Question

One diagonal of a square is the portion of the line $7x+5y=35$ intercepted by the other diagonal.

Answer 1

Slope of diagonal $AB=-7/5$ and hence slop of diagonal $CD$ which is perpendicular to $AB$ is $5/7$.
$\therefore \tan \theta =\frac{5}{7}\Rightarrow \sin \theta =\frac{5}{\sqrt{74}},\cos \theta =\frac{7}{\sqrt{74}}$
Also mid-point $P$ of $AB$ is $(5/2,7/2)$.
Also $AB=CD=\sqrt{25+49}=\sqrt{74}$
The diagonal $CD$ passes through $P$, its equation is
$\frac{x-5/2}{7/\sqrt{74}}=\frac{y-7/2}{5/\sqrt{74}}=\begin{array}{c}r\\ C\end{array},\begin{array}{c}-r\\ D\end{array}$
where $r=\frac{1}{2}CD=\frac{1}{2}\sqrt{74}$
$\therefore C=(\frac{5}{2}+\frac{7}{2},\frac{7}{2}+\frac{5}{2})=(6,6)$
$D=(\frac{5}{2}-\frac{7}{2},\frac{7}{2}-\frac{5}{2})=(-1,1)$.