If (-4,5) is one vertex and 7 x-y+8=0 is one diagonal of a square,...

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 $y=m x+c$ where $m$ is the slope of the line and $c$ is the $\mathrm{y}$-intercept.
Since this line passes through $(-4,5)$


Since this line is at right angle to the line
$7 x-y+8=0$ or $y=7 x+8$, having slope $=7$,
$\therefore 7 \times \mathrm{m}=-1$ or $\mathrm{m}=\frac{-1}{7}$
Putting this value of $m$ in equation (i) we get
$5=-4 \times\left(\frac{-1}{7}\right)+c$
or $5=\frac{4}{7}+c \quad$ or $\quad c=5-\frac{4}{7}=\frac{31}{7}$
Hence equation of the other diagonal is
$y=-\frac{1}{7} x+\frac{31}{7}$ or $7 y=-x+31$ or $x+7 y-31=0$ or $x+7 y=31$