Graphical Solution

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.

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 = $\frac{10}{5}$

⇒ m = 2

Hence, option 2 is the correct option.

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 = $\frac{10}{5}$

⇒ a = 2

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

Hence, option 3 is the correct option.

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 = - $\frac{20}{4}$

⇒ b = - 5

A = (0, -5)

Hence, option 4 is the correct option.

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.

(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:

x	-1	0	1
y	0.6	0	-0.6

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:

x	-1	0	1
y	4.5	3	1.5

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:

x	-2	0	2
y	0.4	0.8	1.2

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:

x	-1	0	1
y	0	-5	-10

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) $\frac{1}{3}x + \frac{1}{5}y = 1$

Step 1:

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

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

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

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

Step 2:

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

x	-3	0	3
y	10	5	0

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) $\frac{2x + 15}{3} = y - 1$

Step 1:

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

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

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

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

Let x = 3, then $\frac{2 \times 3 + 15}{3} = y - 1$ ⇒ y = 8

Step 2:

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

x	-9	-3	0	3
y	0	4	6	8

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).

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:

x	-9	-6	-3	0
y	0	4	8	12

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 = $\frac{1}{2}$ OA x OB

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

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

= 54 square units

Hence, area of triangle = 54 sq. units.

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:

x	-2	0	2	5
y	-3	-1.6	-0.3	1.6

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.

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:

x	-3	0	3	8
y	2	-2	-6	-12.6

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.

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 = $\frac{4}{8}$

⇒ m = $\frac{1}{2}$

substituting the value of m in equation (1),

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

⇒-2 = $\frac{-5}{2}$ + c

⇒-2 + $\frac{5}{2}$ = c

⇒ $\frac{-4}{2} + \frac{5}{2}$ = c

⇒ c = $\frac{1}{2}$

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

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

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 = $\frac{1}{2}$

Substituting the value of m in equation (1),

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

⇒2 = $\frac{3}{2}$ + c

⇒2 - $\frac{3}{2}$ = c

⇒ $\frac{4}{2} - \frac{3}{2}$ = c

⇒ c = $\frac{1}{2}$

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

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

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.

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:

x	0	3	6
y	-6	-5	-4

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.

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

Hence, option 1 is the correct option.

Given equation: y = 20 + 2x

When x = 20

y = 20 + 2 $\times$ 20

= 20 + 40

= 60

Hence, option 2 is the correct option.

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.

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.

y - 2 = 0

⇒ y = 2

And, 2x + 3y = 12

⇒ 2x + 3 x 2 = 12

⇒ 2x + 6 = 12

⇒ 2x = 12 - 6

⇒ 2x = 6

⇒ x = $\frac{6}{2}$

⇒ 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.

(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:

x	2	4	6
y	19	15	11

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:

x	4	6	8
y	-3	5	13

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:

x	0	2	4
y	3.8	3	2.1

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 : $8 - y = \frac{5}{2}x$

Step 1:

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

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

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

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

Step 2:

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

x	0	1	2
y	8	5.5	3

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

Step 1:

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

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

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

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

Step 2:

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

x	0	3	7
y	0.3	-0.2	-1

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

Step 1:

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

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

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

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

Step 2:

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

x	0	7	14
y	-1.6	-1	-0.4

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.

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:

x	0	2	4
y	-2	-1	0

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:

x	0	2	4
y	3	-1	-5

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.

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:

x	0	2	4
y	-1	3	7

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:

x	0	2	4
y	9	5	1

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).

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:

x	0	2	4
y	2.4	1.2	0

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:

x	-2	0	2
y	2.4	3.6	4.8

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.

(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:

x	0	-4	-6
y	-3	1	3

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:

x	0	2	4
y	2	5	8

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.

(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:

x	0	2	4
y	-7	-1	5

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:

x	0	2	4
y	0	-1	-2

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 = $\frac{1}{2}$ x base x height

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

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

= $\frac{21}{2}$

= 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).

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:

x	-3	-2	1
y	4	1	-8

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:

x	-2	1	7
y	1	2	4

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:

x	-7	-2	3
y	3	1	-1

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.

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:

x	-8	-2	4
y	-5	0	5

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:

x	3	4	5
y	0	5	10

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 = $\frac{1}{2}$ x base x height

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

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

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

= 12.5 sq. units

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

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:

x	0	20	40
C.P.	50	110	170

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:

x	0	20	40
S.P.	0	80	160

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.

Given, equation of lines 2x - y = 7 and x = 3

Substituting x = 3 in 2x - y = 7, we get :

⇒ 2 x 3 - y = 7

⇒ 6 - y = 7

⇒ y = 6 - 7

⇒ y = -1.

Hence, option 2 is the correct option.

If the line x + 3y + 2 = 0 passes through the point (4, k), then substituting x = 4 and y = k into the equation of the line,

⇒ 4 + 3 x k + 2 = 0

⇒ 6 + 3k = 0

⇒ 3k = -6

⇒ k = $-\frac{6}{3}$

⇒ k = -2.

Hence, option 4 is the correct option.

⇒ x - 5 = 0

⇒ x = 5

This line represents a vertical line parallel to y-axis.

⇒ y + 3 = 0

⇒ y = ​-3

This represents a horizontal line parallel to x-axis.

Point of intersection of lines x = 5 and y = -3 is (5, -3).

Hence, option 2 is the correct option.

Given,
equation of line 4x - 7y + 6 = 0

Substitute the given value of x (which is 2) into the equation:

⇒ 4 x 2 - 7y + 6 = 0

⇒ 8 - 7y + 6 = 0

⇒ 14 - 7y = 0

⇒ 7y = 14

⇒ y = $\frac{14}{7}$

⇒ y = 2.

Hence, option 1 is the correct option.

Given,

⇒ 2x - 5 = 0

⇒ 2x = 5

⇒ x = $\frac{5}{2}$

This represents a vertical line (parallel to the y-axis) at a distance of $\frac{5}{2}$ units.

This line is at a distance of $\frac{5}{2}$ units to the right of the y-axis because the x-coordinate of every point on the line is $\frac{5}{2}$.

∴ Statement 1 is true.

For equations of the form x = ± k.

These represent vertical lines (parallel to the y-axis) because the value of x remains constant at k or -k for all values of y.

∴ Statement 2 is true.

∴ Both the statements are true.

Hence, option 1 is the correct option.

A line passes through the origin (0, 0) if the coordinates of the origin satisfy the equation of the line.

Substituting x = 0 and y = 0 in L.H.S. of the equation ax ± by = 0, we get :

⇒ a(0) ± b(0)

⇒ 0, which is equal to R.H.S.

Since, (0, 0) satisfies the equation ax ± by = 0, thus lines of the form ax ± by = 0 always pass through the origin.

∴ Both the statements are true.

Hence, option 1 is the correct option.

⇒ y + 5 = 0

⇒ y = -5

This represents a horizontal line (parallel to the x-axis) because the y-coordinate remains constant at -5 for all values of x. The line is 5 units below the x-axis since the y-coordinate of every point is −5, which is 5 units in the negative direction from the x-axis.

∴ Assertion (R) is true.

For all points on the line y = a, the value of ordinate (y-coordinate) = a.

∴ Reason (R) is false.

∴ A is true, but R is false.

Hence, option 1 is the correct option.

Given, equation of line : 3x + 4y = 7

When, y = 0

⇒ 3x + 4 x 0 = 7

⇒ 3x = 7

⇒ x = $\frac{7}{3}$

∴ Reason (R) is true.

So, the abscissa is $\frac{7}{3} \text{ and not } -\frac{3}{4}$.

∴ Assertion (A) is false.

∴ A is false, but R is true.

Hence, option 2 is the correct option.

Plot the point (8, -4) on the graph paper.

Draw a perpendicular line from the point (8, -4) to the y-axis.

From graph it is clear that the distance between the point (8,−4) and the y-axis is 8 units.

Hence, the distance of point (8, -4) from y-axis is 8 units.

Plot the points A(-5, -1), B(3, -1) and C(1, -6) on the graph paper. Join point A with B and B with C.

From the graph, it is clear that the horizontal distance between the points A (-5, -1) and B (3, -1) is 8 units and the vertical distance between the points B (3, -1) and C (1, -6) is 5 units. Therefore, the vertical distance between the points A (-5, -1) and D must be 5 units and the horizontal distance between the points C (1, -6) and D must be 8 units.

Now, complete the parallelogram ABCD and read the coordinates of point D. As shown on the graph, D = (-7, -6).

Hence, the co-ordinates of fourth vertex D are (-7, -6).

Given:

The co-ordinates of B = (2, 0)

The co-ordinates of C = (6, 0)

So, the length of BC = 6 - 2 = 4 units

Since ABC is an equilateral triangle,

The height of the triangle = $\frac{\sqrt3}{2}a$

= $\frac{\sqrt3}{2} \times 4$

= 2 $\sqrt3$ units

The mid-point of BC = $\Big(\frac{2+6}{2}, \frac{0+0}{2}\Big)$

= (4, 0)

Since Δ ABC is equilateral, and BC lies on the x-axis:

• The abscissa (x-coordinate) of A is the same as the midpoint of BC, i.e., x=4.

• The ordinate (y-coordinate) of A is the height of the triangle, $\sqrt{3}$.

Hence, the co-ordinates of A are (4, 2 $\sqrt{3}$).

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 + 2y = 6 ⇒ y = 3

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

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

Step 2:

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

x	0	2	4
y	3	0	-3

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.

The area of the triangle = $\frac{1}{2}$ x base x height

= $\frac{1}{2}$ x OA x OB

= $\frac{1}{2}$ x 2 x 3

= $\frac{6}{2}$ sq. units

= 3 sq. units

Hence, the area of triangle formed by the line drawn and the co-ordinate axes is 3 sq. units.

(i) 2x + 3y = 1

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 + 3y = 1 ⇒ y = 0.3

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

Let x = 4, then 2 $\times$ 4 + 3y = 1 ⇒ y = -2.3

Step 2:

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

x	0	2	4
y	0.3	-1	-2.3

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

Hence, the value of k = 5.

(ii) x - 2y + 1 = 0

Step 1:

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

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

Let x = 0, then 0 - 2y + 1 = 0 ⇒ y = 0.5

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

Step 2:

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

x	-1	0	1
y	0	0.5	1

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 (5, k - 2) lies on the straight line drawn, through x = 5, draw a vertical line which meets the graph at a point, say Q. Through point Q, draw a horizontal line which meets the y-axis at point 3.

k - 2 = 3

⇒ k = 3 + 2

⇒ k = 5

Hence, the value of k = 5.

First equation: 2y - x = 8

Step 1:

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

Let x = -4, then 2y - (-4) = 8 ⇒ y = 2

Let x = -2, then 2y - (-2) = 8 ⇒ y = 3

Let x = 0, then 2y - 0 = 8 ⇒ y = 4

Let x = 2, then 2y - 2 = 8 ⇒ y = 5

Step 2:

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

x	-4	-2	0	2
y	2	3	4	5

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: 5y - x = 14

Step 1:

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

Let x = -4, then 5y - (-4) = 14 ⇒ y = 2

Let x = 0, then 5y - 0 = 14 ⇒ y = 2.8

Let x = 2, then 5y - 2 = 14 ⇒ y = 3.2

Let x = 4, then 5y - 4 = 14 ⇒ y = 3.6

Step 2:

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

x	-4	0	2	4
y	2	2.8	3.2	3.6

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: y - 2x = 1

Step 1:

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

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

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

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

Step 2:

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

x	0	2	4
y	1	5	9

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.

The vertices of the triangle are:

The coordinate of A = (2, 5)

The coordinate of B = (-4, 2)

The coordinate of C = (1, 3)

Hence, the coordinates of the vertices of triangle = (1, 3), (-4, 2) and (2, 5).

First equation: x + y = 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 + y = 0 ⇒ y = 0

Let x = 1, then 1 + y = 0 ⇒ y = -1

Let x = 2, then 2 + y = 0 ⇒ y = -2

Step 2:

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

x	0	1	2
y	0	-1	-2

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 = 10

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 - 2y = 10 ⇒ y = -5

Let x = 1, then 3 $\times$ 1 - 2y = 10 ⇒ y = -3.5

Let x = 2, then 3 $\times$ 2 - 2y = 10 ⇒ y = -2

Step 2:

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

x	0	1	2
y	-5	-3.5	-2

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.

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, -2).

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

First equation: x + 2y = 4

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 ⇒ y = 2

Let x = 2, then 2 + 2y = 4 ⇒ y = 1

Let x = 4, then 4 + 2y = 4 ⇒ y = 0

Step 2:

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

x	0	2	4
y	2	1	0

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

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 - 2y = 4 ⇒ y = -2

Let x = 2, then 3 $\times$ 2 - 2y = 4 ⇒ y = 1

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

Step 2:

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

x	0	2	4
y	-2	1	4

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.

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

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

The area of the triangle = $\frac{1}{2}$ x base x height

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

= $\frac{1}{2}$ x 2.6 x 1

= 1.3 sq. units

Hence, the area of the triangle formed between the lines and the x-axis = 1.3 sq. units.

First equation: y = $\frac{3x + 2}{2}$

Step 1:

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

Let x = -4, then y = $\frac{3 \times (-4) + 2}{2}$ ⇒ y = -5

Let x = -2, then y = $\frac{3 \times (-2) + 2}{2}$ ⇒ y = -2

Let x = 2, then y = $\frac{3 \times 2 + 2}{2}$ ⇒ y = 4

Step 2:

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

x	-4	-2	2
y	-5	-2	4

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 = $\frac{3}{4}x - 2$

Step 1:

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

Let x = -4, then y = $\frac{3}{4} \times (-4) - 2$ ⇒ y = -5

Let x = 4, then y = $\frac{3}{4} \times 4 - 2$ ⇒ y = 1

Let x = 8, then y = $\frac{3}{4} \times 8 - 2$ ⇒ y = 4

Step 2:

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

x	-4	4	8
y	-5	1	4

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.

After plotting both lines on the graph, observe where the two lines intersect. The intersection point represents the value of x where both equations are equal.

From the graph, we see that the lines intersect at x = -4.

Hence, the value of x for which the two equations are equal is -4.

Enemy equation: 2x + 3y = 4

Step 1:

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

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

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

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

Step 2:

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

x	-4	-1	5
y	4	2	-2

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.

Destroyer equation: x - y = 7

Step 1:

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

Let x = 5, then 5 - y = 7 ⇒ y = -2

Let x = 9, then 9 - y = 7 ⇒ y = 2

Let x = 11, then 11 - y = 7 ⇒ y = 4

Step 2:

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

x	5	9	11
y	-2	2	4

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.

Both the straight lines intersect at point P. As it is clear from the graph, co-ordinates of point P are (-2, -1).

Hence, (5, -2) is the point at which the paths of the submarine and the destroyer intersect.

(i) Given,

Equation 1: 7x - 5y + 10 = 0

a₁ = 7, b₁ = -5, c₁ = 10

Equation 2: 6x + 2y - 15 = 0

a₂ = 6, b₂ = 2, c₂ = -15

$\Rightarrow \frac{a_1}{a_2} = \frac{7}{6}, \frac{b_1}{b_2} = \frac{-5}{2}, \frac{c_1}{c_2} = -\frac{10}{15}$

Since $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$, the two lines intersect at a point.

Hence, the two lines intersect at a point.

(ii) Given,

Equation 1: 5x + 2y + 8 = 0

a₁ = 5, b₁ = 2, c₁ = 8

Equation 2: 15x + 6y + 24 = 0

a₂ = 15, b₂ = 6, c₂ = 24

$\Rightarrow \frac{a_1}{a_2} = \frac{5}{15} = \frac{1}{3}, \frac{b_1}{b_2} = \frac{2}{6} = \frac{1}{3}, \frac{c_1}{c_2} = \frac{8}{24} = \frac{1}{3}$

Since $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$, the two lines coincide.

Hence, the two lines coincide.

(iii) Given,

Equation 1: 4x - 8y + 9 = 0

a₁ = 4, b₁ = -8, c₁ = 9

Equation 2: 2x - 4y + 7 = 0

a₂ = 2, b₂ = -4, c₂ = 7

$\Rightarrow \frac{a_1}{a_2} = \frac{4}{2} = 2, \frac{b_1}{b_2} = \frac{-8}{-4} = 2, \frac{c_1}{c_2} = \frac{9}{7}$

Since $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$, the two lines are parallel.

Hence, the two lines are parallel.

(iv) Given,

Equation 1: x - 2y = 0

a₁ = 1, b₁ = -2, c₁ = 0

Equation 2: 3x - 4y - 20 = 0

a₂ = 3, b₂ = -4, c₂ = -20

$\Rightarrow \frac{a_1}{a_2} = \frac{1}{3}, \frac{b_1}{b_2} = \frac{-2}{-4} = \frac{1}{2}, \frac{c_1}{c_2} = 0$

Since $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$, the two lines intersect at a point.

Hence, the two lines intersect at a point.

(v) Given,

Equation 1: 2x + 3y - 9 = 0

a₁ = 2, b₁ = 3, c₁ = -9

Equation 2: 4x + 6y - 18 = 0

a₂ = 4, b₂ = 6, c₂ = -18

$\Rightarrow \frac{a_1}{a_2} = \frac{2}{4} = \frac{1}{2}, \frac{b_1}{b_2} = \frac{3}{6} = \frac{1}{2}, \frac{c_1}{c_2} = \frac{-9}{-18} = \frac{1}{2}$

Since $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$, the two lines coincide.

Hence, the two lines coincide.

(vi) Given,

Equation 1: x + 2y - 4 = 0

a₁ = 1, b₁ = 2, c₁ = -4

Equation 2: 2x + 4y - 12 = 0

a₂ = 2, b₂ = 4, c₂ = -12

$\Rightarrow \frac{a_1}{a_2} = \frac{1}{2}, \frac{b_1}{b_2} = \frac{2}{4} = \frac{1}{2}, \frac{c_1}{c_2} = \frac{-4}{-12} = \frac{1}{3}$

Since $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$, the two lines are parallel.

Hence, the two lines are parallel.