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

Answer 1

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

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

$bx + ay = ab$

$bx + ay – ab = 0 ………………..(1)$

The perpendicular distance (d) of a line$Ax + By + C = 0$ from a point $(x_1, y_1)$ is given by

$d=\frac{\left|Ax_1+By_1+C\right|}{\sqrt{A^2+B^2}}$
On comparing equation (1) to the general equation of line $Ax + By + C = 0$, we obtain $A = b, B = a,$ and $C = -ab.$
Therefore, if p is the length of the perpendicular from point $(x_1, y_1) = (0, 0)$ to line (1), we obtain

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

$⇒ p=\frac{\left|-ab\right|}{\sqrt{a^2+b^2}}$
On squaring both sides, we obtain

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

$⇒ p^2(a^2+b^2)=a^2b^2$

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

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

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