InterviewSolution
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InterviewSolution
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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 show that \(\frac{1}{p^2}\) = \(\frac{1}{a^2}\) + \(\frac{1}{b^2}\) |
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Answer» ∵ Length of perpendicular from point (x1, y1) to line a\(x\) + by + c = 0 = \(\frac{|ax_1+by_1+c|}{\sqrt{a^2+b^2}}\) ∴ Length of perpendicular from (0, 0) to \(\frac{x}{a}\) + \(\frac{y}{b}\) = 1 ⇒ \(\frac{\big|\frac{1}{a}\times0+\frac{1}{b}\times0-1\big|}{\sqrt{\frac{1}{a^2}+\frac{1}{b^2}}}\) = p ⇒ \(\frac{|-1|}{\sqrt{\frac{1}{a^2}+\frac{1}{b^2}}}\) = p ⇒ \(\frac{1}{p}\) = \(\sqrt{\frac{1}{b^2}+\frac{1}{b^2}}\) ⇒ \(\frac{1}{p^2}\) = \(\frac{1}{a^2}\) + \(\frac{1}{b^2}\) |
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