# Class 9 RD Sharma Solutions – Chapter 13 Linear Equation in Two Variable – Exercise 13.3 | Set 2

**Question 11: Draw the graph of the equation 2x + 3y = 12. From the graph, find the coordinates of the point:**

**(i) whose y-coordinates is 3.**

**(ii) whose x-coordinate is −3.**

**Solution:**

Given:

2x + 3y = 12

We get,

Substituting x = 0 in y = we get,

y = 4

Substituting x = 6 in , we get

y = 4

Hence, we have the following table exhibiting the abscissa and ordinates of points on the line represented by the given equation

x 0 6 y 4 0 By plotting the given equation on the graph, we get the point B (0, 4) and C (6,0).

(i) Co-ordinates of the point whose y axis is 3 are A

(ii) Co-ordinates of the point whose x -coordinate is –3 are D (-3, 6)

**Question 12: Draw the graph of each of the equations given below. Also, find the coordinates of the points where the graph cuts the coordinate axes:**

**(i) 6x − 3y = 12**

**(ii) −x + 4y = 8**

**(iii) 2x + y = 6**

**(iv) 3x + 2y + 6 = 0**

**Solution:**

(i) Given:

6x – 3y = 12

We get,

Now, substituting x = 0 in , we get

y = -4

Substituting x = 2 in , we get

y = 0

Hence, we have the following table exhibiting the abscissa and ordinates of points on the line represented by the given equation

x 0 2 y -4 0 Co-ordinates of the points where graph cuts the co-ordinate axes are at y = -4 axis and x = 2

at x axis.

(ii) Given:

-x + 4y = 8

We get,

Now, substituting x = 0 in , we get

y = 2

Substituting x = -8 in , we get

y = 0

Hence, we have the following table exhibiting the abscissa and ordinates of points on the line represented by the given equation

x 0 -8 y 2 0 Co-ordinates of the points where graph cuts the co-ordinate axes are at y = 2 axis and x = -8

at x axis.

(iii) Given:

2x + y = 6

We get,

y = 6 – 2x

Now, substituting x= 0 in y = 6 – 2x w e get

y = 6

Substituting x = 3 in y = 6 – 2x, we get

y = 0

x 0 3 y 6 0 Co-ordinates of the points where graph cuts the co-ordinate axes are y = 6 at y axis and x = 3

at x axis.

(iv) Given:

3x + 2y + 6 = 0

We get,

Now, substituting x = -2 in , we get

y = -3

Substituting x = -2 in , we get

y = 0

x 0 -2 y -3 0 Co-ordinates of the points where graph cuts the co-ordinate axes are y = -3 at y axis and x = -2

at x axis.

### Question 13: Draw the graph of the equation 2x + y = 6. Shade the region bounded by the graph and the coordinate axes. Also, find the area of the shaded region.

**Solution:**

Given:

2x + y = 6

We get,

y = 6 – 2 x

Now, substituting x = 0 in y = 6 – 2x ,we get

y = 6

Substituting x = 3 in y = 6 – 2x ,we get

x 0 3 y 6 0 The region bounded by the graph is ABC which forms a triangle.

AC at y axis is the base of triangle having AC = 6 units on y axis.

BC at x axis is the height of triangle having BC = 3 units on x axis.

Therefore,

Area of triangle ABC, say A is given by

A = 9 sq. units

**Question 14: Draw the graph of the equation** . **Also, find the area of the triangle formed by the line and the coordinates axes.**

**Solution:**

Given:

4x + 3y = 12

We get,

Now, substituting x = 0 in , we get

y = 4

Substituting x = 3 in , we get

y = 0

x 0 3 y 4 0 The region bounded by the graph is ABC which form a triangle.

AC at y axis is the base of triangle having AC = 4 units on y axis.

BC at x axis is the height of triangle having BC = 3 units on x axis.

Therefore,

Area of triangle ABC, say A is given by

A = 6 sq. units

**Question 15: Draw the graph of y = | x |.**

**Solution:**

We are given,

y = |x|

Substituting, x = 1 we get

y = 1

Substituting, x = -1 we get

y = 1

Substituting,x = 2 we get

y = 2

Substituting, x = -2 we get

y = 2

For every value of x, whether positive or negative, we get y as a positive number.

**Question 16: Draw the graph of y = | x | + 2.**

**Solution:**

Given:

y = |x| + 2

Substituting, x = 0 we get

y = 2

Substituting, x = 1 we get

y = 3

Substituting, x = -1 we get

y = 3

Substituting, x = 2 we get

y = 4

Substituting, x = -2 we get

y = 4

For every value of x, whether positive or negative, we get y as a positive number and the minimum value of y is equal to 2 units.

**Question 17: Draw the graphs of the following linear equations on the same graph paper:**

**2x + 3y = 12, x − y = 1**

**Find the coordinates of the vertices of the triangle formed by the two straight lines and the area bounded by these lines and x-axis.**

**Solution:**

Given:

2x + 3y = 12

We get,

Now, substituting x = 0 in , we get

y = 4

Substituting x = 6 in , we get

y = 0

Therefore, we have the following table exhibiting the abscissa and ordinates of points on the line represented by the given equation

x 0 6 y 4 0 Plotting A(0,4) and E(6,0) on the graph and by joining the points , we obtain the graph of equation .

Given:

x – y = 1

We get,

y = x − 1

Now, substituting x = 0 in y = x − 1,we get

y = −1

Substituting x = -1 in y = x − 1,we get

y = −2

Therefore, we have the following table exhibiting the abscissa and ordinates of points on the line represented by the given equation

x 0 -1 y -1 -2 Plotting D(0,1) and E(-1,0) on the graph and by joining the points , we obtain the graph of equation .

By the intersection of lines formed by 2x + 3y = 12 and x – y = 1 on the graph, triangle ABC is formed on y axis.

Therefore,

AC at y axis is the base of triangle ABC having AC = 5 units on y axis.

Draw FE perpendicular from F on y axis.

FE parallel to x axis is the height of triangle ABC having FE = 3 units on x axis.

Therefore,

Area of triangle ABC, say A is given by

**Question 18: Draw the graphs of the linear equations 4x − 3y + 4 = 0 and 4x + 3y −20 = 0. Find the area bounded by these lines and x-axis.**

**Solution:**

Given:

4x – 3y + 4 = 0

We get,

Now, substituting x = 0 in , we get

Substituting x = -1 in , we get

y = 0

x 0 -1 y 0 Plotting E(0, ) and A(-1,0) on the graph and by joining the points, we obtain the graph of equation.

We are given,

4x + 3y – 20 = 0

We get,

Now, substituting x = 0 in , we get

Substituting x = 5 in , we get

y = 0

Therefore, we have the following table exhibiting the abscissa and ordinates of points on the line represented by the given equation

x 0 5 y 0 Plotting D(0, ) and B(5,0) on the graph and by joining the points, we obtain the graph of equation.

By the intersection of lines formed by 4x – 3y + 4 = 0 and 4x + 3y – 20 = 0 on the graph, triangle ABC is formed on x axis.

Therefore,

AB at x axis is the base of triangle ABC having AB = 6 units on x axis.

Draw CF perpendicular from C on x axis.

CF parallel to y axis is the height of triangle ABC having CF = 4 units on y axis.

Therefore,

Area of triangle ABC, say A is given by

**Question 19: The path of a train A is given by the equation 3x + 4y − 12 = 0 and the path of another train B is given by the equation 6x + 8y − 48 = 0. Represent this situation graphically.**

**Solution:**

Given: Path of train A,

3x + 4y – 12 = 0

We get,

Now, substituting x = 0 in ,we get

y = 3

Substituting x = 4 in ,we get

y = 0

x 0 4 y 3 0 Plotting A(4,0) and E(0,3) on the graph and by joining the points , we obtain the graph of equation .

We are given the path of train B,

6x + 8y – 48 = 0

We get,

Now, substituting x = 0 in ,we get

y = 6

Substituting x = 8 in ,we get

y = 0

x 0 8 y 6 0 Plotting C(0,6) and D(8,0) on the graph and by joining the points , we obtain the graph of equation.

**Question 20: Ravish tells his daughter Aarushi, ”Seven years ago, I was seven times as old as you were then. Also, three years from now, I shall be three times as old as you will be”. If present ages of Aarushi and Ravish are x and y years respectively, represent this situation algebraically as well as graphically.**

**Solution:**

We are given the present age of Ravish as y years and Aarushi as x years.

Age of Ravish seven years ago = y – 7

Age of Aarushi seven years ago = x – 7

It has already been said by Ravish that seven years ago he was seven times old then Aarushi was then

So,

y – 7 = 7(x – 7)

y – 7 = 7x – 49

7x – y = -7 + 49

7x – y – 42 = 0

Age of Ravish three years from now = y + 3

Age of Aarushi three years from now = x + 3

It has already been said by Ravish that three years from now he will be three times old then Aarushi will be then

So,

y + 3 = 3(x + 3)

y + 3 = 3x + 9

3x + 9 – y – 3 = 0

3x – y + 6 = 0

(1) and (2) are the algebraic representation of the given statement.

Given:

7x – y – 42 = 0

We get,

y = 7x – 42

Now, substituting x = 0 in y = 7x – 42 ,we get

y = -42

Substituting x = 6 in y = 7x – 42, we get

y = 0

x 0 6 y -42 0 Given:

3x – y + 6 = 0

We get,

y = 3x + 6

Now, substituting x = 0 in y = 3x + 6 ,we get

y = 6

Substituting x = -2 in y = 3x + 6 ,we get

y = 0

x 0 -2 y 6 0 The red -line represents the equation. 7x – y – 42 = 0

The blue-line represents the equation. 3x – y + 6 = 0

**Question 21: Aarushi was driving a car with uniform speed of 60km/h. Draw distance-time graph. From the graph, find the distance traveled by Aarushi in**

**(i)** **Hours**

**(ii)** ** Hour**

**Solution:**

Aarushi is driving the car with the uniform speed of 60 km/h.

We represent time on X-axis and distance on Y-axis

Now, graphically

We are given that the car is travelling with a uniform speed 60 km/hr. This means car travels 60 km distance each hour. Thus the graph we get is of a straight line.

Also, we know when the car is at rest, the distance traveled is 0 km, speed is 0 km/hr and the time is also 0 hr.

Therefore, the given straight line will pass through O(0,0) and M(1,60).

Join the points O and M and extend the line in both directions.

Now, we draw a dotted line parallel to y-axis from x = that meets the straight line graph at L from which we draw a line parallel to x-axis that crosses y-axis at 30. Thus, in hr, distance traveled by the car is 30 km.

Now, we draw a dotted line parallel to y-axis from x = that meets the straight line graph at N from which we draw a line parallel to x-axis that crosses y-axis at 150. Thus, in hr, distance traveled by the car is 150 km.

(i) Distance = Speed × Time

Distance traveled in hours is given by

Distance =

Distance =

Distance = 150 km

(ii) Distance = Speed × Time

Distance traveled in hours is given by

Distance =

Distance = 30 km