Two sides of a square are along the lines $5 x-12 y+39=0$ and $5 x-12 y+78=0$, then area of the square is
Two sides of a square are along the lines $5 x-12 y+39=0$ and $5 x-12 y+78=0$, then area of the square is
9 sq. units.
$\frac{1}{3}$ sq. units.
18 sq. units.
3 sq. units.
Solution
Given equations of lines are $5 x-12 y+39=0$ and $5 x-12 y+78=0$.
Slope of $5 x-12 y+39=0$ is $\frac{5}{12}$
Slope of $5 x-12 y+78=0$ is $\frac{5}{12}$
$\therefore \quad$ Lines are parallel.
$\therefore \quad$ Distance between two parallel lines $=\left|\frac{\mathrm{c}_1-\mathrm{c}_2}{\sqrt{\mathrm{a}^2+\mathrm{b}^2}}\right|$
$\begin{aligned}
& =\left|\frac{39-78}{\sqrt{5^2+(-12)^2}}\right| \\
& =3 \text { units }
\end{aligned}$
$\therefore \quad$ Side of the square $=3$ units
$\therefore \quad$ Area of square $=3^2=9$ sq. units