If P is the length of perpendicular from the origin to the line whose intercepts on the axes a and b, then show that, $\frac{1}{p^{2}} = \frac{1}{a^{2}} + \frac{1}{b^{2}}$ .

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Solution

It is known that the equation of a line whose intercepts on the axes are a and b is

$\frac{x}{a} + \frac{y}{b} = 1$

Or $b x + a y = a b$

Or $b x + a y - a b = 0$ ___(1)

The perpendicular distance (d) of a line $A x + B y + C = 0$ from a point $(x_{2}, y_{1})$ is given by

$d = \frac{| A x_{1} + B y_{1} + C |}{\sqrt{A^{2} + B^{2}}}$

On comparing equation (1) to the general equation of line $A x - B y + C = 0$ , we obtain $A = b, B = a,$ and $C = - a b$

Therefore, if p is the length of the perpendicular from point $(x_{1}, y_{1}) = (0, 0)$ to line (1),

We obtain

$p = \frac{| A (0) + B (0) - a b |}{\sqrt{b^{2} + a^{2}}}$

$\Rightarrow p = \frac{| - a b |}{\sqrt{b^{2} + a^{2}}}$

On squaring both sides, we obtain

$p^{2} = \frac{(- a b)^{2}}{a^{2} + b^{2}}$

$\Rightarrow p^{2} (a^{2} + b^{2}) = a^{2} b^{2}$

$\Rightarrow \frac{a^{2} + b^{2}}{a^{2} b^{2}} = \frac{1}{p^{2}}$

$\Rightarrow \frac{1}{p^{2}} = \frac{1}{a^{2}} + \frac{1}{b^{2}}$

Hence, we showed that $\frac{1}{p^{2}} = \frac{1}{a^{2}} + \frac{1}{b^{2}}$

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(I.I.T. 1990)

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

If P is the length of perpendicular from the origin to the line whose intercepts on the axes a and b, then show that, $\frac{1}{p^{2}} = \frac{1}{a^{2}} + \frac{1}{b^{2}}$ .