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 $x+7 y=31$.