The graphs of the equations $6 x - 2 y + 9 = 0 and 3 x - y + 12 = 0$ are two lines which are

(a) coincident
(b) parallel
(c) intersecting exactly at one point
(d) perpendicular to each other

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Solution

The correct option is (b).

The given equations are as follows:
$6 x - 2 y + 9 = 0 and 3 x - y + 12 = 0$
They are of the following form:
a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0
Here, a₁ = 6, b₁ = −2, c₁ = 9 and a₂ = 3, b₂ = −1 and c₂ = 12
∴ $\frac{a_{1}}{a_{2}} = \frac{6}{3} = \frac{2}{1}, \frac{b_{1}}{b_{2}} = \frac{- 2}{- 1} = \frac{2}{1} and \frac{c_{1}}{c_{2}} = \frac{9}{12} = \frac{3}{4}$
∴ $\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}$
The given system has no solution.
Hence, the lines are parallel.

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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 $6 x - 2 y + 9 = 0 and 3 x - y + 12 = 0$ are two lines which are

(a) coincident
(b) parallel
(c) intersecting exactly at one point
(d) perpendicular to each other