NIMCET 2013 Question Paper with solutions

Slope of line $2\mathit{x}-3\mathit{y}=7$ is $\frac{2}{3}$
Since, required line is parallel to this line, so slope of the line will be $\frac{2}{3}$
So, the mid-point of line segment joining the points $\left(1,3\right)$ and $\left(1,-7\right)$ is $\left(1,-2\right)$
∴ Required line is $\mathit{y}+2=\frac{2}{3}\left(\mathit{x}-1\right)$
$\Rightarrow 3\mathit{y}+6=2\mathit{x}-2$
$\Rightarrow 2\mathit{x}-3\mathit{y}-8=0$