Graphical Solution — Exercise 24(A)

Point (k, 2) lies on the line x - 4y = 2; the value of k is :

1. 10

2. -6

3. -10

4. 6

Answer

Given that the point (k, 2) lies on the line x - 4y = 2, substitute x = k and y = 2 into the equation:

⇒ k - 4 $\times$ 2 = 2

⇒ k - 8 = 2

⇒ k = 2 + 8

⇒ k = 10

Hence, option 1 is the correct option.

The line y = mx - 8 passes through the point (5, 2); the value of m is :

1. 1

2. 2

3. -2

4. -1

Answer

Since the line y = mx - 8 passes through the point (5, 2), substitute x = 5 and y = 2 into the equation:

⇒ 2 = m $\times$ 5 - 8

⇒ 2 = 5m - 8

⇒ 5m = 2 + 8

⇒ 5m = 10

⇒ m = $\dfrac{10}{5}$

⇒ m = 2

Hence, option 2 is the correct option.

The line 5x - 2y - 10 = 0 intersects x-axis at point P. The co-ordinates of point P are:

1. (0, 2)

2. (0, -2)

3. (2, 0)

4. (-2, 0)

Answer

It is given that the line 5x - 2y - 10 = 0 intersects x-axis at point P.

At the x-axis, the y-coordinate is 0. Let P = (a, 0), where x = a and y = 0.

⇒ 5 $\times$ a - 2 $\times$ 0 - 10 = 0

⇒ 5a - 0 - 10 = 0

⇒ 5a = 10

⇒ a = $\dfrac{10}{5}$

⇒ a = 2

The coordinates of point P are (2, 0).

Hence, option 3 is the correct option.

The line 5x - 4y - 20 = 0 intersects y-axis at point A. The co-ordinates of point A are :

1. (-5, 0)

2. (5, 0)

3. (0, 5)

4. (0, -5)

Answer

It is given that the line 5x - 4y - 20 = 0 intersects y-axis at point A.

At the y-axis, the x-coordinate is 0. Let A = (0, b) means x = 0 and y = b.

⇒ 5 $\times$ 0 - 4 $\times$ b - 20 = 0

⇒ 0 - 4b - 20 = 0

⇒ 4b + 20 = 0

⇒ 4b = - 20

⇒ b = - $\dfrac{20}{4}$

⇒ b = - 5

A = (0, -5)

Hence, option 4 is the correct option.

The point (0, 0) lies on :

1. x-axis

2. y-axis

3. x-axis or y-axis

4. x-axis and y-axis both

Answer

The axes are two perpendicular lines (x-axis and y-axis) that intersect at the origin. The point (0,0), known as the origin, lies on both the x-axis and the y-axis because:

• The x-coordinate of (0,0) is 0, which satisfies the condition for a point on the y-axis.

• The y-coordinate of (0,0) is 0, which satisfies the condition for a point on the x-axis.

Thus, the point (0,0) lies on both the x-axis and y-axis.

Hence, option 4 is the correct option.

Draw the graph for each equation, given below :

(i) x = 5

(ii) x + 5 = 0

(iii) y = 7

(iv) y + 7 = 0

(v) 2x + 3y = 0

(vi) 3x + 2y = 6

(vii) x - 5y + 4 = 0

(viii) 5x + y + 5 = 0

Answer

(i) x = 5

(ii) x + 5 = 0

x = -5

(iii) y = 7

(iv) y + 7 = 0

y = -7

(v) 2x + 3y = 0

Step 1:

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

Let x = -1, then 2 $\times$ (-1) + 3y = 0 ⇒ y = 0.6

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

Let x = 1, then 2 $\times$ 1 + 3y = 0 ⇒ y = - 0.6

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.

(vi) 3x + 2y = 6

Step 1:

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

Let x = -1, then 3 $\times$ (-1) + 2y = 6 ⇒ y = 4.5

Let x = 0, then 3 $\times$ 0 + 2y = 6 ⇒ y = 3

Let x = 1, then 3 $\times$ 1 + 2y = 6 ⇒ y = 1.5

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.

(vii) x - 5y + 4 = 0

Step 1:

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

Let x = -2, then (-2) - 5y + 4 = 0 ⇒ y = 0.4

Let x = 0, then 0 - 5y + 4 = 0 ⇒ y = 0.8

Let x = 2, then 2 - 5y + 4 = 0 ⇒ y = 1.2

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.

(viii) 5x + y + 5 = 0

Step 1:

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

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

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

Let x = 1, then 5 $\times$ 1 + y + 5 = 0 ⇒ 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.

Draw the graph for each equation given below; hence find the co-ordinates of the points where the graph drawn meets the co-ordinate axes :

(i) $\dfrac{1}{3}x + \dfrac{1}{5}y = 1$

(ii) $\dfrac{2x + 15}{3} = y - 1$

Answer

(i) $\dfrac{1}{3}x + \dfrac{1}{5}y = 1$

Step 1:

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

Let x = -3, then $\dfrac{1}{3} \times (-3) + \dfrac{1}{5}y = 1$ ⇒ y = 10

Let x = 0, then $\dfrac{1}{3} \times 0 + \dfrac{1}{5}y = 1$ ⇒ y = 5

Let x = 3, then $\dfrac{1}{3} \times 3 + \dfrac{1}{5}y = 1$ ⇒ 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.

Hence, the co-ordinates of the points where the graph drawn meets the co-ordinate axes are (0, 5) and (3, 0).

(ii) $\dfrac{2x + 15}{3} = y - 1$

Step 1:

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

Let x = -9, then $\dfrac{2 \times (-9) + 15}{3} = y - 1$ ⇒ y = 0

Let x = -3, then $\dfrac{2 \times (-3) + 15}{3} = y - 1$ ⇒ y = 4

Let x = 0, then $\dfrac{2 \times 0 + 15}{3} = y - 1$ ⇒ y = 6

Let x = 3, then $\dfrac{2 \times 3 + 15}{3} = y - 1$ ⇒ 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.

Hence, the co-ordinates of the points where the graph drawn meets the co-ordinate axes are (0, 6) and (-9, 0).

Draw the graph of the straight line given by the equation 4x - 3y + 36 = 0

Calculate the area of the triangle formed by the line drawn and the co-ordinate axes.

Answer

Step 1:

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

Let x = -9, then 4 $\times$ (-9) - 3y + 36 = 0 ⇒ y = 0

Let x = -6, then 4 $\times$ (-6) - 3y + 36 = 0 ⇒ y = 4

Let x = -3, then 4 $\times$ (-3) - 3y + 36 = 0 ⇒ y = 8

Let x = 0, then 4 $\times$ 0 - 3y + 36 = 0 ⇒ y = 12

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.

Area of triangle OAB = $\dfrac{1}{2}$ OA x OB

= $\dfrac{1}{2}$ x 9 x 12 square units

= $\dfrac{1}{2}$ x 108 square units

= 54 square units

Hence, area of triangle = 54 sq. units.

Draw the graph of the equation

2x - 3y - 5 = 0

From the graph, find :

(i) x₁, the value of x, when y = 7

(ii) x₂, the value of x, when y = -5

Answer

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) - 3y - 5 = 0 ⇒ y = -3

Let x = 0, then 2 $\times$ 0 - 3y - 5 = 0 ⇒ y = -1.6

Let x = 2, then 2 $\times$ 2 - 3y - 5 = 0 ⇒ y = -0.3

Let x = 5, then 2 $\times$ 5 - 3y - 5 = 0 ⇒ y = 1.6

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 AB passing through the points plotted on the graph.

(i) To find x₁, the value of x, when y = 7:

Through the point y = 7, draw a horizontal straight line which meets the line AB at point C.

Through point C, draw a vertical line which meets the x-axis at x = 13.

Hence, the value of x, when y = 7 is 13 , i.e, x₁ = 13.

(ii) To find x₂, the value of x, when y = -5:

Through the point y = -5, draw a horizontal straight line which meets the line AB at point D.

Through point D, draw a vertical line which meets the x - axis at x = -5.

Hence, the value of x, when y = -5 is -5, i.e, x₂ = -5.

Draw the graph of the equation

4x + 3y + 6 = 0

From the graph, find :

(i) y₁, the value of y, when x = 12

(ii) y₂, the value of y, when x = -6

Answer

Step 1:

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

Let x = -3, then 4 $\times$ (-3) + 3y + 6 = 0 ⇒ y = 2

Let x = 0, then 4 $\times$ 0 + 3y + 6 = 0 ⇒ y = -2

Let x = 3, then 4 $\times$ 3 + 3y + 6 = 0 ⇒ y = -6

Let x = 8, then 4 $\times$ 8 + 3y + 6 = 0 ⇒ y = -12.6

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 AB passing through the points plotted on the graph.

(i) To find y₁, the value of y, when x = 12:

Through the point x = 12, draw a vertical straight line which meets the line AB at point C.

Through point C, draw a horizontal line which meets the y-axis at y = -18.

Hence, the value of y, when x = 12 is -18 , i.e, y₁ = -18.

(ii) To find y₂, the value of y, when x = -6:

Through the point x = -6, draw a vertical straight line which meets the line AB at point D.

Through point D, draw a horizontal line which meets the y-axis at y = 6.

Hence, the value of x, when y = -5 is 6, i.e, y₂ = 6.

Use the table given below to draw the graph.

From your graph, find the values of 'a' and 'b'. State a linear relation between the variables x and y.

Answer

Plot the given points (-5, -2), (3, 2) and (13, 7) on a graph paper.

Draw a straight line passing through these points.

To find the value of 'a':

From the graph, y = 0 when x = −1.

∴ a = 0

To find the value of 'b':

Through y = 5, draw a horizontal line which meets the graph at a point, say Q. Through Q, draw a vertical line which meets the x-axis at x = 9.

∴ b = 9

Let the linear relation between the variable x and y be y = mx + c.

Since, the graph passes through the point (-5, -2); substitute x = -5 and y = -2 in y = mx + c.

This gives -2 = -5m + c ...............(1)

Again, the graph passes through the point (3, 2); substitute x = 3 and y = 2 in y = mx + c

This gives 2 = 3m + c ...............(2)

Subtracting (2) from (1),

-2 - 2 = -5m + c -3m - c

⇒ -4 = -8m

⇒ m = $\dfrac{4}{8}$

⇒ m = $\dfrac{1}{2}$

substituting the value of m in equation (1),

-2 = -5 $\times \dfrac{1}{2}$ + c

⇒-2 = $\dfrac{-5}{2}$ + c

⇒-2 + $\dfrac{5}{2}$ = c

⇒ $\dfrac{-4}{2} + \dfrac{5}{2}$ = c

⇒ c = $\dfrac{1}{2}$

∴ Required relation is : y = mx + c i.e. y = $\dfrac{x + 1}{2}$

Hence, a = 0 and b = 9. Linear relation : y = $\dfrac{x + 1}{2}$.

Draw the graph obtained from the table below :

Use the graph to find the values of a, b and c. State a linear relation between the variables x and y.

Answer

Plot the given points (3, 2), (5, 3) and (-1, 0) on a graph paper.

Draw a straight line passing through these points.

To find the value of 'a':

Through y = -1, draw a horizontal line which meets the graph at a point, say P. Through P, draw a vertical line which meets the x-axis at x = -3.

∴ a = -3

To find the value of 'b':

Similarly, through x = -5, draw a vertical line which meets the graph at a point, say Q. Through Q, draw a horizontal line which meets the y-axis at y = -2.

∴ b = -2

To find the value of 'c':

Similarly, through y = 4, draw a horizontal line which meets the graph at a point, say R. Through R, draw a vertical line which meets the x-axis at x = 7.

∴ c = 7

Let the linear relation between the variable x and y be y = mx + c.

Since, the graph passes through the point (3, 2); substitute x = 3 and y = 2 in y = mx + c.

This gives 2 = 3m + c ...............(1)

Again, the graph passes through the point (5, 3); substitute x = 5 and y = 3 in y = mx + c

This gives 3 = 5m + c ...............(2)

Subtracting (2) from (1),

2 - 3 = 3m + c -5m - c

⇒ -1 = -2m

⇒ m = $\dfrac{1}{2}$

Substituting the value of m in equation (1),

2 = 3 $\times \dfrac{1}{2}$ + c

⇒2 = $\dfrac{3}{2}$ + c

⇒2 - $\dfrac{3}{2}$ = c

⇒ $\dfrac{4}{2} - \dfrac{3}{2}$ = c

⇒ c = $\dfrac{1}{2}$

∴ Required relation is : y = mx + c i.e. x = 2y - 1

Hence, a = -3, b = -2 and c = 7. Linear relation : x = 2y - 1.

A straight line passes through the points (2, 4) and (5, -2). Taking 1 cm = 1 unit; mark these points on a graph paper and draw the straight line through these points. If points (m, -4) and (3, n) lie on the line drawn; find the values of m and n.

Answer

Plot the given points (2, 4) and (5, -2) on a graph paper.

Draw a straight line AB passing through these points.

Since, point (m, -4) lies on the straight line drawn, through y = -4 draw a horizontal line which meets the straight line AB at point P. Through P, draw a vertical line which meets the x-axis at point 6.

∴ m = 6

Also, as (3, n) lies on the straight line drawn, through x = 3, draw a vertical line which meets the straight line at point Q. Through point Q, draw a horizontal line which meets the y-axis at point 2.

∴ n = 2

Hence, m = 6 and n = 2.

Draw the graph (straight line) given by equation x - 3y = 18. If the straight line drawn passes through the points (m, -5) and (6, n); find the values of m and n.

Answer

Given equation: x - 3y = 18

Step 1:

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

Let x = 0, then 0 - 3y = 18 ⇒ y = - 6

Let x = 3, then 3 - 3y = 18 ⇒ y = - 5

Let x = 6, then 6 - 3y = 18 ⇒ 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.

Since, point (m, -5) lies on the straight line drawn, through y = -5, draw a horizontal line which meets the graph at a point, say P. Through P, draw a vertical line which meets the x-axis at x = 3.

Also, (6, n) also lies on the straight line drawn, through x = 6, draw a vertical line which meets the graph at a point, say Q. Through Q, draw a horizontal line which meets the y-axis at x = -4.

Hence, m = 3 and n = -4.