MHT CET · Maths · Straight Lines
The length of the perpendicular from the point \(P(a, b)\) to the line \(\frac{x}{a}+\frac{y}{b}=1\) is
- A \(\mid \frac{\sqrt{a^{2}+b^{2}}}{a b}\) units
- B \(\left|\frac{a b}{\sqrt{a^{2}+b^{2}}}\right|\) units
- C \(\left|\frac{b^{2}}{\sqrt{a^{2}+b^{2}}}\right|\) units
- D \(\left|\frac{a^{2}}{\sqrt{a^{2}+b^{2}}}\right|\) units
Answer & Solution
Correct Answer
(B) \(\left|\frac{a b}{\sqrt{a^{2}+b^{2}}}\right|\) units
Step-by-step Solution
Detailed explanation
We have line \(b x+a y-a b=0\)
Length of \(\perp e r\) from \(P(a, b)\) on the given line is
\(
\left|\frac{b a+a b-a b}{\sqrt{b^{2}+a^{2}}}\right|=\left|\frac{a b}{\sqrt{a^{2}+b^{2}}}\right|
\)
Length of \(\perp e r\) from \(P(a, b)\) on the given line is
\(
\left|\frac{b a+a b-a b}{\sqrt{b^{2}+a^{2}}}\right|=\left|\frac{a b}{\sqrt{a^{2}+b^{2}}}\right|
\)
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