MHT CET · Maths · Area Under Curves
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
- A 36 sq. units
- B 16 sq. units
- C 4 sq. units
- D 49 sq. units
Answer & Solution
Correct Answer
(D) 49 sq. units
Step-by-step Solution
Detailed explanation
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
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
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