The document RD Sharma Solutions Ex-13.3, (Part -3), Linear Equation In Two Variables, Class 9, Maths Class 9 Notes | EduRev is a part of the Class 9 Course RD Sharma Solutions for Class 9 Mathematics.

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

**Ans.** 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.

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

**Ans.** We are given,

Y = |x|+2

Substituting x = 0 we get y=2

Substituting x = I , 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.

**Q 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 y-axis. Also, find the area of the triangle.**

**Ans.** We are given,

2x +3y = 12

We get,

Now, substituting x = 0 in y , we get

Substituting x = 6 in we get

y = 0

Thus, 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

2x+3y = 12.

We are given,

x – y = 1

We get, y = x – 1

Now, substituting x = 0 in y = x-1,

we get y = -1

Substituting x in y = x-1,

we get y = -2

Thus, 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,) and E(-1,0) on the graph and by joining the points , we obtain the graph of equation x— y = I .

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 A = (Base x Height)/2 =(AC x FE)/2 = (5×3)/2 ⇒15/2 = 7.5 sq. units

**Q 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.**

**Ans.** We are given, 4x – 3y + 4 = 0

We get,

Now, substituting x = 0 in , we get

Substituting x = -I in

we get y =0

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

x | 0 | -1 |

y | 4/3 | 0 |

Plotting E(0, 4/3) and A (-1, 0) on the graph and by joining the points, we obtain the graph of equation

4x – 3y + 4 = 0.

We are given, 4x + 3y – 20 = 0

We get,

Now, substituting x = 0 in , we get

y = 7

Substituting x = 5 in ,we get

y = 0

X | 0 | 5 |

Y | 20/3 | 0 |

Plotting D (0,20/3) and B(5,0) on the graph and by joining the points , we obtain the graph of equation 4x +3y – 20 = 0.

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

A = (Base x Height)/2

A = (AB x CF)/2

A = (6 x 4)/2

k = 12 sq. units

**Q19 : 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.**

**Ans.** We are given the 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 3x+4y-12 = 0.

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 6x+8y-48 = 0

**Q 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.**

**Ans.** 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 = 7 x – 49

7x – y = – 7 + 49

7x – y – 42 = 0 ——(1)

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 older then Aarushi will be then So,

Y + 3 = 3 (x + 3)

y + 3 = 3x + 9

3x + 9 – y – 3 = 0

3x – y + 6 = 0 —–(2)

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

We are 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 |

We are 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.

**Q21: Aarushi was driving a car with uniform speed of 60 km/h. Draw distance-time graph From the graph, find the distance travelled by Aarushi in **

**(i) 212 Hours (ii) 12 Hour**

**Ans.** 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 travelled is 0 km, speed is 0 km/hr and the time is also 0 hr. Thus, the given straight line will pass through O (0 , 0) and M (1, 60).

Join the points 0 and M and extend the line in both directions.

Now, we draw a dotted line parallel to y-axis from x = 12 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 12hr, distance travelled by the car is 30 km.

Now, we draw a dotted line parallel to y-axis from x = 212 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 212hr, distance travelled by the car is 150 km.

(i) Distance = Speed x Time Distance travelled in hours is given by

Distance=60 x

Distance = 60 x 5/2

Distance = 150 Km

(ii) Distance = Speed x Time Distance travelled in 12 hour is given by

Distance=60 x

Distance = 30 km