Question

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

Answer 1

In the fig. the line makes intercept a and b on the axes, thus making a right triangle ABC with origin and axes. The perpendicular drawn from origin to the line meets the line at D.

Then in right triangle ACD,

• $\smallAD=\sqrt{AC^2-CD^2}=\sqrt{b^2-p^2}$

and in right triangle CBD,

• $\smallBD=\sqrt{BC^2-CD^2}=\sqrt{a^2-p^2}$

Method 1:-

We see triangles ACD and CBD are similar with ∠CAD = ∠BCD, ∠ACD = ∠CBD and ∠ADC = ∠CDB. Therefore,

$\small\longrightarrow\dfrac{AD}{AC}=\dfrac{CD}{CB}$

$\small\text{ \longrightarrow\dfrac{\sqrt{b^2-p^2}}{b}=\dfrac{p}{a} }$

Squaring both sides,

$\small\longrightarrow\dfrac{b^2-p^2}{b^2}=\dfrac{p^2}{a^2}$

By rule of alternendo (i),

$\small\longrightarrow\dfrac{b^2-p^2}{p^2}=\dfrac{b^2}{a^2}$

By rule of componendo (ii),

$\small\longrightarrow\dfrac{b^2-p^2+p^2}{p^2}=\dfrac{b^2+a^2}{a^2}$

$\small\longrightarrow\dfrac{b^2}{p^2}=\dfrac{a^2+b^2}{a^2}$

Dividing both sides by b²,

$\small\longrightarrow\dfrac{1}{p^2}=\dfrac{a^2+b^2}{a^2b^2}$

$\small\text{ \longrightarrow\underline{\underline{\dfrac{1}{p^2}=\dfrac{1}{a^2}+\dfrac{1}{b^2}}} }$

Method 2:-

Since $\smallAD=\sqrt{b^2-p^2}$ and $\smallBD=\sqrt{a^2-p^2},$ the length of AB,

$\small\longrightarrow AB=AD+BD=\sqrt{a^2-p^2}+\sqrt{b^2-p^2}\quad\dots(1)$

But in triangle ABC, applying Pythagoras' Theorem,

$\small\longrightarrow AB=\sqrt{AC^2+BC^2}=\sqrt{a^2+b^2}\quad\dots(2)$

From (1) and (2),

$\small\longrightarrow\sqrt{a^2-p^2}+\sqrt{b^2-p^2}=\sqrt{a^2+b^2}$

Squaring both sides,

$\small\longrightarrow\big(\sqrt{a^2-p^2}+\sqrt{b^2-p^2}\big)^2=\big(\sqrt{a^2+b^2}\big)^2$

$\small\longrightarrow a^2-p^2+b^2-p^2+2\sqrt{(a^2-p^2)(b^2-p^2)}=a^2+b^2$

$\small\longrightarrow a^2+b^2-2p^2+2\sqrt{(a^2-p^2)(b^2-p^2)}=a^2+b^2$

$\small\longrightarrow-2p^2+2\sqrt{(a^2-p^2)(b^2-p^2)}=0$

$\small\longrightarrow2\sqrt{(a^2-p^2)(b^2-p^2)}=2p^2$

$\small\longrightarrow\sqrt{(a^2-p^2)(b^2-p^2)}=p^2$

Squaring both sides,

$\small\longrightarrow(a^2-p^2)(b^2-p^2)=p^4$

$\small\longrightarrow a^2b^2-p^2(a^2+b^2)+p^4=p^4$

$\small\longrightarrow a^2b^2-p^2(a^2+b^2)=0$

$\small\longrightarrow a^2b^2=p^2(a^2+b^2)$

$\small\longrightarrow\dfrac{a^2b^2}{p^2}=a^2+b^2$

$\small\longrightarrow\dfrac{1}{p^2}=\dfrac{a^2+b^2}{a^2b^2}$

$\small\text{ \longrightarrow\underline{\underline{\dfrac{1}{p^2}=\dfrac{1}{a^2}+\dfrac{1}{b^2}}} }$

Hence Proved!