The graphs of the equations 6x - 2y + 9 = 0 and 3x - y + 12 = 0 are two lines which are (a) coincident (b) parallel (c) intersecting exactly at one point (d) perpendicular to each other

Home » The graphs of the equations 6x – 2y + 9 = 0 and 3x – y + 12 = 0 are two lines which are (a) coincident (b) parallel (c) intersecting exactly at one point (d) perpendicular to each other

The graphs of the equations 6x – 2y + 9 = 0 and 3x – y + 12 = 0 are two lines which are
(a) coincident
(b) parallel
(c) intersecting exactly at one point
(d) perpendicular to each other

The graphs of the equations 6x – 2y + 9 = 0 and 3x – y + 12 = 0 are two lines which are
(a) coincident
(b) parallel
(c) intersecting exactly at one point
(d) perpendicular to each other

Answer: (b) parallel

Solution:
The given system of equations are as follows:
$6 x-2 y+9=0$ and $3 x-y+12=0$
They are of the following form:
$a_{1} x+b_{1} y+c_{1}=0$ and $a_{2} x+b_{2} y+c_{2}=0$
Here, $a_{1}=6, b_{1}=-2, c_{1}=9$ and $a_{2}=3, b_{2}=-1$ and $c_{2}=12$
$\therefore \frac{a_{1}}{a_{2}}=\frac{6}{3}=\frac{2}{1}, \frac{b_{1}}{b_{2}}=\frac{-2}{-1}=\frac{2}{1}$ and $\frac{c_{1}}{c_{2}}=\frac{9}{12}=\frac{3}{4}$
$\therefore \frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}$
The given system has no solution.
As a result, the lines are parallel.