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

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

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.