Given the linear equation 2x + 3y - 8 = 0, write another linear e…

Question

Given the linear equation $2x + 3y - 8 = 0$, write another linear equation in two variables such that the geometrical representation of the pair so formed is:

Intersecting lines.
Parallel lines.
Coincident lines.

✓

Answer

We have, $2x + 3y - 8 = 0$
Let another equation of line is:
$4x + 9y - 4 = 0$
Here, $a_1 = 2, b_1 = 3, c_1 = -8$
$a_2= 4, b_2 = 9, c_2 = -4$
Now, $\frac{\text{a}_1}{\text{a}_2}=\frac{2}{4}=\frac{1}{2}$
$\frac{\text{b}_1}{\text{b}_2}=\frac{3}{9}=\frac{1}{3}$
And $\frac{\text{c}_1}{\text{c}_2}=\frac{-8}{-4}=\frac{2}{1}$
$\therefore\frac{\text{a}_1}{\text{a}_2}\neq\frac{\text{b}_1}{\text{b}_2}$
$\therefore$ $2x + 3y - 8 = 0$ and $4x + 9y - 4 = 0$ intersect each other at one point.
Hence, required equation of line is $4x + 9y - 4 = 0$
We have, $2x + 3y - 8 = 0$
Let another equation of line is:
$4x + 6y - 4 = 0$
Here, $a_1 = 2, b_1 = 3, c_1 = -8$
$a_2= 4, b_2 = 9, c_2 = -4$
$$Now, $\frac{\text{a}_1}{\text{a}_2}=\frac{2}{4}=\frac{1}{2}$
$\frac{\text{b}_1}{\text{b}_2}=\frac{3}{6}=\frac{1}{2}$
And $\frac{\text{c}_1}{\text{c}_2}=\frac{-8}{-4}=\frac{2}{1}$
$\therefore\frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}\neq\frac{\text{c}_1}{\text{c}_2}$
$\therefore$ Lines are parallel to each other.
Hence, required equation of line is $4x + 6y - 4 = 0.$
Let another equation of line is:
$4x + 6y - 16 = 0$
Now, $\frac{\text{a}_1}{\text{a}_2}=\frac{2}{4}=\frac{1}{2}$
$\frac{\text{b}_1}{\text{b}_2}=\frac{3}{6}=\frac{1}{2}$
and $\frac{\text{c}_1}{\text{c}_2}=\frac{-8}{-16}=\frac{1}{2}$
$\therefore\frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}=\frac{\text{c}_1}{\text{c}_2}$
$\therefore$ Lines are coincident to each other.