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.

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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 – 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.

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 – 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)

We know that

2a/3 – 1 = a/2

2a/3 – a/2 = 1

Taking LCM

(4a – 3a)/ 6 = 1

a = 6

Similarly

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)

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

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)

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 – 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)

(ii)

(iii)

(iv)

(v)

(vi)

(vii)

(viii)

(ix)

(x)

(xi)

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)

(ii)

(iii)

(iv)

(v)

(vi)

(v)

(viii)

(ix)

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

(i) y = 2x + 3

Solution:

(i)

(ii)

(iii)

(iv)

(v)

(vi)

(vii) We can write the equation as

2x – 3y = 8

(viii) We can write the equation as

5x – 2y = 17

(ix)

4. Draw the graph for each equation given below:

(i) 3x + 2y = 6

(ii) 2x – 5y = 10

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

Solution:

(i)

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

(ii)

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

(iii)

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

(iv)

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)

We know that

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

= ½ × 4 × 12

= 24 sq. units

(ii)

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:

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.