Question

Line $L$has intercepts $a\&b$ on the coordinate axes. When the axes are rotated through a given angle, keeping the origin fixed, the same line $L$ has intercepts $p\&q$, then

• A. $a^2+b^2=p^2+q^2$
• B. $a^2+p^2=b^2+q^2$
• C. $\frac{1}{a^2}+\frac{1}{b^2}=\frac{1}{p^2}+\frac{1}{q^2}$
• D. none of these

Answer 1

The correct option is C

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

Explanation for the correct option:

Given that the line $L$ has intercepts $a\&b$.

$\Rightarrow \frac{x}{a}+\frac{y}{b}=1...\left(i\right)$

The axes are rotated through angle $\alpha$ at the origin then the same line $L$ has intercepts $p\&q$

$\Rightarrow \frac{x}{p}+\frac{y}{q}=1...\left(ii\right)$

We Know that line equation is $ax+by+c=0$and the point having the coordinate $\left(x_1,y_1\right)$ then the distance is

$\Rightarrow d=\left|\frac{a{x}_1+b{y}_1+c}{\sqrt{a^2+b^2}}\right|$

Now consider the equation (i) and find the value of $d_1$

$$\begin{gathered} \Rightarrow d_1=\frac{\left|0+0-1\right|}{\sqrt{{\displaystyle \frac{1}{a^2}}+{\displaystyle \frac{1}{b^2}}}} \\ \Rightarrow d_1=\frac{-1}{\sqrt{{\displaystyle \frac{1}{a^2}+\frac{1}{b^2}}}} \end{gathered}$$

Also, consider the equation $\left(ii\right)$and find the value of $d_2$

$$\begin{gathered} \Rightarrow d_2=\frac{\left|0+0-1\right|}{\sqrt{{\displaystyle \frac{1}{p^2}+\frac{1}{q^2}}}} \\ \Rightarrow d_2=\frac{-1}{\sqrt{{\displaystyle \frac{1}{p^2}+\frac{1}{q^2}}}} \end{gathered}$$

Equating $d_1\&d_2$

$$\begin{gathered} \Rightarrow \frac{-1}{\sqrt{{\displaystyle \frac{1}{a^2}+\frac{1}{b^2}}}}=\frac{-1}{\sqrt{{\displaystyle \frac{1}{p^2}+\frac{1}{q^2}}}} \\ \Rightarrow \frac{1}{\frac{1}{a^2}+\frac{1}{b^2}}=\frac{1}{\frac{1}{p^2}+\frac{1}{q^2}} \\ \Rightarrow \frac{1}{a^2}+\frac{1}{b^2}=\frac{1}{p^2}+\frac{1}{q^2} \end{gathered}$$

Hence, option (C) is the correct answer.