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Question:
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If $p$ is the length of the perpendicular drawn from the
origin to the line $\frac{x}{a}+\frac{y}{b}=1$, then which one of the following
is correct?
Options:
origin to the line $\frac{x}{a}+\frac{y}{b}=1$, then which one of the following
is correct?
Solution:
2006 Upvotes
Verified Answer
The correct answer is:
$\frac{1}{\mathrm{p}^{2}}=\frac{1}{\mathrm{a}^{2}}+\frac{1}{\mathrm{~b}^{2}}$
Given equation of line is $\frac{x}{a}+\frac{y}{b}=1$
$\Rightarrow b x+a y-a b=0$
$p=\frac{|b \cdot 0+a \cdot 0-a b|}{\sqrt{b^{2}+a^{2}}}$
$p=\frac{a b}{\sqrt{b^{2}+a^{2}}}$
on squaring both side, we get
$p^{2}=\frac{a^{2} b^{2}}{a^{2}+b^{2}} \Rightarrow \frac{1}{p^{2}}=\frac{a^{2}+b^{2}}{a^{2} b^{2}}=\frac{1}{b^{2}}+\frac{1}{a^{2}}$
$\Rightarrow b x+a y-a b=0$
$p=\frac{|b \cdot 0+a \cdot 0-a b|}{\sqrt{b^{2}+a^{2}}}$
$p=\frac{a b}{\sqrt{b^{2}+a^{2}}}$
on squaring both side, we get
$p^{2}=\frac{a^{2} b^{2}}{a^{2}+b^{2}} \Rightarrow \frac{1}{p^{2}}=\frac{a^{2}+b^{2}}{a^{2} b^{2}}=\frac{1}{b^{2}}+\frac{1}{a^{2}}$
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