Question

If two sides of a square are along $5x-12y+26=0$ and $5x-12y-65=0$ then find its area (in sq.units).

Answer 1

$5x-12y+26=0$

$5x-12y-65=0$

The lines are parallel .

Distance between them $=\frac{|26-(-65)|}{\sqrt{5^2+{12}^2}}$

$=\frac{91}{13}$

$=7$

$=$ side of the square.

$\therefore$ Area $=7^2$

$=49$ sq.units.