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:-Download Here

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Exercise 26A page: 315

1. For each equation given below; name the dependent and independent variables. Concise Selina Solutions for Class 9 Maths Chapter 26 Image 1

Solution:

(i)

Concise Selina Solutions for Class 9 Maths Chapter 26 Image 2

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)

Concise Selina Solutions for Class 9 Maths Chapter 26 Image 3

x is the dependent variable

y is the independent variable

(iv)

Concise Selina Solutions for Class 9 Maths Chapter 26 Image 4

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

Concise Selina Solutions for Class 9 Maths Chapter 26 Image 5

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)

Solution:

We know that two ordered pairs are equal.

(i) (x – 1, y + 3) = (4, 4)

It can be written as

x – 1 = 4 and y + 3 = 4

x = 5 and y = 1

(ii) (3x + 1, 2y – 7) = (9, -9)

It can be written as

3x + 1 = 9 and 2y – 7 = -9

3x = 8 and 2y = -2

x = 8/3 and y = -1

(iii) (5x – 3y, y – 3x) = (4, -4)

It can be written as

5x – 3y = 4 ….. (1)

y – 3x = -4 ….. (2)

By multiplying equation (2) by 3

3y – 9x = – 12 …… (3)

Now add equations (1) and (3)

(5x – 3y) + (3y – 9x) = 4 + (-12)

– 4x = -8

x = 2

Substituting the value of x in equation (2)

y – 3x = -4

y = 3x – 4

y = 3 (2) – 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.

Concise Selina Solutions for Class 9 Maths Chapter 26 Image 6

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:

Concise Selina Solutions for Class 9 Maths Chapter 26 Image 7

Solution:

(i)

Concise Selina Solutions for Class 9 Maths Chapter 26 Image 8

We know that

3 – 2x = 7

3 – 7 = 2x

– 4 = 2x

x = – 2

Similarly

2y + 1 – 10 – 2 ½ y

2y + 1 = 10 – 5/2 y

By cross multiplication

4y + 2 = 20 – 5y

4y + 5y = 20 – 2

9y = 18

y = 2

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

(ii)

Concise Selina Solutions for Class 9 Maths Chapter 26 Image 9

We know that

2a/3 – 1 = a/2

2a/3 – a/2 = 1

Taking LCM

(4a – 3a)/ 6 = 1

a = 6

Similarly

Concise Selina Solutions for Class 9 Maths Chapter 26 Image 10

By taking LCM

45 – 12b = 14b – 7

45 + 7 = 14b + 12b

52 = 26b

b = 2

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

(iii)

Concise Selina Solutions for Class 9 Maths Chapter 26 Image 11

We know that

5x – (5 – x) = ½ (3 – x)

It can be written as

(5x + x) – 5 = ½ (3 – x)

By cross multiplication

12x – 10 = 3 – x

12x + x = 3 + 10

13x = 13

x = 1

Similarly

Concise Selina Solutions for Class 9 Maths Chapter 26 Image 12

By cross multiplication

12 – 9y = 4 + y

12 – 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)

Concise Selina Solutions for Class 9 Maths Chapter 26 Image 13

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).

Concise Selina Solutions for Class 9 Maths Chapter 26 Image 14

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).

Concise Selina Solutions for Class 9 Maths Chapter 26 Image 15

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)

Concise Selina Solutions for Class 9 Maths Chapter 26 Image 16

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

Concise Selina Solutions for Class 9 Maths Chapter 26 Image 17

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.

Concise Selina Solutions for Class 9 Maths Chapter 26 Image 18

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:

Concise Selina Solutions for Class 9 Maths Chapter 26 Image 19

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 – 5 = 0

(iv) 2x – 7 = 0

(v) y = 4

(vi) y + 6 = 0

(vii) y – 2 = 0

(viii) 3y + 5 = 0

(ix) 2y – 5 = 0

(x) y = 0

(xi) x = 0

Solution:

(i)

x

3

3

3

y

-1

0

1

Concise Selina Solutions for Class 9 Maths Chapter 26 Image 20

(ii)

x

-3

-3

-3

y

-1

0

1

Concise Selina Solutions for Class 9 Maths Chapter 26 Image 21

(iii)

x

5

5

5

y

-1

0

1

Concise Selina Solutions for Class 9 Maths Chapter 26 Image 22

(iv)

x

7/2

7/2

7/2

y

-1

0

1

Concise Selina Solutions for Class 9 Maths Chapter 26 Image 23

(v)

x

-1

0

-1

y

4

4

4

Concise Selina Solutions for Class 9 Maths Chapter 26 Image 24

(vi)

x

-1

0

1

y

-6

-6

-6

Concise Selina Solutions for Class 9 Maths Chapter 26 Image 25

(vii)

x

-1

0

1

y

2

2

2

Concise Selina Solutions for Class 9 Maths Chapter 26 Image 26

(viii)

x

-1

0

1

y

-6

-6

-6

Concise Selina Solutions for Class 9 Maths Chapter 26 Image 27

(ix)

x

-1

0

1

y

5/2

5/2

5/2

Concise Selina Solutions for Class 9 Maths Chapter 26 Image 28

(x)

x

-1

0

1

y

0

0

0

Concise Selina Solutions for Class 9 Maths Chapter 26 Image 29

(xi)

x

0

0

0

y

-1

0

1

Concise Selina Solutions for Class 9 Maths Chapter 26 Image 30

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 – y = 0

(viii) 3x + 2y = 0

(ix) x = -2y

Solution:

(i)

x

-1

0

1

y

-3

0

3

Concise Selina Solutions for Class 9 Maths Chapter 26 Image 31

(ii)

x

-1

0

1

y

1

0

-1

Concise Selina Solutions for Class 9 Maths Chapter 26 Image 32

(iii)

x

-1

0

1

y

2

0

-2

Concise Selina Solutions for Class 9 Maths Chapter 26 Image 33

(iv)

x

-1

0

1

y

-1

0

1

Concise Selina Solutions for Class 9 Maths Chapter 26 Image 34

(v)

x

-1

0

1

y

5

0

-5

Concise Selina Solutions for Class 9 Maths Chapter 26 Image 35

(vi)

x

-1

0

1

y

1/2

0

  • ½

Concise Selina Solutions for Class 9 Maths Chapter 26 Image 36

(v)

x

-1

0

1

y

-4

0

4

Concise Selina Solutions for Class 9 Maths Chapter 26 Image 37

(viii)

x

-1

0

1

y

3/2

0

-3/2

Concise Selina Solutions for Class 9 Maths Chapter 26 Image 38

(ix)

x

-1

0

1

y

½

0

  • ½

Concise Selina Solutions for Class 9 Maths Chapter 26 Image 39

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

(i) y = 2x + 3

Concise Selina Solutions for Class 9 Maths Chapter 26 Image 40

Solution:

(i)

x

-1

0

1

y

-5/3

3

5

Concise Selina Solutions for Class 9 Maths Chapter 26 Image 41

(ii)

x

-1

0

1

y

-5/3

-1

-1/3

Concise Selina Solutions for Class 9 Maths Chapter 26 Image 42

(iii)

x

-1

0

1

y

5

4

3

Concise Selina Solutions for Class 9 Maths Chapter 26 Image 43

(iv)

x

-1

0

1

y

-13/2

-5/2

3/2

Concise Selina Solutions for Class 9 Maths Chapter 26 Image 44

(v)

x

-1

0

1

y

-5/6

2/3

13/6

Concise Selina Solutions for Class 9 Maths Chapter 26 Image 45

(vi)

x

-1

0

1

y

-2

-4/3

-2/3

Concise Selina Solutions for Class 9 Maths Chapter 26 Image 46

(vii) We can write the equation as

2x – 3y = 8

x

-1

0

1

y

-10/3

-8/3

-2

Concise Selina Solutions for Class 9 Maths Chapter 26 Image 47

(viii) We can write the equation as

5x – 2y = 17

x

-1

0

1

y

-11

-17/2

-6

Concise Selina Solutions for Class 9 Maths Chapter 26 Image 48

(ix)

x

-1

0

1

y

-1/5

-2/5

-3/5

Concise Selina Solutions for Class 9 Maths Chapter 26 Image 49

4. Draw the graph for each equation given below:

(i) 3x + 2y = 6

(ii) 2x – 5y = 10

Concise Selina Solutions for Class 9 Maths Chapter 26 Image 50

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

Concise Selina Solutions for Class 9 Maths Chapter 26 Image 51

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

Concise Selina Solutions for Class 9 Maths Chapter 26 Image 52

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

Concise Selina Solutions for Class 9 Maths Chapter 26 Image 53

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

Concise Selina Solutions for Class 9 Maths Chapter 26 Image 54

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 – (5 – y) = 7

(ii) 7 – 3 (1 – y) = – 5 + 2x

Solution:

(i)

Concise Selina Solutions for Class 9 Maths Chapter 26 Image 55

We know that

Area of the right triangle obtained = ½ × base × altitude

= ½ × 4 × 12

= 24 sq. units

(ii)

Concise Selina Solutions for Class 9 Maths Chapter 26 Image 56

We know that

Area of the right triangle obtained = ½ × base × altitude

= ½ × 9/2 × 3

= 27/4

= 6.75 sq. units

Exercise 26C page: 323

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

Concise Selina Solutions for Class 9 Maths Chapter 26 Image 57

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 θ = 450

(ii) The inclination of line AB is θ = 1350

(iii) The inclination of line AB is θ = 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 θ = 00.

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

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

(iv) The inclination of a line which is perpendicular to y-axis θ = 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 θ if θ is the inclination of a line.

Here slope is usually denoted by the letter m.

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

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

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

Therefore, the slope of the line is m = tan θ = tan 300 = 1/ √3

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

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

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

Therefore, the slope of the line is m = tan θ = tan 600 = √3

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

(i) 0

(ii) 1

(iii) √3

(iv) 1/√3

Solution:

If tan θ is the slope of a line; then the inclination of the line is θ

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

tan θ = 0

tan θ = tan 00

θ = 00

Hence, the inclination of the given line is θ = 00.

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

tan θ = 1

tan θ = tan 450

θ = 450

Hence, the inclination of the given line is θ = 450.

(iii) If the slope of the line is √3; then tan θ = √3

tan θ = √3

tan θ = tan 600

θ = 600

Hence, the inclination of the given line is θ = 600.

(iv) If the slope of the line is 1/√3; then tan θ = 1/√3

tan θ = 1/√3

tan θ = tan 300

θ = 300

Hence, the inclination of the given line is θ = 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 θ = 00

So the slope (m) = tan θ = tan 00 = 0

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

So the slope (m) = tan θ = tan 900 = ∞ (not defined)

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

So the slope (m) = tan θ = tan 900 = ∞ (not defined)

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

So the slope (m) = tan θ = 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

26.6 Quadrants and Sign Convention

26.7 Plotting of Points

26.8 Graphing a linear equation

26.10 Inclination And Slope

26.11 Y-Intercept

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

The branch of Mathematics that deals with questions related to shape, size, relative position of figures, and the properties of space is known as Geometry. Further, the branch of geometry in which the position of the points on the plane is defined by coordinates are known as Co-ordinate Geometry. Read and learn the Chapter 26 of Selina textbook to learn more about Co-ordinate Geometry along with the concepts covered in it. Solve the Selina Solutions for Class 9 effectively to score high in the examination.

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