Question

On comparing the ratios $\frac{a_1}{a_2},\frac{b_1}{b_2}$ and $\frac{c_1}{c_2}$, and without drawing them, find out whether the lines representing the following pairs of linear equation intersect at a point, are parallel or coincide:

(i) $5x-4y+8=0$

$7x+6y-9=0$

(ii) $9x+3y+12=0$

$18x+6y+24=0$

(iii) $6x-3y+10=0$

$2x-y+9=0$

Answer 1

(i) $5x-4y+8=0$

$7x+6y-9=0$

$\because a=$ coefficient of $x,b=$ coefficient of $y$, and $c=$ constant term]

Here, $\frac{a_1}{a_2}=\frac{5}{7}$, $\frac{b_1}{b_2}=\frac{-4}{6}=\frac{-2}{3}$, $\frac{c_1}{c_2}=\frac{8}{-9}$

When $\frac{a_1}{a_2}\ne \frac{b_1}{b_2}$, then the given lines are Intersecting.

(ii) $9x+3y+12=0$

$18x+6y+24=0$

Here, $\frac{a_1}{a_2}=\frac{9}{18}=\frac{1}{2}$, $\frac{b_1}{b_2}=\frac{3}{6}=\frac{1}{2}$, and $\frac{c_1}{c_2}=\frac{12}{24}=\frac{1}{2}$

When $\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$, then the lines are Coincident.

(iii) $6x-3y+10=0$

$2x-y+9=0$

Here, $\frac{a_1}{a_2}=\frac{6}{2}=\frac{3}{1},$ $\frac{b_1}{b_2}=\frac{-3}{-1}=\frac{3}{1}$, and $\frac{c_1}{c_2}=\frac{10}{9}$

When $\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne \frac{c_1}{c_2}$, then the lines are Parallel.