(ix) y −2 5 = 0 (x) y = 0

(xi) x = 0 (xii) y x −= 2 1

(xiii) y x− = 5

Ans. (i) The graph of 3 0,x + = is a straight line CD || to y-axis, at a distance equal to

3− units from it.


(ii) 5 0 5x x− = ⇒ =

It is a line parallel to y-axis at a distance of 5 units.

(iii) 4y =

It is a line parallel to x-axis at a distance of 4 units.


(iv) The graph of 2 7 0 7 2x x− = ⇒ = is a straight line GH || to y-axis, at a

distance equal to 7

2 i.e. 3.5 units from it.

(v) The graph of 6 0 6y y+ = ⇒ = − is a straight line KL || to x-axis, at a

distance equal to 6− units from it.


(vi) 7 1

2 7 0 2 7 32 2

y y y− = ⇒ = ⇒ = =

It is a line parallel to x-axis at a distance of 1

32

units.

(vii) Giving some suitable values of x, we get corresponding values of y as given

below :

x 1 2 1− y 1− 2− 1

Now plot the points (1, 1), (2, 2)− − and ( 1,1)− on the graph and join them

to form a line.


(viii) The graph of 3 5 0,y + = is a straight line. OP || to x-axis, at a distance

equal to 5

,3

y = − i.e. 2

13

− units from it.

(ix) The graph of 2 5 0y − = is a straight line AB || to x-axis at a distance equal to

5

2 i.e.

12 .

2


(x) The graph of 0.y =

(xi) The graph of 0.x =


(xii) 2 1y x= −

Giving some suitable values of x, we get corresponding values of y, as given

below:

x 1 0 2

y 1 1− 3

Now plot the points (1, 1), (0, 1)− and (2, 3) on the graph and join them to

form a line.

(xiii) 5 5y x y x− = ⇒ = +

Now giving some suitable values to x, we get corresponding values of y as

given below :

x 0 1− 2− y 5 4 3


Q.16. Draw the graph of the equation, x y2 + = 6. Find the co-ordinates of the

points, where the graph meets the co-ordinates axis.

Ans. 2 6 6 2x y y x+ = ⇒ = −

Now giving some suitable values to x, we get corresponding values of y as

given below :

x 1 2 3

y 4 2 0

Now plot the points (1, 4), (2, 2) and (3, 0) on the graph and join them to get a

line which meet x-axis at (3, 0) and y-axis at (0, 6).


Q.17. Draw the graph of the equation x y−2 3 = 5.

(i) Find the value of y, when x = 4.

(ii) Find the value of x, when y = 3.

Ans. 2 3 5x y− = ⇒ 2 5 3x y= + ⇒5 3

2

yx

+=

Giving some suitable values to y, we get the corresponding values of x as given

below :

x 4 1 7

y 1 1− 3

Now plot the points (4,1), (1, 1)− and (7, 3) on the graph and join them to get a

line.

(i) If 4,x = then 1.y =

(ii) If 3,y = then 7.x =

Q.18. Draw the graphs of the equations y x −= 2 3 and y x −= 3 1. Find the co-

ordinates of the points of intersection of the two lines.

Ans. 2 3y x= −

Giving some different values to x, we get the corresponding values of y as given

below :

x 0 1 2

y 3− 1− 1

Now, plot the points (0, 3), (1, 1)− − and (2, 1) on the graph and join them to get

a line as shown 3 1y x= − .


Giving some different value to x, we get corresponding values of y as given

x 0 1 2

y 1− 2 5

Now, plot the points (0, 1), (1, 2)− and (2, 5) on the graph and join them to get

another line.

We see that these two lines intersect each other at the point P.

Hence, co-ordinates of P are ( 2, 7)− − .

Q.19. Draw the graph of the equations x y+ = 1 and x y3 + 2 = 6 . Find the co-

ordinates of the point of intersection of the two lines.

Ans. 1 1x y x y+ = ⇒ = −

Giving some suitable values to y, we get corresponding values of x as given

below :

x 0 2 1

y 1 1− 0

Now plot the points (0, 1), (2, 1)− and (1, 0) on the graph and join them to get

a line.

Again, 3 2 6x y+ =

2 6 3y x= −

6 3

2

xy

−=

Now giving some suitable values to x, we get the corresponding values of y as

given below :

x 0 2 4

y 3 0 3−

Now plot the points (0, 3), (2, 0) and (4, 3)− on the graph and join them to get

a line.


We see that these two lines intersect each other at the point (4, 3).−

∴ Co-ordinates are (4, 3).−

Q.20. Draw the graph of the equation : x y

+ = 1.2 3

Use the graph to find :

(i) the value of y, when x = 0.

(ii) the value of x, when y = 0.

Ans. 12 3

x y+ =

⇒ 3 2 6x y+ =

⇒ 3 6 2x y= −

⇒ 6 2

3

yx

−=

Now giving some suitable values to y, we get the corresponding values of x, as

given below :

x 2 0 4

y 0 3 3−

Now plot the point (2, 0), (0, 3) and (4, 3)− on the graph and join them to get

a line.


From the graph, we see that

(i) If 0,x = then 3.y =

(ii) and if 0,y = then 2.x =

Q.21. Draw the graphs of each of the equations y x y x y x− − −2 = 8, 5 = 14, 2 = 1.

Obtain the vertices of the triangle so formed.

Ans. 2 8 2 8y x x y− = ⇒ = −

Giving some different values of y, we get corresponding values of x as given

below:

x 4− 2− 0

y 2 3 4

Now, plot the points ( 4, 2), ( 2, 3)− − and (0, 4) on the graph and join them to get

a line. 5 14y x− =

⇒ 5 14x y= −

Now, giving some different values to y, we get corresponding values of x as

given

x 4− 1 6

y 2 3 4

Now, plot the points ( 4, 2), (1, 3)− and (6, 4) on the graph and join them to get

second line. 2 1y x− = ⇒ 2 1y x= +

Now, giving some different values to x, we get the corresponding values of y as

given:

x 0 1 2

y 1 3 5


Now, plot the points (0, 1), (1, 3) and (2, 5) on the graph and join them to get

third line.

We see that these lines intersect at each other at A, B and C.

Now, the co-ordinates of A are ( 4, 2),− of B are (1, 3) and of C are (2, 5).

Q.22. Draw the graph for each linear equation given below :

(i) y x= 2 + 3 (ii) y x −2

= 13

(iii) y x−= + 4 (iv) y x −5

= 42

(v) y x3 2

= +2 3

(vi) x y−2 3 = 4

(vii) x y−

−1 + 2

= 03 2

(viii) x y−2

3 = ( + 1)5

(ix) 5 2 0x y+ + =

Ans. (i) 2 3y x= +

When 1,x = then 5y =

When, 1,x = − 1y =

When, 2,x = − 1y = −

The table of 2 3y x= + valus are

x 1 1− 2− y 5 1 1−


The required graph of 2 3,y x= + is shown in the above figure.

(ii) 2

13

y x= −

When, 3,x = then 1y =

When, 3,x = − then 3y = −


The table of 2

1,3

y x= − values are

x 3 3− 6

y 1 3− 3

The required graph of 2

1,3

y x= − is shown in the above figure.


(iii) 4y x= − + or 4y x= −


When, 1,x = − then 5y =


The table of 4y x= − + values are

x 1 1− 2

y 3 5 2

The required graph of 4y x= − + is shown in the above figure.

(iv) 5

42

y x= − or 8 5

2

xy

−=

When, 0,x = then 5

2.52

y = − = −

When, 1,x = then 3

1.52

y = =

When, 2,x = then 11

5.52

y = =

The table of 5

4,2

y x= = − values are

x 0 1 2

y 2.5− 1.5 5.5


The required graph of 5

4 ,2

y x= − is shown in the adjoining figure.

(v) 3 2

2 3y x= +

When, 0,x = then 2

3y =

When, 2,x = then 3 2 2 2

2 3 32 3 3 3

y = × + = + =

When, 2

,3

x = then 3 2 2 2

12 3 3 3

y = × + =

The table of 3 2

2 3y x= + values are

x 0 2 2

3

y 2

3

23

3

21

3


The required graph of 3 2

2 3y x= + is shown in the above figure.

(vi) 2 3 4x y− =

⇒ 3 2 4y x= −

⇒ 2 4

3

xy

−=

When, 1,x = − then 2 ( 1) 4 6

23 3

y× − − −

= = − = −

When, 2,x = then 2 2 4 0

03 3

y× −

= = − =

When, 4,x = − then 2 ( 4) 4 12

43 3

y× − − −

= = − = −

The table of 2 3 4x y− = values are

x 1− 2 4− y 2− 0 4−


The required graph of 2 3 4x y− = is shown in the above figure.

(vii) 1 2

03 2

x y− +− =

⇒ 2( 1) 3( 2)

06

x y− − +=

⇒ 2 2 3 6 0x y− − − =

⇒ 2 3 8 0x y− − =

⇒ 2 8 3x y− =

⇒ 3 2 8y x= −

⇒ 2 8

3

xy

−=

When, 1,x = then 2 1 8 2 8

23 3

y× − −

= = = −

When, 2,x = − then 2 ( 2) 8 4 8 12

43 3 3

y× − − − −

= = = − = −

When, 4,x = then 2 4 8

03

y× −

= =

The table of 1 2

0,3 2

x y− +− = values are

x 1 2− 4

y 2− 4− 0


The required graph of 1 2

03 2

x y− +− = is shown in the above figure.

Documents

CO-ORDINATE GEOMETRY - Testlabz.com