VITEEE 2021 Paper

One vertex of square is $(-4,5)$ and equation of one diagonal is $7x - y + 8 = 0$ Diagonal of a square are perpendicular and bisect each other Let the equation of the other diagonal be $y = m x$ $+c$ where $m$ is the slope of the line and $c$ is the $y-$ intercept. Since this line passes through $(-4,5)$ ∴ $5 = -4m + c \dots$ (i) Since this line is at right angle to the line $7x - y + 8 = 0$ or $y = 7x + 8$, having slope $=7$, ∴ $7 \times m = -1$ or $m = \frac{-1}{7}$ Putting this value of $m$ in equation (i) we get $5 = -4 \times \left(\frac{-1}{7}\right) + c$ or $5 = \frac{4}{7} + c$ or $c = 5 - \frac{4}{7} = \frac{31}{7}$ Hence equation of the other diagonal is $y = -\frac{1}{7} x + \frac{31}{7}$ or $7y = -x + 31$ or $x + 7y - 31 = 0$ or $x + 7y = 31$.