Co-ordinate Geometry Class-9 Concise ICSE Maths Selina Solutions Chapter-26. We provide step by step Solutions of Exercise / lesson-26 Co-ordinate Geometry for ICSE Class-9 Concise Selina Mathematics by RK Bansal.

Our Solutions contain all type Questions with Exe-26 A, Exe-26 B, Exe-26 C to develop skill and confidence. Visit official Website CISCE for detail information about ICSE Board Class-9 Mathematics .

## Co-ordinate Geometry Class-9 Concise ICSE Maths Selina Solutions Chapter-26

–: Select Topics :–

Exe-26 A,

Exe-26 B,

Exe-26 C,

### Exe-26 A, Co-ordinate Geometry Class-9 Concise ICSE Maths Selina Solutions

#### Question. 1

For the equation given below; name the dependent and independent variables

……………….

#### Question. 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)

On the graph paper, let us draw the co-ordinate axes XOX’ and YOY’ intersecting at the origin O. With proper scale, mark the numbers on the two co-ordinate axes.
Now for the point A(8,7)
Step I
Starting from origin O, move 8 units along the positive direction of X axis, to the right of the origin O
Step II
Now from there, move 7 units up and place a dot at the point reached. Label this point as A(8,7)
Similarly plotting the other points

#### Question. 3

Find the values of x and y if:

(i). (x – 1, y + 3) = (4, 4)

(ii).  (3x + 1, 2y – 7) = (9, – 9)

(iii). (5x – 3y, y – 3x) = (4, -4)

#### Question. 4

Use the graph given alongside, to find the coordinates 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.

(i) The abscissa is 2
Now using the given graph the co-ordinate of the given point A is given by (2,2)
(ii) The ordinate is 0
Now using the given graph the co-ordinate of the given point B is given by (5,0)
(iii) The ordinate is 3
Now using the given graph the co-ordinate of the given point C and E is given by (-4,3)& (6,3)
(iv) The ordinate is -4
Now using the given graph the co-ordinate of the given point D is given by (2,-4)
(v) The abscissa is 5
Now using the given graph the co-ordinate of the given point H, B and G is given by (5,5) ,(5,0) & (5,-3)
(vi)The abscissa is equal to the ordinate.
Now using the given graph the co-ordinate of the given point I,A & H is given by (4,4),(2,2) & (5,5)
(vii)The ordinate is half of the abscissa
Now using the given graph the co-ordinate of the given point E is given by (6,3)

#### Question. 5

State, true or false:
(i)The ordinate of a point is its x-co-ordinate.
(ii)The origin is in the first quadrant.
(iii)The y-axis is the vertical number line.
(iv)Every point is located in one of the four quadrants.
(v)If the ordinate of a point is equal to its abscissa; the point lies either in the first quadrant or in the second quadrant.
(vi)The origin (0,0) lies on the x-axis.
(vii)The point (a,b) lies on the y-axis if b=0.

(i)The ordinate of a point is its x-co-ordinate.
False.
(ii)The origin is in the first quadrant.
False.
(iii)The y-axis is the vertical number line.
True.
(iv)Every point is located in one of the four quadrants.
True.
(v)If the ordinate of a point is equal to its abscissa; the point lies either in the first quadrant or in the second quadrant.
False.
(vi)The origin (0,0) lies on the x-axis.
True.
(vii)The point (a,b) lies on the y-axis if b=0.
False

#### Question. 6

In the following, find the coordinates of the point whose abscissa is the solution of the first equation and ordinate is the solution of the second equation
(i) 3 – 2x = 7; 2y + 1 = 10 – 2 1/2y.
(ii)…………….
(iii)………………

#### Question. 7

In the following, the coordinates of the three vertices of a rectangle ABCD are given. By plotting the given points; find, in case, the coordinates 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).

After plotting the given points A(2,0), B(8,0) and C(8,4) on a graph paper; joining A with B and B with C. From the graph it is clear that the vertical distance between the points B(8,0) and C(8,4) is 4 units, therefore the vertical distance between the points A(2,0) and D must be 4 units. Now complete the rectangle ABCD
As is clear from the graph D(2,4)
(ii)A(4,2), B(-2,-2) and D(4,-2)

After plotting the given points A(4,2), B(-2,2) and D(4,-2) on a graph paper; joining A with B and A with D. From the graph it is clear that the vertical distance between the points A(4,2) and D(4,-2) is 4 units and the horizontal distance between the points A(4,2) and B(-2,2) is 6 units , therefore the vertical distance between the points B(-2,2)and C must be 4 units and the horizontal distance between the points B(-2,2) and C must be 6 units. Now complete the rectangle ABCD
As is clear from the graph C(-2,2)

#### Question. 8

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, form the same graph, state the co-ordinates of the mid-points of the sides AB and CD

After plotting the given points A(2,-2), B(8,2) and C(4,-4) on a graph paper; joining B with C and B with A . Now complete the parallelogram ABCD.
As is clear from the graph D(-6,4)
Now from the graph we can find the mid points of the sides AB and CD.
Therefore the co-ordinates of the mid-point of AB is E(3,2) and the co-ordinates of the mid-point of CD is F(-1,-4)

#### Question. 9

A (-2, 4), C(4, 10) and D(-2, 10) are the vertices of a square ABCD. Use the graphical method to find the co-ordinates of the fourth vertex B. Also, find:
(i) The co-ordinates of the mid-point of BC;
(ii) The co-ordinates of the mid-point of CD and
(iii) The co-ordinates of the point of intersection of the diagonals of the square ABCD.

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