If two sides of a square are $4 x+3 y-20=0$ and $4 x+3 y+15=0$, then the area of the square is
If two sides of a square are $4 x+3 y-20=0$ and $4 x+3 y+15=0$, then the area of the square is
36 sq. units
16 sq. units
4 sq. units
49 sq. units
Solution
Given equations of lines are $4 x+3 y-20=0$ and $4 x+3 y+15=0$
Slope of $4 x+3 y-20=0$ is $\frac{-4}{3}$.
Slope of $4 x+3 y+15=0$ is $\frac{-4}{3}$
$\therefore \quad$ Lines are parallel.
$\therefore \quad$ Distance between two parallel lines
$\begin{aligned}
& =\left|\frac{c_1-c_2}{\sqrt{a^2+b^2}}\right| \\
& =\left|\frac{-20-15}{\sqrt{4^2+3^2}}\right| \\
& =\frac{35}{5}=7 \text { units }
\end{aligned}$
$\therefore \quad$ Area of square $=7^2=49$ sq. units