#### Chapter 13 Linear Equations in Two Variables R.D. Sharma Solutions for Class 9th MCQ's

**Exercise 13.3**

**1. If (4, 19) is a solution of the equation y = ax + 3, then a=**

**(a) 3**

**(b) 4**

**(c) 5**

**(d) 6**

**Solution**

We are given (4,19)as the solution of equation.

y = ax+3

Substituting x= 4 and y= 19, we get

19 = 4a+3

⇒ 4a=19-3

⇒ 4a=16

⇒ a=4

Therefore, the correct answer is (b).

**2. If (a, 4) lies on the graph of 3x + y = 10, then the value of a is**

**(a) 3**

**(b) 1**

**(c) 3**

**(d) 4**

**Solution**

We are given (a, 4) lies on the graph of linear equation 3x+ y= 10.

So, the given co-ordinates are the solution of the equation 3x+ y= 10.

Therefore, we can calculate the value of a by substituting the value of given co-ordinates in equation 3x+ y= 10.

Substituting x= a and y= 4 in equation 3x+ y= 10, we get

3×a + 4 = 10

⇒ 3a = 10-4

⇒ 3a=6

⇒ a=6/3

⇒ a=2

No, option is correct.

**3. The graph of the linear equation 2x − y = 4 cuts x- axis at**

**(a) (2, 0)**

**(b) (−2, 0)**

**(c) (0, −4)**

**(d) (0, 4)**

**Solution**

Given,

2x- y = 4

we get,

y = 2x-4

We will substitute y =0 in y = 2x-4 to get the co-ordinates for the graph of 2x- y = 4 at x axis

0 = 2x-4

⇒ 2x=0+4

⇒ 2x=4 x=2

Co-ordinates for the graph of 2x- y = 4 are (2,0).

Therefore, the correct answer is (a).

**4. How many linear equations are satisfied by x = 2 and y = −3?**

**(a) Only one**

**(b) Two**

**(c) Three**

**(d) Infinitely many**

**Solution**

There are infinite numbers of linear equations that are satisfied by x= 2 and y = -3 as

(i) Every solution of the linear equation represent a point on the line.

(ii) Every point on the line is the solution of the linear equation.

Therefore, the correct answer is (d).

5. The equation

5. The equation

*x*− 2 = 0 on number line is represented by**(a) a line**

**(b) a point**

**(c) infinitely many lines**

**(d) two lines**

**Solution**

The equation x - 2 = 0 is represented by a point on the number line.

Therefore, the correct answer is (b).

**6. x = 2, y = −1 is a solution of the linear equation**

**(a) x + 2y = 0**

**(b) x + 2y = 4**

**(c) 2x + y = 0**

**(d) 2x + y = 5**

**Solution**

**We are given x=2; y = -1 as the solution of linear equation, which we have to find.**

The equation is x+2y = 0 which can he proved by

Substituting x=2 and y= -1 in the equation x+2y= 0, we get

2 + 2×(-1) = 0

2-2 = 0

0 = 0

RHS = LHS

Therefore, the correct answer is (a).

**7. If (2k − 1, k) is a solution of the equation 10x − 9y = 12, then k =**

**(a) 1**

**(b) 2**

**(c) 3**

**(d) 4**

**Solution**

We are given (2k-1, k)as the solution of equation

10x - 9y = 12

Substituting x=2k-1 and y = k , we get

10 ×(2k-1)- 9×k =12

⇒ 20k - 10-9k =12

⇒ 11k = 12+10

⇒ 11k = 22

⇒ k = 22/11

⇒ k = 2

Therefore, the correct answer is (b).

8. The distance between the graph of the equations

*x*= −3 and

*x*= 2 is

(a) 1

(b) 2

(c) 3

(d) 5

**Solution**

Distance between the graph of equations x= -3 and x = 2.

D = Distance of co-ordinate on negative side of x axis + Distance of co-ordinate on positive side of x axis

Distance of co-ordinate on negative side of x axis = x = 3 units

Distance of co-ordinate on positive side of x axis = x = 2 units

D = 3+2

D = 5 units

Therefore, the correct answer is d

**9. The distance between the graphs of the equations y = −1 and y = 3 is**

**(a) 2**

**(b) 4**

**(c) 3**

**(d) 1**

**Solution**

Here, you can see, the distance is 4 units. The distance can be calculated by subtracting y coordinates. 3-(-1) =4. So, (b) is the correct answer.

**10. If the graph of the equation 4x + 3y = 12 cuts the coordinate axes at A and B, then hypotenuse of right triangle AOB is of length**

**(a) 4 units**

**(b) 3 units**

**(c) 5 units**

**(d) none of these**

**Solution**