# If p is the length of the perpendicular from the origin to the line $$x\over a$$ + $$y\over b$$ = 1, then prove that $$1\over p^2$$ = $$1\over a^2$$ + $$1\over b^2$$

## Solution :

The given line is bx + ay – ab = 0 ………….(i)

It is given that

p = Length of the perpendicular from the origin to line (i)

$$\implies$$ p = $$|b(0) + a(0) – ab|\over {\sqrt{b^2+a^2}}$$ = $$ab\over \sqrt{a^2+b^2}$$

$$\implies$$ $$p^2$$ = $$a^2b^2\over a^2+b^2$$ $$\implies$$ $$1\over p^2$$ = $$a^2+b^2\over a^2b^2$$ $$\implies$$ $$1\over p^2$$ = $$1\over a^2$$ + $$1\over b^2$$

Hence Proved.

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