If $(-4,5)$ is one vertex and $7 x-y+8=0$ is one diagonal of a square, then the equation of second diagonal is

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