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

Selina Solutions for Class 9 Maths Chapter 26 Co-ordinate Geometry are provided here. Since these concepts are continued in Class 10, it is highly important to understand the concepts taught in Class 9 in depth. To score good marks in Class 9 Mathematics examination, students are advised to solve questions provided in each exercise in the book by Selina publication. This Selina solutions for Class 9 Maths helps students in understanding the concepts better. Download pdf of Class 9 Maths Chapter 26 Selina Solutions from the link given below.

## Download PDF of Selina Solutions for Class 9 Maths Chapter 26:-

Exercise 26A page: 315

1. For each equation given below; name the dependent and independent variables.

Solution:

(i)

y is the dependent variable

x is the independent variable

(ii) x = 9y + 4

x is the dependent variable

y is the independent variable

(iii)

x is the dependent variable

y is the independent variable

(iv)

y is the dependent variable

x is the independent variable

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)

Solution:

Consider the points as

(i) (8, 7) = A

(ii) (3, 6) = B

(iii) (0, 4) = C

(iv) (0, -4) = D

(v) (3, -2) = E

(vi) (-2, 5) = F

(vii) (-3, 0) = G

(viii) (5, 0) = H

(ix) (-4, -3) = I

3. Find the values of x and y if:

(i) (x â€“ 1, y + 3) = (4, 4)

(ii) (3x + 1, 2y â€“ 7) = (9, -9)

(iii) (5x â€“ 3y, y â€“ 3x) = (4, -4)

Solution:

We know that two ordered pairs are equal.

(i) (x â€“ 1, y + 3) = (4, 4)

It can be written as

x â€“ 1 = 4 and y + 3 = 4

x = 5 and y = 1

(ii) (3x + 1, 2y â€“ 7) = (9, -9)

It can be written as

3x + 1 = 9 and 2y â€“ 7 = -9

3x = 8 and 2y = -2

x = 8/3 and y = -1

(iii) (5x â€“ 3y, y â€“ 3x) = (4, -4)

It can be written as

5x â€“ 3y = 4 â€¦.. (1)

y â€“ 3x = -4 â€¦.. (2)

By multiplying equation (2) by 3

3y â€“ 9x = – 12 â€¦â€¦ (3)

Now add equations (1) and (3)

(5x â€“ 3y) + (3y â€“ 9x) = 4 + (-12)

– 4x = -8

x = 2

Substituting the value of x in equation (2)

y â€“ 3x = -4

y = 3x â€“ 4

y = 3 (2) â€“ 4

y = 2

Therefore, x = 2 and y = 2.

4. Use the graph given alongside, to find the coordinates of 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.

Solution:

(i) The abscissa is 2.

Based on the graph,

The co-ordinate of the point A is given by (2, 2).

(ii) The ordinate is 0.

Based on the graph,

The co-ordinate of the point B is given by (5, 0).

(iii) The ordinate is 3.

Based on the graph,

The co-ordinates of the points C and E are given by (-4, 3) and (6, 3).

(iv) The ordinate is -4.

Based on the graph,

The co-ordinate of the point D is given by (2, -4).

(v) The abscissa is 5.

Based on the graph,

The co-ordinates of the points H, B and G are given by (5, 5), (5, 0) and (5, -3).

(vi) The abscissa is equal to the ordinate.

Based on the graph,

The co-ordinates of the points I, A and H are given by (4, 4), (2, 2) and (5, 5).

(vii) The ordinate is half of the abscissa.

Based on the graph,

The co-ordinate of the point E is given by (6, 3).

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.

Solution:

(i) False

(ii) False

(iii) True

(iv) True

(v) False

(vi) True

(vii) False

6. In each of the following, 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:

Solution:

(i)

We know that

3 â€“ 2x = 7

3 â€“ 7 = 2x

– 4 = 2x

x = – 2

Similarly

2y + 1 â€“ 10 â€“ 2 Â½ y

2y + 1 = 10 â€“ 5/2 y

By cross multiplication

4y + 2 = 20 â€“ 5y

4y + 5y = 20 â€“ 2

9y = 18

y = 2

Hence, the co-ordinates of the point are (-2, 2).

(ii)

We know that

2a/3 â€“ 1 = a/2

2a/3 â€“ a/2 = 1

Taking LCM

(4a â€“ 3a)/ 6 = 1

a = 6

Similarly

By taking LCM

45 â€“ 12b = 14b â€“ 7

45 + 7 = 14b + 12b

52 = 26b

b = 2

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

(iii)

We know that

5x â€“ (5 â€“ x) = Â½ (3 â€“ x)

It can be written as

(5x + x) â€“ 5 = Â½ (3 â€“ x)

By cross multiplication

12x â€“ 10 = 3 â€“ x

12x + x = 3 + 10

13x = 13

x = 1

Similarly

By cross multiplication

12 â€“ 9y = 4 + y

12 â€“ 4 = y + 9y

8 = 10y

y = 8/10

y = 4/5

Hence, the co-ordinates of the point are (1, 4/5).

7. In each of the following, 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).

Solution:

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

From the graph the co-ordinates of the fourth vertex is D (2, 4).

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

From the graph the co-ordinates of the fourth vertex is C (-2, 2).

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

From the graph the co-ordinates of the fourth vertex is B (6, -6).

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

From the graph the co-ordinates of the fourth vertex is A (10, -2).

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.

Solution:

It is given that

A (2, -2), B (8, 2) and C (4, -4) are the vertices of the parallelogram ABCD

By joining A, B, C and D we get the parallelogram ABCD.

From the graph, we get D (-6, 4)

Using the graph,

The co-ordinates of the mid-point of AB is E (3, 2)

The co-ordinates of the mid-point of CD is F (-1, -4)

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.

Solution:

It is given that

A (-2, 4), C (4, 10) and D (-2, 10) are the vertices of a square ABCD.

From the graph, we get B (4, 4)

Using the graph,

The co-ordinates of the mid-point of BC is E (4, 7)

The co-ordinates of the mid-point of CD is F (1, 10)

The co-ordinates of the diagonals of the square is G (1, 7)

10. By plotting the following points on the same graph paper. Check whether they are collinear or not:

(i) (3, 5), (1, 1) and (0, -1)

(ii) (-2, -1), (-1, -4) and (-4, 1)

Solution:

After plotting the points, we clearly see from the graph that

(i) A (3, 5), B (1, 1) and C (0, -1) are collinear

(ii) P (-2, -1), Q (-1, -4) and R (-4, 1) are non-collinear.

Exercise 26B page: 320

1. Draw the graph for each linear equation given below:

(i) x = 3

(ii) x + 3 = 0

(iii) x â€“ 5 = 0

(iv) 2x â€“ 7 = 0

(v) y = 4

(vi) y + 6 = 0

(vii) y â€“ 2 = 0

(viii) 3y + 5 = 0

(ix) 2y â€“ 5 = 0

(x) y = 0

(xi) x = 0

Solution:

(i)

 x 3 3 3 y -1 0 1

(ii)

 x -3 -3 -3 y -1 0 1

(iii)

 x 5 5 5 y -1 0 1

(iv)

 x 7/2 7/2 7/2 y -1 0 1

(v)

 x -1 0 -1 y 4 4 4

(vi)

 x -1 0 1 y -6 -6 -6

(vii)

 x -1 0 1 y 2 2 2

(viii)

 x -1 0 1 y -6 -6 -6

(ix)

 x -1 0 1 y 5/2 5/2 5/2

(x)

 x -1 0 1 y 0 0 0

(xi)

 x 0 0 0 y -1 0 1

2. Draw the graph for each linear equation given below:

(i) y = 3x

(ii) y = – x

(iii) y = – 2x

(iv) y = x

(v) 5x + y = 0

(vi) x + 2y = 0

(vii) 4x â€“ y = 0

(viii) 3x + 2y = 0

(ix) x = -2y

Solution:

(i)

 x -1 0 1 y -3 0 3

(ii)

 x -1 0 1 y 1 0 -1

(iii)

 x -1 0 1 y 2 0 -2

(iv)

 x -1 0 1 y -1 0 1

(v)

 x -1 0 1 y 5 0 -5

(vi)

 x -1 0 1 y 1/2 0 Â½

(v)

 x -1 0 1 y -4 0 4

(viii)

 x -1 0 1 y 3/2 0 -3/2

(ix)

 x -1 0 1 y Â½ 0 Â½

3. Draw the graph for each linear equation given below:

(i) y = 2x + 3

Solution:

(i)

 x -1 0 1 y -5/3 3 5

(ii)

 x -1 0 1 y -5/3 -1 -1/3

(iii)

 x -1 0 1 y 5 4 3

(iv)

 x -1 0 1 y -13/2 -5/2 3/2

(v)

 x -1 0 1 y -5/6 2/3 13/6

(vi)

 x -1 0 1 y -2 -4/3 -2/3

(vii) We can write the equation as

2x â€“ 3y = 8

 x -1 0 1 y -10/3 -8/3 -2

(viii) We can write the equation as

5x â€“ 2y = 17

 x -1 0 1 y -11 -17/2 -6

(ix)

 x -1 0 1 y -1/5 -2/5 -3/5

4. Draw the graph for each equation given below:

(i) 3x + 2y = 6

(ii) 2x â€“ 5y = 10

In each case, find the co-ordinates of the points where the graph (line) drawn meets the co-ordinates axes.

Solution:

(i)

 x -2 0 2 y 6 3 0

From the graph, the line intersects x-axis at (2, 0) and y-axis at (0, 3).

(ii)

 x -1 0 1 y -12/5 -2 -8/5

From the graph, the line intersects x-axis at (5, 0) and y-axis at (0, -2).

(iii)

 x -1 0 1 y 5.25 4.5 3.75

From the graph, the line intersects x-axis at (10, 0) and y-axis at (0, 7.5).

(iv)

 x -1 0 1 y -3 1/3 11/3

From the graph, the line intersects x-axis at (-1/10, 0) and y-axis at (0, 4.5).

5. For each linear equation, given above, draw the graph and then use the graph drawn (in each case) to find the area of a triangle enclosed by the graph and the co-ordinates axes:

(i) 3x â€“ (5 â€“ y) = 7

(ii) 7 â€“ 3 (1 â€“ y) = – 5 + 2x

Solution:

(i)

We know that

Area of the right triangle obtained = Â½ Ã— base Ã— altitude

= Â½ Ã— 4 Ã— 12

= 24 sq. units

(ii)

We know that

Area of the right triangle obtained = Â½ Ã— base Ã— altitude

= Â½ Ã— 9/2 Ã— 3

= 27/4

= 6.75 sq. units

Exercise 26C page: 323

1. In each of the following, find the inclination of line AB:

Solution:

The angle which is a straight line makes with the positive direction of x-axis (measured in anticlockwise direction) is called as inclination of the line.

(i) The inclination of line AB is Î¸ = 450

(ii) The inclination of line AB is Î¸ = 1350

(iii) The inclination of line AB is Î¸ = 300

2. Write the inclination of a line which is:

(i) Parallel to x-axis.

(ii) Perpendicular to x-axis.

(iii) Parallel to y-axis.

(iv) Perpendicular to y-axis.

Solution:

(i) The inclination of a line which is parallel to x-axis is Î¸ = 00.

(ii) The inclination of a line which is perpendicular to x-axis is Î¸ = 900.

(iii) The inclination of a line which is parallel to y-axis is Î¸ = 900.

(iv) The inclination of a line which is perpendicular to y-axis Î¸ = 00.

3. Write the slope of the line whose inclination is:

(i) 00

(ii) 300

(iii) 450

(iv) 600

Solution:

The slope of the line is tan Î¸ if Î¸ is the inclination of a line.

Here slope is usually denoted by the letter m.

(i) The inclination of a line is 00 then Î¸ = 00.

Therefore, the slope of the line is m = tan 00 = 0

(ii) The inclination of a line is 300 then Î¸ = 300.

Therefore, the slope of the line is m = tan Î¸ = tan 300 = 1/ âˆš3

(iii) The inclination of a line is 450 then Î¸ = 450.

Therefore, the slope of the line is m = tan Î¸ = tan 450 = 1

(iv) The inclination of a line is 600 then Î¸ = 600.

Therefore, the slope of the line is m = tan Î¸ = tan 600 = âˆš3

4. Find the inclination of the line whose slope is:

(i) 0

(ii) 1

(iii) âˆš3

(iv) 1/âˆš3

Solution:

If tan Î¸ is the slope of a line; then the inclination of the line is Î¸

(i) If the slope of the line is 0; then tan Î¸ =0

tan Î¸ = 0

tan Î¸ = tan 00

Î¸ = 00

Hence, the inclination of the given line is Î¸ = 00.

(ii) If the slope of the line is 1; then tan Î¸ = 1

tan Î¸ = 1

tan Î¸ = tan 450

Î¸ = 450

Hence, the inclination of the given line is Î¸ = 450.

(iii) If the slope of the line is âˆš3; then tan Î¸ = âˆš3

tan Î¸ = âˆš3

tan Î¸ = tan 600

Î¸ = 600

Hence, the inclination of the given line is Î¸ = 600.

(iv) If the slope of the line is 1/âˆš3; then tan Î¸ = 1/âˆš3

tan Î¸ = 1/âˆš3

tan Î¸ = tan 300

Î¸ = 300

Hence, the inclination of the given line is Î¸ = 300.

5. Write the slope of the line which is:

(i) Parallel to x-axis.

(ii) Perpendicular to x-axis.

(iii) Parallel to y-axis.

(iv) Perpendicular to y-axis.

Solution:

(i) We know that the inclination of line parallel to x-axis Î¸ = 00

So the slope (m) = tan Î¸ = tan 00 = 0

(ii) We know that the inclination of line perpendicular to x-axis Î¸ = 900

So the slope (m) = tan Î¸ = tan 900 = âˆž (not defined)

(iii) We know that the inclination of line parallel to y-axis Î¸ = 900

So the slope (m) = tan Î¸ = tan 900 = âˆž (not defined)

(iv) We know that the inclination of line perpendicular to y-axis Î¸ = 00

So the slope (m) = tan Î¸ = tan 00 = 0

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

The Chapter 26, Co-ordinate Geometry, contains 3 exercises and the Selina Solutions given here contains the answers for all the questions present in these exercises. Let us have a look at some of the topics that are being discussed in this chapter.

26.1 Introduction

26.2 Dependent and Independent Variable

26.3 Ordered Pair

26.4 Cartesian Plane

26.5 Co-ordinates of Points