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