Question

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

We know,

$\sf Equation \: of \: line \: making \: intercepts \: on \: x \: and \: y-axis,$

= x/a + y/b = 1

• Comparing with Ax + By + C = 0,
• A = 1/x
• B = 1/b
• C = -1

Also,

$\sf Perpendicular \: Distance \: of \: line \: Ax + By + C = 0 \: from \: the \: point \: (x,y),$

$D = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2+b^2} }$

Where,

• (x₁, y₁) = (0,0)  (Given)
• D = p (Given)
• A = 1/x
• B = 1/b
• C = -1

Substituting values,

$\sf D = \dfrac{|Ax_1 + By_1 + C|}{\sqrt{A^2+b^2}} \\ \\ \sf p = \dfrac{|\frac{1}{a} \times 0 + \frac{1}{b} \times 0 + (-1)|}{\sqrt{(\frac{1}{a})^2+(\frac{1}{b})^2}} \\ \\ \sf p = \dfrac{|-1|}{\sqrt{\frac{1}{a^2}+\frac{1}{b^2}}} \\ \\ \sf p = \dfrac{1}{\sqrt{\frac{1}{a^2}+\frac{1}{b^2}}}$

Squaring both sides,

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

Taking reciprocal,

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

Hence proved !