The graphs of the equations $2 \mathrm{x}+3 \mathrm{y}-2=0$ and $\mathrm{x}-2 \mathrm{y}-8=0$ are two lines which are (a) coincident (b) parallel (c) intersecting exactly at one point (d) perpendicular to each other

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The graphs of the equations $2 \mathrm{x}+3 \mathrm{y}-2=0$ and $\mathrm{x}-2 \mathrm{y}-8=0$ 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 $2 \mathrm{x}+3 \mathrm{y}-2=0$ and $\mathrm{x}-2 \mathrm{y}-8=0$ are two lines which are
(a) coincident
(b) parallel
(c) intersecting exactly at one point
(d) perpendicular to each other

Answer:

Solution:
The given system of equations are as follows:
$2 x+3 y-2=0$ and $x-2 y-8=0$
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, $a_{1}=2, b_{1}=3, c_{1}=-2$ and $a_{2}=1, b_{2}=-2$ and $c_{2}=-8$
$\therefore \frac{a_{1}}{a_{2}}=\frac{2}{1}, \frac{b_{1}}{b_{2}}=\frac{3}{-2}$ and $\frac{c_{1}}{c_{2}}=\frac{-2}{-8}=\frac{1}{4}$
$\therefore \frac{a_{1}}{a_{2}} \neq \frac{h_{1}}{b_{2}}$
Given system has a unique solution.
As a result, the lines intersect exactly at one point.