RD Sharma Solutions Ex-13.3, (Part -2), Linear Equation In Two Variables, Class 9, Maths | RD Sharma Solutions for Class 9 Mathematics PDF Download

Q8: From the choices given below, choose the equations whose graph is given in fig

(i) y = x
 (ii) x + y = 0
 (iii) y = 2x
 (iv) 2 + 3y = 7x

Ans: We are given co-ordinates (1, – 1) and (-1, 1) as the solution of one of the following equations.

We will substitute the value of both co-ordinates in each of the equation and find the equation which satisfies the given co-ordinates.

(i) We are given, y = x

Substituting x =I and y = -1 ,

we get; 1 ≠ -1

L.H.S ≠ R.H.S

Substituting x = -1  and y = 1 ,

we get; -1 ≠1

L.H.S ≠R.H.S

Therefore, the given equation y = x does not represent the graph in the figure.

(ii) We are given,

x + y = 0

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

⇒1 + (-1) = 0

⇒ 0 = 0

L.H.S = R.H.S

Substituting x = —1 and y = 1 ,we get

(-1)+ 1 = 0

0 = 0

L.H.S = R.H.S

Therefore, the given solutions satisfy this equation.

Thus, it is the equation whose graph is given.

Q9: From the choices given below, choose the equation whose graph is given fig:

(i) y = x + 2
 (ii) y = x – 2
 (iii)y = – x + 2
 (iv) x + 2y = 6

Ans. We are given co-ordinates (-1, 3) and (2, 0) as the solution of one of the following equations.

We will substitute the value of both co-ordinates in each of the equation and find the equation which satisfies the given co-ordinates.

(i) We are given, y = x+2

Substituting x = – 1 and y = 3 ,we get

3 ≠ – 1 + 2

L.H.S ≠ R.H.S

Substituting x = 2 and y = 0 ,we get

0 ≠ 4

L.H.S ≠R.H.S

Therefore, the given solution does not satisfy this equation.

(ii) We are given, y = x – 2

Substituting x = —1 and y = 3 ,we get

3 = – 1 – 2

L.H.S ≠ R.H.S

Substituting x = 2 and y = 0 ,we get

0 = 0

L.H.S = R.H.S

Therefore, the given solutions does not completely satisfy this equation.

(iii) We are given, y = – x + 2

Substituting x = – 1 and y = 3,we get

3 = – (– 1) + 2

L.H.S = R.H.S

Substituting x = 2 and y = 0 ,we get

0 = -2 + 2

0 = 0

L.H.S = R.H.S

Therefore, the given solutions satisfy this equation.

Thus, it is the equation whose graph is given.

Q 10 : If the point (2, -2) lies on the graph of linear equation, 5x + 4y = 4, find the value of k.

Ans. It is given that the point (2,-2) lies on the given equation,

5x + ky = 4

Clearly, the given point is the solution of the given equation.

Now, Substituting x = 2 and y = – 2 in the given equation, we get 5x + ky = 4

5 x 2 + (– 2) k = 4

2k = 10 – 4

2k = 6

k = 6/2

k = 3

Q 11 : Draw the graph of equation 2x + 3y = 12. From the graph, find the co ordinates of the point:

(i) whose y-coordinate is 3
 (ii) whose x coordinate is -3

Ans. We are given,

2x +3y =12

Substituting, x = 0 in 

y = 4

Substituting x = 6 in 

y = 0

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

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 (3/2, 3)

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

Q 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

Ans. (i) We are given,

6x – 3y = 12 We get,

y = (6x —12) /3

Now, substituting x = 0 in y = – (6x – 12)/3 we get

y =- 4

Substituting x = 2 in y = (- 6x —12)/3, 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

Co-ordinates of the points where graph cuts the co-ordinate axes are y = – 4 at y axis and x = 2 at x axis. (ii) We are given,

– x + 4y = 8

We get,

Now, substituting x = 0 in   we get

y = 2

Substituting x = -8 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

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

(iii) We are given,

2x + y = 6

We get, y = 6 – 2x

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

y = 6

Substituting x = 3 in y = 6-2x, 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

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) We are given,

3x+2y+6 = 0

We get,

Now, substituting x = 0 in 

y= – 3

Substituting x = —2 in 

y = 0

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

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

Q 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.

Ans. We are given,

2x + y = 6

We get,

y = 6 – 2x

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

we get y = 6

Substituting x =3 in y = 6— 2x,

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

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 = (Base x Height)/2

A = (AC x BC)/2

A = (6 x 3)/2

A = 9 sq. units

Q 14 : Draw the graph of the equation Also, find the area of the triangle formed by 3 4 the line and the coordinates axes.

Ans. We are given.

4x +3y = 12

We get,

Now, substituting x = 0 in  ,we get 

y = 4

Substituting x = 3 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

The region bounded by the graph is ABC which forms 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 = (Base x Height)/2

A= (AC x BC)/2

A = (4 x 3)/2

A = 6 sq. units