Question

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If p is the length of perpendicular from the origin to the line whose intercepts on the axes are a and b, then show that 1/p² = 1/a² + 1/b².

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Answer 1

Answer:

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Answer 2

Step-by-step explanation:

Clearly , a right angle triangle is formed by the line and the coordinate axes

Let ABC be a triangle with A(a,0) ,B (0,0) and C ( 0,b )

Draw BM perpendicular to AC , such that BM = p

We have , a^2 + b^2 = (AC) ^2. { By Pythagoras theorem}....(1)

Now , Area of triangle ABC = (1/2) a b = (1/2) p • AC

→ ab = AC • p

→ a^2b^2 = (AC) ^2• p^2

→ (1/p)^2 = (a^2+b^2) / (ab)^2 { by equation (1) }

→ 1/p² = 1/a² + 1/ b²

Hence , proved