Question

Heyo! ʕ´•ᴥ•`ʔ

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.

Answer 2

Given that,

the intercepts are a and b.

Therefore, equation of the line is

$\displaystyle{\sf{ \dfrac{x}{a} + \frac{x}{b} = 1.}}$

And the perpendicular distance (d) of the line$\displaystyle{\sf{ax + by + c = 0}}$ from a point $\displaystyle{\sf{(x_1 , y_1)}}$ is given by

$\displaystyle{\boxed{\sf{\red{ d\:=\: \frac{|Ax_1 + By_1 + C|}{ \sqrt{A^{2} + B^{2}}}}}}}$

After that,

$\displaystyle{\sf{ \dfrac{x}{a} + \dfrac{x}{b} = 1.}}$

• Compare it with $\displaystyle{\sf{ax + by + c = 0}}$

$\displaystyle{\sf{ \dfrac{1}{a}x + \dfrac{1}{b}y - 1 = 0}}$

Therefore, a = $\displaystyle{\sf{ \frac{1}{a}}}$, b = $\displaystyle{\sf{ \frac{1}{b}}}$, c = $\displaystyle{\sf{ -1}}$

Again, the distance from origin (0, 0) to the line $\displaystyle{\sf{ \dfrac{x}{a} + \dfrac{x}{b} = 1.}}$ is p.

So, distance (d) = p &$\displaystyle{\sf{x_1 = 0\:,\: y_1 =0}}$

• Substitute the values in $\displaystyle{\sf{ d\:=\: \dfrac{|Ax_1 + By_1 + C|}{ \sqrt{A^{2} + B^{2}}}}}$ and simplify the equation.

:$\implies{\sf{p\:=\: \dfrac{ \bigg| (0) \dfrac{1}{a} + (0) \dfrac{1}{b} - 1 \bigg|}{ \sqrt{ \dfrac{1}{a}^{2} + \dfrac{1}{b}^{2}}}}}$

:$\implies{\sf{p\:=\: \dfrac{ | 0+ 0 -1|}{ \sqrt{ (\dfrac{1}{a})^{2} + (\dfrac{1}{b})^{2}}}}}$

:$\implies{\sf{p\:=\: \dfrac{ | -1|}{ \sqrt{ \dfrac{1}{a^{2}}+ \dfrac{1}{b^{2}}}}}}$

:$\implies{\sf{p\:=\: \dfrac{ 1}{ \sqrt{ \dfrac{1}{a^{2}}+ \dfrac{1}{b^{2}}}}}}$

:$\implies{\sf{ \dfrac{1}{p}\:=\: \sqrt{ \dfrac{1}{a^{2}}+ \dfrac{1}{b^{2}}}}}$

• Squaring both the sides.

:$\implies{\sf{ \bigg( \dfrac{1}{p} \bigg)^{2}\:=\: \bigg( \sqrt{ \dfrac{1}{a^{2}}+ \dfrac{1}{b^{2}}} \bigg)^{2}}}$

:$\large\implies{\boxed{\bf{\purple{ \dfrac{1}{p^{2}} = \dfrac{1}{a^{2}} + \dfrac{1}{b^{2}}}}}}$

Hence, proved!

And we are done! :D

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