Co-ordinate Geometry — Exercise 23(B)

Line y + 7 = 0 is :

1. parallel to x-axis

2. parallel to y-axis

3. not parallel to x-axis

4. not parallel to y-axis

Answer

Given:

y + 7 = 0

⇒ y = -7

The graph of y = -7 is a straight line that is parallel to the x-axis and at a distance of -7 units from it.

Hence, option 1 is the correct option.

A line is parallel to y-axis and at a distance of 5 units on the positive side of the x-axis. The equation of the line is :

1. y = 5

2. y + 5 = 0

3. x = 5

4. x + 5 = 0

Answer

A line is parallel to y-axis means the equation of the line is in the form x = + a or -a units, where a is the distance from the y-axis.

If the line is at a distance of 5 units from the y-axis on the positive side, the equation will be x = 5.

Hence, option 3 is the correct option.

6x - 5y = 7 is the equation of a line. If x = 2 then the value of y will be :

1. 1

2. -1

3. 5

4. -5

Answer

Given:

6x - 5y = 7

If x = 2,

⇒ 6 $\times$ 2 - 5y = 7

⇒ 12 - 5y = 7

⇒ 12 - 7 = 5y

⇒ 5 = 5y

⇒ y = $\dfrac{5}{5}$

⇒ y = 1

Hence, option 1 is the correct option.

For equation $\dfrac{x}{3}-\dfrac{y}{2}$ = 1, the value of y for x = 9 is :

1. -4

2. 6

3. 4

4. -6

Answer

Given:

$\dfrac{x}{3} - \dfrac{y}{2} = 1$

When x = 9,

$⇒ \dfrac{9}{3} - \dfrac{y}{2} = 1\\[1em] ⇒ 3 - \dfrac{y}{2} = 1\\[1em] ⇒ 3 - 1 = \dfrac{y}{2}\\[1em] ⇒ 2 = \dfrac{y}{2}\\[1em] ⇒ y = 2 \times 2\\[1em] ⇒ y = 4$

Hence, option 3 is the correct option.

Lines x - 4 = 0 and 3y = 1 intersect each other at point P. The co-ordinates of point P are:

1. $\Big(4,-\dfrac{1}{3}\Big)$

2. $\Big(4,\dfrac{1}{3}\Big)$

3. $\Big(-4,\dfrac{1}{3}\Big)$

4. $\Big(-4,-\dfrac{1}{3}\Big)$

Answer

x - 4 = 0 ⇒ x = 4

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

3y = 1 ⇒ y = $\dfrac{1}{3}$

This represents a horizontal line parallel to x-axis.

Since the first line has x = 4 and the second line has y = $\dfrac{1}{3}$, the point od intersection P is (4, $\dfrac{1}{3}$).

Hence, option 2 is the correct option.

Draw the graph for the linear equation given below :

x = 3

Answer

x = 3

The graph of x = 3 is the straight line which is parallel to the y-axis at a distance of 3 units from it.

Draw the graph for the linear equation given below :

x + 3 = 0

Answer

x + 3 = 0

⇒ x = -3

The graph of x + 3 = 0 is the straight line which is parallel to the y-axis at a distance of -3 units from it.

Draw the graph for the linear equation given below :

x - 5 = 0

Answer

x - 5 = 0

⇒ x = 5

The graph of x - 5 = 0 is the straight line which is parallel to the y-axis at a distance of 5 units from it.

Draw the graph for the linear equation given below :

2x - 7 = 0

Answer

2x - 7 = 0

⇒ 2x = 7

⇒ x = $\dfrac{7}{2}$

⇒ x = 3.5

The graph of 2x - 7 = 0 is the straight line which is parallel to the y-axis at a distance of 3.5 units from it.

Draw the graph for the linear equation given below :

y = 4

Answer

y = 4

The graph of y = 4 is the straight line which is parallel to the x-axis at a distance of 4 units from it.

Draw the graph for the linear equation given below :

y + 6 = 0

Answer

y + 6 = 0

⇒ y = - 6

The graph of y + 6 = 0 is the straight line which is parallel to the x-axis at a distance of -6 units from it.

Draw the graph for the linear equation given below :

y - 2 = 0

Answer

y - 2 = 0

⇒ y = 2

The graph of y - 2 = 0 is the straight line which is parallel to the x-axis at a distance of 2 units from it.

Draw the graph for the linear equation given below :

3y + 5 = 0

Answer

3y + 5 = 0

⇒ 3y = - 5

⇒ y = - $\dfrac{5}{3}$

⇒ y = - 1.6

The graph of 3y + 5 = 0 is the straight line which is parallel to the x-axis at a distance of - 1.6 units from it.

Draw the graph for the linear equation given below :

2y - 5 = 0

Answer

2y - 5 = 0

⇒ 2y = 5

⇒ y = $\dfrac{5}{2}$

⇒ y = 2.5

The graph of 2y - 5 = 0 is the straight line which is parallel to the x-axis at a distance of 2.5 units from it.

Draw the graph for the linear equation given below :

y = 0

Answer

y = 0

The graph of y = 0 is the straight line which is on x-axis.

Draw the graph for the linear equation given below :

x = 0

Answer

x = 0

The graph of x = 0 is the straight line which is on y-axis.

Draw the graph for the linear equation given below :

y = 3x

Answer

y = 3x

Step 1:

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

Let x = -1, then y = 3 $\times$ (-1) = -3

Let x = 0, then y = 3 $\times$ 0 = 0

Let x = 1, then y = 3 $\times$ 1 = 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.

Draw the graph for the linear equation given below :

y = - x

Answer

y = - x

Step 1:

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

Let x = -1, then y = - (-1) = 1

Let x = 0, then y = 0 = 0

Let x = 1, then y = - 1 = - 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.

Draw the graph for the linear equation given below :

y = - 2x

Answer

y = - 2x

Step 1:

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

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

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

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

Draw the graph for the linear equation given below :

y = x

Answer

y = x

Step 1:

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

Let x = -1, then y = -1

Let x = 0, then y = 0

Let x = 1, then 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.

Draw the graph for the linear equation given below :

5x + y = 0

Answer

5x + y = 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 = 0 ⇒ y = 5

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

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

Draw the graph for the linear equation given below :

x + 2y = 0

Answer

x + 2y = 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 = 0 ⇒ y = 0.5

Let x = 0, then 0 + 2y = 0 ⇒ y = 0

Let x = 1, then 1 + 2y = 0 ⇒ y = -0.5

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.

Draw the graph for the linear equation given below :

4x - y = 0

Answer

4x - y = 0

Step 1:

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

Let x = -1, then 4 $\times$ (-1) - y = 0 ⇒ y = -4

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

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

Draw the graph for the linear equation given below :

3x + 2y = 0

Answer

3x + 2y = 0

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 = 0 ⇒ y = 1.5

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

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

Draw the graph for the linear equation given below :

x = - 2y

Answer

x = - 2y

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 ⇒ y = 0.5

Let x = 0, then 0 = - 2y ⇒ y = 0

Let x = 1, then 1 = - 2y ⇒ y = - 0.5

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.

Draw the graph for the linear equation given below :

y = 2x + 3

Answer

y = 2x + 3

Step 1:

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

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

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

Let x = 1, then y = 2 $\times$ 1 + 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.

Draw the graph for the linear equation given below :

$y = \dfrac{2}{3} x - 1$

Answer

$y = \dfrac{2}{3} x - 1$

Step 1:

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

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

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

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

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.

Draw the graph for the linear equation given below :

y = -x + 4

Answer

y = -x + 4

Step 1:

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

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

Let x = 0, then y = 0 + 4 ⇒ y = 4

Let x = 1, then y = -1 + 4 ⇒ y = 3

Step 2:

Make a table i(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 the linear equation given below :

$y = 4x - \dfrac{5}{2}$

Answer

$y = 4x - \dfrac{5}{2}$

Step 1:

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

Let x = -$\dfrac{3}{2} = -1.5$, then $y = 4 \times \Big(\dfrac{-3}{2}\Big) - \dfrac{5}{2} ⇒ y = -\dfrac{17}{2} = -8.5$

Let x = $\dfrac{1}{2} = 0.5$, then $y = 4 \times \dfrac{1}{2} - \dfrac{5}{2} ⇒ y = -\dfrac{1}{2} = -0.5$

Let x = $\dfrac{3}{2} = 1.5$, then $y = 4 \times \dfrac{3}{2} - \dfrac{5}{2} ⇒ y = \dfrac{7}{2} = 3.5$

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.

Draw the graph for the linear equation given below :

$y =\dfrac{3}{2}x + \dfrac{2}{3}$

Answer

$y =\dfrac{3}{2}x + \dfrac{2}{3}$

Step 1:

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

Let x = -2, then $y =\dfrac{3}{2} \times (-2) + \dfrac{2}{3} ⇒ y = -2.3$

Let x = 0, then $y =\dfrac{3}{2} \times 0 + \dfrac{2}{3} ⇒ y = 0.6$

Let x = 2, then $y =\dfrac{3}{2} \times 2 + \dfrac{2}{3} ⇒ y = 3.6$

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.

Draw the graph for the linear equation given below :

2x - 3y = 4

Answer

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

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:

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 the linear equation given below :

$\dfrac{x-1}{3} - \dfrac{y+2}{2} = 0$

Answer

$\dfrac{x-1}{3} - \dfrac{y+2}{2} = 0$

Step 1:

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

Let x = -5, then $\dfrac{-5-1}{3} - \dfrac{y+2}{2} = 0 ⇒ y = -6$

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

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

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.

Draw the graph for the linear equation given below :

$x - 3 = \dfrac{2}{5}(y + 1)$

Answer

$x - 3 = \dfrac{2}{5}(y + 1)$

Step 1:

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

Let x = 0, then $0 - 3 = \dfrac{2}{5}(y + 1) ⇒ y = -8.5$

Let x = 1, then $1 - 3 = \dfrac{2}{5}(y + 1) ⇒ y = -6$

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

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.

Draw the graph for the linear equation given below :

x + 5y + 2 = 0

Answer

x + 5y + 2 = 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 + 2 = 0 ⇒ y = 0

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

Let x = 2, then 2 + 5y + 2 = 0 ⇒ y = - 0.8

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.

Draw the graph for the equation given below :

3x + 2y = 6

Find the co-ordinates of the points where the graph (line) drawn meets the co-ordinate axes.

Answer

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 = $\dfrac{9}{2} = 4.5$

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

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

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.

From the graph we get that the line meets x-axis at (2, 0) and y-axis at (0, 3)

Draw the graph for the equation given below :

2x - 5y = 10

Find the co-ordinates of the points where the graph (line) drawn meets the co-ordinate axes.

Answer

2x - 5y = 10

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) - 5y = 10 ⇒ y = $- \dfrac{12}{5} = -2.4$

Let x = 0, then 2 $\times$ 0 - 5y = 10 ⇒ y = $-\dfrac{10}{5} = -2$

Let x = 1, then 2 $\times$ 1 - 5y = 10 ⇒ y = $-\dfrac{8}{5} = -1.6$

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 passing through the points plotted on the graph.

From the graph we get that the line meets x-axis at (5, 0) and y-axis at (0, -2)

Draw the graph for the equation given below :

$\dfrac{1}{2}x + \dfrac{2}{3}y = 5$

Find the co-ordinates of the points where the graph (line) drawn meets the co-ordinate axes.

Answer

$\dfrac{1}{2}x + \dfrac{2}{3}y = 5$

Step 1:

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

Let x = -2, then $\dfrac{1}{2} \times (-2) + \dfrac{2}{3}y = 5 ⇒ y = \dfrac{6 \times 3}{2} = 9$

Let x = 0, then $\dfrac{1}{2} \times 0 + \dfrac{2}{3}y = 5 ⇒ y = \dfrac{15}{2} = 7.5$

Let x = 2, then $\dfrac{1}{2} \times 2 + \dfrac{2}{3}y = 5 ⇒ y = \dfrac{4 \times 3}{2} = 6$

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.

From the graph we get that the line meets x-axis at (10, 0) and y-axis at (0, 7.5)

Draw the graph for the equation given below :

$\dfrac{2x - 1}{3} - \dfrac{y - 2}{5} = 0$

Find the co-ordinates of the points where the graph (line) drawn meets the co-ordinate axes.

Answer

$\dfrac{2x - 1}{3} - \dfrac{y - 2}{5} = 0$

Step 1:

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

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

Let x = 0, then $\dfrac{2 \times 0 - 1}{3} - \dfrac{y - 2}{5} = 0 ⇒ y = 0.3$

Let x = 1, then $\dfrac{2 \times 1 - 1}{3} - \dfrac{y - 2}{5} = 0 ⇒ y = 3.6$

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.

From the graph we get that the line meets x-axis at (-0.1, 0) and y-axis at (0, 0.3)

For the linear equation, given below, draw the graph and then use the graph drawn to find the area of a triangle enclosed by the graph and the co-ordinate axes :

3x - (5 - y) = 7

Answer

3x - (5 - y) = 7

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) - (5 - y) = 7 ⇒ y = 15

Let x = 0, then 3 $\times$ 0 - (5 - y) = 7 ⇒ y = 12

Let x = 1, then 3 $\times$ 1 - (5 - y) = 7 ⇒ y = 9

Let x = 4, then 3 $\times$ 4 - (5 - y) = 7 ⇒ 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.

The area of the triangle ABO will be = $\dfrac{1}{2} \times base \times$ altitude

= $\dfrac{1}{2} \times OA \times$ OB

= $\dfrac{1}{2} \times 4 \times$ 12 square unit

= 24 square unit

Hence, the area of triangle = 24 square unit.

For the linear equation, given below, draw the graph and then use the graph drawn to find the area of a triangle enclosed by the graph and the co-ordinate axes :

7 - 3 (1 - y) = - 5 + 2x.

Answer

7 - 3 (1 - y) = - 5 + 2x

Step 1:

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

Let x = 0, then 7 - 3 (1 - y) = - 5 + 2 $\times$ 0 ⇒ y = -3

Let x = 4.5, then 7 - 3 (1 - y) = - 5 + 2 $\times$ 4.5 ⇒ y = 0

Let x = 6, then 7 - 3 (1 - y) = - 5 + 2 $\times$ 6 ⇒ 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.

The area of the triangle ABO will be = $\dfrac{1}{2} \times base \times$ altitude

= $\dfrac{1}{2}$ $\times$ OA $\times$ OB

= $\dfrac{1}{2}$ $\times$ 3 $\times$ 4.5

= 6.75 square unit

Hence, the area of triangle = 6.75 square unit.

For the pair of linear equations given below, draw graphs and then state, whether the lines drawn are parallel or perpendicular to each other.

y = 3x - 1
y = 3x + 2

Answer

First equation = y = 3x - 1

Step 1:

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

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

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

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

Second equation = y = 3x + 2

Step 1:

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

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

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

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

Hence, the two lines are parallel to each other.

For the pair of linear equations given below, draw graphs and then state, whether the lines drawn are parallel or perpendicular to each other.

y = x - 3
y = - x + 5

Answer

First equation = y = x - 3

Step 1:

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

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

Let x = 0, then y = 0 - 3 ⇒ y = - 3

Let x = 1, then y = 1 - 3 ⇒ y = - 2

Let x = 4, then y = 4 - 3 ⇒ 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 = y = - x + 5

Step 1:

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

Let x = -1, then y = - (-1) + 5 ⇒ y = 6

Let x = 0, then y = - 0 + 5 ⇒ y = 5

Let x = 1, then y = - 1 + 5 ⇒ y = 4

Let x = 4, then y = - 4 + 5 ⇒ 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.

Hence, the two lines are perpendicular to each other.

For the pair of linear equations given below, draw graphs and then state, whether the lines drawn are parallel or perpendicular to each other.

2x - 3y = 6

$\dfrac{x}{2} + \dfrac{y}{3} = 1$

Answer

First equation = 2x - 3y = 6

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 = 6 ⇒ y = - 2.6

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

Let x = 1, then 2 $\times$ 1 - 3y = 6 ⇒ y = - 1.3

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

Step 1:

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

Let x = -1, then $\dfrac{-1}{2} + \dfrac{y}{3} = 1$ ⇒ y = 4.5

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

Let x = 1, then $\dfrac{1}{2} + \dfrac{y}{3} = 1$ ⇒ y = 1.5

Let x = 2, then $\dfrac{2}{2} + \dfrac{y}{3} = 1$ ⇒ 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 two lines are perpendicular to each other.

For the pair of linear equations given below, draw graphs and then state, whether the lines drawn are parallel or perpendicular to each other.

3x + 4y = 24

$\dfrac{x}{4} + \dfrac{y}{3} = 1$

Answer

First equation = 3x + 4y = 24

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) + 4y = 24 ⇒ y = 7.5

Let x = 0, then 3 $\times$ 0 + 4y = 24 ⇒ y = 6

Let x = 2, then 3 $\times$ 2 + 4y = 24 ⇒ y = 4.5

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 = $\dfrac{x}{4} + \dfrac{y}{3} = 1$

Step 1:

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

Let x = -1, then $\dfrac{-1}{4} + \dfrac{y}{3} = 1$ ⇒ y = 3.7

Let x = 0, then $\dfrac{0}{4} + \dfrac{y}{3} = 1$ ⇒ y = 3

Let x = 1, then $\dfrac{1}{4} + \dfrac{y}{3} = 1$ ⇒ y = 2.2

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.

Hence, the two lines are parallel to each other.

On the same graph paper, plot the graph of y = x - 2, y = 2x + 1 and y = 4 from x = -4 to 3.

Answer

The given equations are y = x - 2, y = 2x + 1 and y = 4.

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

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, plot the graphs of y = 2x - 1, y = 2x and y = 2x + 1 from x = - 2 to x = 4. Are the graphs (lines) drawn parallel to each other ?

Answer

The given equations are y = 2x - 1, y = 2x and y = 2x + 1.

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

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

Yes, the lines drawn are parallel to each other.

The graph of 3x + 2y = 6 meets the x-axis at point P and the y-axis at point Q. Use the graphical method to find the co-ordinates of points P and Q.

Answer

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) + 2y = 6 ⇒ y = 6

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

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

The graph intersects the x-axis at P(2,0).

The graph intersects the y-axis at Q(0,3).

Hence, the coordinates of P = (2, 0) and Q = (0, 3).