Question

The following lines are the sides of squares $\left(1\right)$ $2x-y-4=0$ $\left(2\right)$ $x+2y-7=0$
$\left(3\right)$ $x+2y+3=0$
What will be the equation of the fourth side of the squares ?

Answer 1

Distance between parallel sides.
$=\frac{3+7}{\sqrt{5}}=\frac{10}{\sqrt{5}}=2\sqrt{5}$
Let the fourth sides parallel to $2x-y-4=0$ be $2x-y+k=0$
Distance between these two
$=\frac{k+4}{\sqrt{5}}=\pm 2\sqrt{5}$
$\therefore k=-4\pm 10=-14$ or $6$
Hence the other two fourth sides are
$2x-y-14=0$
Or $2x-y+6=0$