Question

If the length of the perpendicular drawn from the origin to the line whose intercepts on the axes are $a$ and $b$ be $p$, then

• A. $a^2+b^2=p^2$
• B. $a^2+b^2=\frac{1}{p^2}$
• C. $\frac{1}{a^2}+\frac{1}{b^2}=\frac{2}{p^2}$
• D. $\frac{1}{a^2}+\frac{1}{b^2}=\frac{1}{p^2}$

Answer 1

The correct option is D

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

Explanation for the correct option:

The intercepts on the axes of the straight line are $a$ and $b$.

Thus, the equation of the straight line can be given by $\frac{x}{a}+\frac{y}{b}=1$.

$$\begin{gathered} \Rightarrow \frac{bx+ay}{ab}=1 \\ \Rightarrow bx+ay=ab \\ \Rightarrow bx+ay-ab=0 \end{gathered}$$

The length of the perpendicular drawn from the origin to the straight line represented by $ax+by+c=0$ is $\frac{\left|c\right|}{\sqrt{a^2+b^2}}$.

Thus, the length of the perpendicular drawn from the origin to the given straight line is $p=\frac{\left|-ab\right|}{\sqrt{b^2+a^2}}$.

$$\begin{gathered} \Rightarrow p^2=\frac{a^2{b}^2}{{\left(\sqrt{b^2+a^2}\right)}^2} \\ \Rightarrow \frac{1}{a^2}+\frac{1}{b^2}=\frac{1}{p^2} \end{gathered}$$

Hence, option (D) is the correct answer.