## RS Aggarwal Class 9 Solutions Chapter 6 Coordinate Geometry Ex 6A

These Solutions are part of RS Aggarwal Solutions Class 9. Here we have given RS Aggarwal Class 9 Solutions Chapter 6 Coordinate Geometry Ex 6A.

**Question 1.**

**Write down the coordinates of each of the points A, B, C, D, E shown below :**

**Solution:**

Co-ordinates of

A are ( – 6, 5)

B are (5, 4)

C are ( – 3, 2)

D are (2, – 2)

E are ( – 1, – 4)

Ans.

**Question 2.**

**Draw the lines X’OX and YOY’ as the coordinates axes on a paper and plot the following points on it.**

**(i) P (7, 4)**

**(ii) Q ( – 5, 3)**

**(iii) R ( – 6, – 3)**

**(iv) S (3, – 7)**

**(v) A (6, 0)**

**(vi) B (0, 9)**

**(vii) O(0,0)**

**(viii) C ( – 3, – 3)**

**Solution:**

The given points have been plotted as shown in the adjoining graph.

Where X’OX and YOY’ are the axis:

**Question 3.**

**On which axis do the following points lie?**

**(i) (7, 0)**

**(ii) (0, – 5)**

**(iii) (0, 1)**

**(iv) ( – 4, 0)**

**Solution:**

(i) (7, 0) lies on x-axis as its ordinate is (0)

(ii) (0, – 5) lies on y-axis as its abscissa is (0)

(iii) (0, 1) lies on y-axis as its abscissa is (0)

(iv) ( – 4, 0) lies on jc-axis as its ordinate is (0)

Ans.

**Question 4.**

**In which quadrant do the given points lie ?**

**(i) ( – 6, 5)**

**(ii) ( – 3, – 2)**

**(iii) (2, – 9)**

**Solution:**

(i) ( – 6, 5) lies in second quadrant because A is of the type (-, +)

(ii) ( – 3, – 2) lies in third quadrant because A is of the type (-, -)

(iii) (2, – 9) lies in fourth quadrant because it is of the type (+, -).

**Question 5.**

**Draw the graph of the equation y = x+1.**

**Solution:**

In the given equation, .

y = x+1

Put x = 0, then y = 0 + 1 = 1

x = 1, then, y = 1 + 1=2

x = 2, then, y = 2 + 1 = 3

Now, plot the points as given in the table given below on the graph, and join them as shown.

**Question 6.**

**Draw the graph of the equation, y = 3x+2.**

**Solution:**

In the given equation

y = 3x + 2

Put x = 0,

then y = 3 x 0 + 2 = 0 + 2 = 2

x = 1, then

y = 3 x 1 + 2 = 3 + 2 = 5

and x = – 2, then

y = 3 x ( – 2) + 2 = – 6 + 2 = – 4

**Question 7.**

**Draw the graph of the equation, y = 5x – 3**

**Solution:**

In the given equation,

y = 5x – 3

Put x = 1,y = 5 x 1 – 3 = 5 – 3 = 2

x = 0 then,

y = 5 x 0 – 3 = 0 – 3 = – 3

and x = 2, then

y = 5 x 2 – 3 = 10 – 3 = 7

**Question 8.**

**Draw the graph of the equation y = 3x.**

**Solution:**

In the given equation,

y = 3x,

Put x = 0,

then y = 3 x 0 = 0

Put x = 1, then

y = 3 x 1 = 3

Put x = – 1, then

y = 3 ( – 1) = – 3

Now, plot the points as given in the table below

and join them as shown.

**Question 9.**

Draw the graph of the equation y = – x,

**Solution:**

In the given equation, y = – x,

Put x = 1,

then y = – 1

Put x = 2, then

y = – 2

Put x = – 2, then

y = – ( – 2) = 2

Hope given RS Aggarwal Class 9 Solutions Chapter 6 Coordinate Geometry Ex 6A are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

## Leave a Reply