Graphical Solution — Exercise 24(B)

The point of intersection of lines x = 8 and y - 8 = 0 is :

1. (8, 8)

2. (8, -8)

3. (-8, 8)

4. (-8, -8)

Answer

The lines x = 8 and y = 8 intersect at point (8, 8).

Hence, option 1 is the correct option.

For line y = 20 + 2x, if x = 20, the value of y is :

1. 40

2. 60

3. 80

4. 0

Answer

Given equation: y = 20 + 2x

When x = 20

y = 20 + 2 $\times$ 20

= 20 + 40

= 60

Hence, option 2 is the correct option.

A point that lies on the line 2x - 3y = 4 can be taken as :

1. (-5, -2)

2. (-5, 2)

3. (5, 2)

4. (5, -2)

Answer

Given equation: 2x - 3y = 4

1. For (-5, -2):

Substitute x = -5 and y = -2

L.H.S. : 2 $\times$ (-5) - 3 $\times$ (-2)

= -10 + 6

= -4

R.H.S. = 4

Since, L.H.S. ≠ R.H.S., (-5, -2) does not lie on the line.

2. For (-5, 2):

Substitute x = -5 and y = 2

L.H.S. : 2 $\times$ (-5) - 3 $\times$ 2

= -10 - 6

= - 16

R.H.S. = 4

Since, L.H.S. ≠ R.H.S., (-5, 2) does not lie on the line.

3. For (5, 2):

Substitute x = 5 and y = 2

L.H.S. : 2 $\times$ 5 - 3 $\times$ 2

= 10 - 6

= 4

R.H.S. = 4

Since, L.H.S. = R.H.S., (5, 2) lies on the line.

4. For (5, -2):

Substitute x = 5 and y = -2

L.H.S. : 2 $\times$ 5 - 3 $\times$ (-2)

= 10 + 6

= 16

R.H.S. = 4

Since, L.H.S. ≠ R.H.S., (5, -2) does not lie on the line.

Hence, option 3 is the correct option.

The line x + 5 = 0 :

1. is parallel to x-axis

2. is parallel to y-axis

3. passes through (0, 0)

4. passes through (5, 0)

Answer

x + 5 = 0

⇒ x = - 5

This equation represents a vertical line that is parallel to the y-axis, where the x-coordinate is always −5.

Hence, option 2 is the correct option.

The lines y - 2 = 0 and 2x + 3y = 12 cut each other at point :

1. (-3, 2)

2. (3, -2)

3. (-3, -2)

4. (3, 2)

Answer

y - 2 = 0

⇒ y = 2

And, 2x + 3y = 12

⇒ 2x + 3 x 2 = 12

⇒ 2x + 6 = 12

⇒ 2x = 12 - 6

⇒ 2x = 6

⇒ x = $\dfrac{6}{2}$

⇒ x = 3

Hence, the lines y - 2 = 0 and 2x + 3y = 12 cut each other at point (3, 2).

Hence, option 4 is the correct option.

Solve, graphically, the following pairs of equations :

(i)

x - 5 = 0
y + 4 = 0

(ii)

2x + y = 23
4x - y = 19

(iii)

3x + 7y = 27

$8 - y = \dfrac{5}{2}x$

(iv)

$\dfrac{x + 1}{4} = \dfrac{2}{3}(1 - 2y)$

$\dfrac{2 + 5y}{3} = \dfrac{x}{7} - 2$

Answer

(i) x - 5 = 0

⇒ x = 5

and, y + 4 = 0

⇒ y = - 4

On the same graph paper, draw the graph for each given equation.

Both the straight lines drawn meet at point P. As it is clear from the graph, co-ordinates of the common point are (5, -4).

Solution of the given equations is : x = 5 and y = -4.

(ii)

First equation : 2x + y = 23

Step 1:

Give at least three suitable values to the variable x and find the corresponding values of y.

Let x = 2, then 2 $\times$ 2 + y = 23 ⇒ y = 19

Let x = 4, then 2 $\times$ 4 + y = 23 ⇒ y = 15

Let x = 6, then 2 $\times$ 6 + y = 23 ⇒ y = 11

Step 2:

Make a table (as given below) for the different pairs of the values of x and y:

Step 3:

Plot the points, from the table, on a graph paper and then draw a straight line passing through the points plotted on the graph.

Second equation : 4x - y = 19

Step 1:

Give at least three suitable values to the variable x and find the corresponding values of y.

Let x = 4, then 4 $\times$ 4 - y = 19 ⇒ y = -3

Let x = 6, then 4 $\times$ 6 - y = 19 ⇒ y = 5

Let x = 8, then 4 $\times$ 8 - y = 19 ⇒ y = 13

Step 2:

Make a table (as given below) for the different pairs of the values of x and y:

Step 3:

Plot the points, from the table, on a graph paper and then draw a straight line passing through the points plotted on the graph.

On the same graph paper, draw the graph for each given equation.

Both the straight lines drawn meet at point P. As it is clear from the graph, co-ordinates of the common point are (7, 9).

Solution of the given equations is : x = 7 and y = 9.

(iii)

First equation : 3x + 7y = 27

Step 1:

Give at least three suitable values to the variable x and find the corresponding values of y.

Let x = 0, then 3 $\times$ 0 + 7y = 27 ⇒ y = 3.8

Let x = 2, then 3 $\times$ 2 + 7y = 27 ⇒ y = 3

Let x = 4, then 3 $\times$ 4 + 7y = 27 ⇒ y = 2.1

Step 2:

Make a table (as given below) for the different pairs of the values of x and y:

Step 3:

Plot the points, from the table, on a graph paper and then draw a straight line passing through the points plotted on the graph.

Second equation : $8 - y = \dfrac{5}{2}x$

Step 1:

Give at least three suitable values to the variable x and find the corresponding values of y.

Let x = 0, then $8 - y = \dfrac{5}{2} \times 0$ ⇒ y = 8

Let x = 1, then $8 - y = \dfrac{5}{2} \times 1$ ⇒ y = 5.5

Let x = 2, then $8 - y = \dfrac{5}{2} \times 2$ ⇒ y = 3

Step 2:

Make a table (as given below) for the different pairs of the values of x and y:

Step 3:

Plot the points, from the table, on a graph paper and then draw a straight line passing through the points plotted on the graph.

On the same graph paper, draw the graph for each given equation.

Both the straight lines drawn meet at point P. As it is clear from the graph, co-ordinates of the common point are (2, 3).

Solution of the given equations is : x = 2 and y = 3.

(iv)

First equation : $\dfrac{x + 1}{4} = \dfrac{2}{3}(1 - 2y)$

Step 1:

Give at least three suitable values to the variable x and find the corresponding values of y.

Let x = 0, then $\dfrac{0 + 1}{4} = \dfrac{2}{3}(1 - 2y)$ ⇒ y = 0.3

Let x = 3, then $\dfrac{3 + 1}{4} = \dfrac{2}{3}(1 - 2y)$ ⇒ y = -0.2

Let x = 7, then $\dfrac{7 + 1}{4} = \dfrac{2}{3}(1 - 2y)$ ⇒ y = -1

Step 2:

Make a table (as given below) for the different pairs of the values of x and y:

Step 3:

Plot the points, from the table, on a graph paper and then draw a straight line passing through the points plotted on the graph.

Second equation : $\dfrac{2 + 5y}{3} = \dfrac{x}{7} - 2$

Step 1:

Give at least three suitable values to the variable x and find the corresponding values of y.

Let x = 0, then $\dfrac{2 + 5y}{3} = \dfrac{0}{7} - 2$ ⇒ y = -1.6

Let x = 7, then $\dfrac{2 + 5y}{3} = \dfrac{7}{7} - 2$ ⇒ y = -1

Let x = 14, then $\dfrac{2 + 5y}{3} = \dfrac{14}{7} - 2$ ⇒ y = -0.4

Step 2:

Make a table (as given below) for the different pairs of the values of x and y:

Step 3:

Plot the points, from the table, on a graph paper and then draw a straight line passing through the points plotted on the graph.

On the same graph paper, draw the graph for each given equation.

Both the straight lines drawn meet at point P. As it is clear from the graph, co-ordinates of the common point are (7, -1).

Solution of the given equations is : x = 7 and y = -1.

Solve graphically the simultaneous equations given below. Take the scale as 2 cm = 1 unit on both the axes.

x - 2y - 4 = 0
2x + y = 3

Answer

First equation: x - 2y - 4 = 0

Step 1:

Give at least three suitable values to the variable x and find the corresponding values of y.

Let x = 0, then 0 - 2y - 4 = 0 ⇒ y = -2

Let x = 2, then 2 - 2y - 4 = 0 ⇒ y = -1

Let x = 4, then 4 - 2y - 4 = 0 ⇒ y = 0

Step 2:

Make a table (as given below) for the different pairs of the values of x and y:

Step 3:

Plot the points, from the table, on a graph paper and then draw a straight line passing through the points plotted on the graph.

Second equation: 2x + y = 3

Step 1:

Give at least three suitable values to the variable x and find the corresponding values of y.

Let x = 0, then 2 $\times$ 0 + y = 3 ⇒ y = 3

Let x = 2, then 2 $\times$ 2 + y = 3 ⇒ y = -1

Let x = 4, then 2 $\times$ 4 + y = 3 ⇒ y = -5

Step 2:

Make a table (as given below) for the different pairs of the values of x and y:

Step 3:

Plot the points, from the table, on a graph paper and then draw a straight line passing through the points plotted on the graph.

On the same graph paper, draw the graph for each given equation.

Both the straight line drawn meet the point P. As it is clear from the graph, co-ordinates of the common point P are (2, -1).

Solution of the given equation x = 2 and y = -1.

Use graph paper for this question. Draw the graph of 2x - y - 1 = 0 and 2x + y = 9 on the same axes. Use 2 cm = 1 unit on both axes and plot only 3 points per line.

Write down the co-ordinates of the point of intersection of the two lines.

Answer

First equation: 2x - y - 1 = 0

Step 1:

Give at least three suitable values to the variable x and find the corresponding values of y.

Let x = 0, then 2 $\times$ 0 - y - 1 = 0 ⇒ y = -1

Let x = 2, then 2 $\times$ 2 - y - 1 = 0 ⇒ y = 3

Let x = 4, then 2 $\times$ 4 - y - 1 = 0 ⇒ y = 7

Step 2:

Make a table (as given below) for the different pairs of the values of x and y:

Step 3:

Plot the points, from the table, on a graph paper and then draw a straight line passing through the points plotted on the graph.

Second equation: 2x + y = 9

Step 1:

Give at least three suitable values to the variable x and find the corresponding values of y.

Let x = 0, then 2 $\times$ 0 + y = 9 ⇒ y = 9

Let x = 2, then 2 $\times$ 2 + y = 9 ⇒ y = 5

Let x = 4, then 2 $\times$ 4 + y = 9 ⇒ y = 1

Step 2:

Make a table (as given below) for the different pairs of the values of x and y:

Step 3:

Plot the points, from the table, on a graph paper and then draw a straight line passing through the points plotted on the graph.

On the same graph paper, draw the graph for each given equation.

Both the straight line drawn meet the point P. As it is clear from the graph, co - ordinate of the common point P are (2.5, 4).

Solution of the given equation x = 2.5 and y = 4.

Hence, point of intersection = (2.5, 4).

Use graph paper for this question. Take 2 cm = 2 units on x-axis and 2 cm = 1 unit on y-axis.

Solve graphically the following equations :

3x + 5y = 12; 3x - 5y + 18 = 0

(Plot only three points per line)

Answer

First equation: 3x + 5y = 12

Step 1:

Give at least three suitable values to the variable x and find the corresponding values of y.

Let x = 0, then 3 $\times$ 0 + 5y = 12 ⇒ y = 2.4

Let x = 2, then 3 $\times$ 2 + 5y = 12 ⇒ y = 1.2

Let x = 4, then 3 $\times$ 4 + 5y = 12 ⇒ y = 0

Step 2:

Make a table (as given below) for the different pairs of the values of x and y:

Step 3:

Plot the points, from the table, on a graph paper and then draw a straight line passing through the points plotted on the graph.

Second equation: 3x - 5y + 18 = 0

Step 1:

Give at least three suitable values to the variable x and find the corresponding values of y.

Let x = -2, then 3 $\times$ (-2) - 5y + 18 = 0 ⇒ y = 2.4

Let x = 0, then 3 $\times$ 0 - 5y + 18 = 0 ⇒ y = 3.6

Let x = 2, then 3 $\times$ 2 - 5y + 18 = 0 ⇒ y = 4.8

Step 2:

Make a table (as given below) for the different pairs of the values of x and y:

Step 3:

Plot the points, from the table, on a graph paper and then draw a straight line passing through the points plotted on the graph.

On the same graph paper, draw the graph for each given equation.

Both the straight line drawn meet the point P. As it is clear from the graph, co-ordinates of the common point P are (-1, 3).

Solution of the given equation x = -1 and y = 3.

Use graph paper for this question. Take 2 cm = 1 unit on both the axes.

(i) Draw the graphs of x + y + 3 = 0 and 3x - 2y + 4 = 0. Plot only three points per line.

(ii) Write down the co-ordinates of the point of intersection of the lines.

(iii) Measure and record the distance of the point of intersection of the lines from the origin in cm.

Answer

(i)

First equation: x + y + 3 = 0

Step 1:

Give three suitable values to the variable x and find the corresponding values of y.

Let x = 0, then 0 + y + 3 = 0 ⇒ y = -3

Let x = -4, then -4 + y + 3 = 0 ⇒ y = 1

Let x = -6, then -6 + y + 3 = 0 ⇒ y = 3

Step 2:

Make a table (as given below) for the different pairs of the values of x and y:

Step 3:

Plot the points, from the table, on a graph paper and then draw a straight line passing through the points plotted on the graph.

Second equation: 3x - 2y + 4 = 0

Step 1:

Give three suitable values to the variable x and find the corresponding values of y.

Let x = 0, then 3 $\times$ 0 - 2y + 4 = 0 ⇒ y = 2

Let x = 2, then 3 $\times$ 2 - 2y + 4 = 0 ⇒ y = 5

Let x = 4, then 3 $\times$ 4 - 2y + 4 = 0 ⇒ y = 8

Step 2:

Make a table (as given below) for the different pairs of the values of x and y:

Step 3:

Plot the points, from the table, on a graph paper and then draw a straight line passing through the points plotted on the graph.

(ii) Both the straight line drawn meet at the point A. As it is clear from the graph, co-ordinates of the common point A are (-2, -1).

Hence, co-ordinates of the point of intersection of the lines are (-2, -1).

(iii) In triangle OAB,

Using pythagoras theorem,

OA² = AB² + OB²

= 2² + 1²

= 4 + 1

= 5

OA = $\sqrt5$

OA = 2.2 cm

Hence, the distance of the point of intersection of the lines from the origin is 2.2 cm.

The sides of a triangle are given by the equations y - 2 = 0; y + 1 = 3 (x - 2) and x + 2y = 0.

Find, graphically :

(i) the area of triangle;

(ii) the co-ordinates of the vertices of the triangle.

Answer

(i)

First equation: y - 2 = 0

y = 2

Second equation: y + 1 = 3 (x - 2)

Step 1:

Give at least three suitable values to the variable x and find the corresponding values of y.

Let x = 0, then y + 1 = 3 (0 - 2) ⇒ y = -7

Let x = 2, then y + 1 = 3 (2 - 2) ⇒ y = -1

Let x = 4, then y + 1 = 3 (4 - 2) ⇒ y = 5

Step 2:

Make a table (as given below) for the different pairs of the values of x and y:

Step 3:

Plot the points, from the table, on a graph paper and then draw a straight line passing through the points plotted on the graph.

Third equation: x + 2y = 0

Step 1:

Give at least three suitable values to the variable x and find the corresponding values of y.

Let x = 0, then 0 + 2y = 0 ⇒ y = 0

Let x = 2, then 0 + 2y = 0 ⇒ y = -1

Let x = 4, then 4 + 2y = 0 ⇒ y = -2

Step 2:

Make a table (as given below) for the different pairs of the values of x and y:

Step 3:

Plot the points, from the table, on a graph paper and then draw a straight line passing through the points plotted on the graph.

(i) The area of the triangle formed by the lines = $\dfrac{1}{2}$ x base x height

= $\dfrac{1}{2}$ x AB x CD

= $\dfrac{1}{2}$ x 7 x 3

= $\dfrac{21}{2}$

= 10.5 sq. units

Hence, area of triangle = 10.5 sq. units.

(ii) The co-ordinates of A = (-4, 2)

The co-ordinates of B = (3, 2)

The co-ordinates of C = (2, -1)

Hence, the co-ordinates of the vertices of the triangle are (-4, 2), (3, 2) and (2, -1).

By drawing a graph for each of the equations 3x + y + 5 = 0; 3y - x = 5 and 2x + 5y = 1 on the same graph paper; show that the lines given by these equations are concurrent (i.e. they pass through the same point).

Take 2 cm = 1 unit on both the axes.

Answer

First equation: 3x + y + 5 = 0

Step 1:

Give at least three suitable values to the variable x and find the corresponding values of y.

Let x = -3, then 3 $\times$ (-3) + y + 5 = 0 ⇒ y = 4

Let x = -2, then 3 $\times$ (-2) + y + 5 = 0 ⇒ y = 1

Let x = 1, then 3 $\times$ 1 + y + 5 = 0 ⇒ y = -8

Step 2:

Make a table (as given below) for the different pairs of the values of x and y:

Step 3:

Plot the points, from the table, on a graph paper and then draw a straight line passing through the points plotted on the graph.

Second equation: 3y - x = 5

Step 1:

Give at least three suitable values to the variable x and find the corresponding values of y.

Let x = -2, then 3y - (-2) = 5 ⇒ y = 1

Let x = 1, then 3y - 1 = 5 ⇒ y = 2

Let x = 7, then 3y - 7 = 5 ⇒ y = 4

Step 2:

Make a table (as given below) for the different pairs of the values of x and y:

Step 3:

Plot the points, from the table, on a graph paper and then draw a straight line passing through the points plotted on the graph.

Third equation: 2x + 5y = 1

Step 1:

Give at least three suitable values to the variable x and find the corresponding values of y.

Let x = -7, then 2 $\times$ (-7) + 5y = 1 ⇒ y = 3

Let x = -2, then 2 $\times$ (-2) + 5y = 1 ⇒ y = 1

Let x = 3, then 2 $\times$ 3 + 5y = 1 ⇒ y = -1

Step 2:

Make a table (as given below) for the different pairs of the values of x and y:

Step 3:

Plot the points, from the table, on a graph paper and then draw a straight line passing through the points plotted on the graph. 3

From the graph, it is clear that all three lines intersect at a common point (-2, 1), confirming that the lines are concurrent.

Using a scale of 1 cm to 1 unit for both the axes, draw the graphs of the following equations : 6y = 5x + 10, y = 5x - 15.

From the graph find :

(i) the co-ordinates of the point where the two lines intersect;

(ii) the area of the triangle between the lines and the x-axis.

Answer

First equation: 6y = 5x + 10

Step 1:

Give at least three suitable values to the variable x and find the corresponding values of y.

Let x = -8, then 6y = 5 $\times$ (-8) + 10 ⇒ y = -5

Let x = -2, then 6y = 5 $\times$ (-2) + 10 ⇒ y = 0

Let x = 4, then 6y = 5 $\times$ 4 + 10 ⇒ y = 5

Step 2:

Make a table (as given below) for the different pairs of the values of x and y:

Step 3:

Plot the points, from the table, on a graph paper and then draw a straight line passing through the points plotted on the graph.

Second equation: y = 5x - 15

Step 1:

Give at least three suitable values to the variable x and find the corresponding values of y.

Let x = 3, then y = 5 $\times$ 3 - 15 ⇒ y = 0

Let x = 4, then y = 5 $\times$ 4 - 15 ⇒ y = 5

Let x = 5, then y = 5 $\times$ 5 - 15 ⇒ y = 10

Step 2:

Make a table (as given below) for the different pairs of the values of x and y:

Step 3:

Plot the points, from the table, on a graph paper and then draw a straight line passing through the points plotted on the graph.

(i) Both the straight line drawn meet at the point A. As it is clear from the graph, co-ordinates of the common point A are (4, 5).

Hence, the co-ordinates of the point where the two lines intersect = (4, 5).

(ii) The area of the triangle = $\dfrac{1}{2}$ x base x height

= $\dfrac{1}{2}$ x BC x AD

= $\dfrac{1}{2}$ x 5 x 5

= $\dfrac{25}{2}$ sq. units

= 12.5 sq. units

Hence, the area of the triangle between the lines and the x-axis = 12.5 sq. units.

The cost of manufacturing x articles is ₹ (50 + 3x). The selling price of x articles is ₹ 4x.

On a graph sheet, with the same axes, and taking suitable scales draw two graphs, first for the cost of manufacturing against no. of articles and the second for the selling price against number of articles.

Use your graph to determine :

(i) No. of articles to be manufactured and sold to breakeven (no profit and no loss),

(ii) The profit or loss made when

(a) 30 (b) 60 articles are manufactured and sold.

Answer

Given:

The cost of manufacturing x articles = ₹ (50 + 3x).

C.P. = ₹ (50 + 3x)

Step 1:

Give at least three suitable values to the variable x and find the corresponding values of C.P.

Let x = 0, then C.P. = ₹ (50 + 3 $\times$ 0) ⇒ C.P. = ₹ 50

Let x = 20, then C.P. = ₹ (50 + 3 $\times$ 20) ⇒ C.P. = ₹ 110

Let x = 40, then C.P. = ₹ (50 + 3 $\times$ 40) ⇒ C.P. = ₹ 170

Step 2:

Make a table (as given below) for the different pairs of the values of x and y:

Step 3:

Plot the points, from the table, on a graph paper and then draw a straight line passing through the points plotted on the graph.

And, the selling price of x articles is ₹ 4x

S.P. = ₹ 4x

Step 1:

Give at least three suitable values to the variable x and find the corresponding values of S.P.

Let x = 0, then S.P. = ₹ 4 $\times$ 0 ⇒ S.P. = ₹ 0

Let x = 20, then S.P. = ₹ 4 $\times$ 20 ⇒ S.P. = ₹ 80

Let x = 40, then S.P. = ₹ 4 $\times$ 40 ⇒ S.P. = ₹ 160

Step 2:

Make a table (as given below) for the different pairs of the values of x and y:

Step 3:

Plot the points, from the table, on a graph paper and then draw a straight line passing through the points plotted on the graph.

(i) The above figure shows the graphs of C.P. and S.P. Since the two straight lines meet at x = 50, it shows that the C.P. of 50 articles is the same as their selling price.

Hence, no. of articles to be manufactured and sold to breakeven point (no profit and no loss) = 50.

(ii)

(a) Draw the vertical line through x = 30, which meets graph for C.P. at ₹ 140 and graph for S.P. at ₹ 120.

C.P. > S.P.

Therefore, loss = C.P. - S.P.

= ₹ 140 - ₹ 120

= ₹ 20

Hence, the loss = ₹ 20.

(b) Draw the vertical line through x = 60, which meets graph for C.P. at ₹ 230 and graph for S.P. at ₹ 240.

C.P. = ₹ 230 and S.P. = ₹ 240

S.P. > C.P.

Therefore, profit = S.P. - C.P.

= ₹ 240 - ₹ 230

= ₹ 10

Hence, the profit = ₹ 10.