AP ICET 2015 Paper

Given : $\left(-4,5\right)$ is the coordinate of one vertex and $7\mathit{x}-\mathit{y}+8=0$ is the equation of one of the diagonals of the square.
We know that the diagonals of a square bisect at right angle.∴ Equation of the other diagonal can be written as,
$\mathit{x}+7\mathit{y}+\mathit{k}=0$
We can check that $\left(-4,5\right)$ does not lie on
$7\mathit{x}-\mathit{y}+8=0$ as $7\times \left(-4\right)-\left(5\right)+8\ne 0$
It means $\left(-4,5\right)$ must lie on the other diagonal.
$\therefore \mathit{–}4+7\times 5+\mathit{k}=0$
$\Rightarrow \mathit{k}=\mathit{–}31$
∴ Equation of the other diagonal is $\mathit{x}+7\mathit{y}-31=0$