Question

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

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

Answer 1

The correct option is B

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

Explanation for the correct answer:

Finding the intercepts:

Line $L$ has intercepts $a$ and $b$, then the equation of the line be $\frac{x}{a}+\frac{y}{b}=1$$\to \left(i\right)$

After rotating the axis, the same line $L$ has intercepts changed into $p$ and $q$, then the equation of the line be $\frac{x}{p}+\frac{y}{q}=1$$\to \left(ii\right)$

Keeping the origin fixed that is the perpendicular distance between the origin remains the same

$\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$

So

$$\begin{gathered} \sqrt{{\left(\frac{1}{a}-0\right)}^2+{\left(\frac{1}{b}-0\right)}^2}=\sqrt{{\left(\frac{1}{p}-0\right)}^2+{\left(\frac{1}{q}-0\right)}^2} \\ \sqrt{{\left(\frac{1}{a}\right)}^2+{\left(\frac{1}{b}\right)}^2}=\sqrt{{\left(\frac{1}{p}\right)}^2+{\left(\frac{1}{q}\right)}^2} \\ \text{squaring on both sides} \\ \frac{1}{a^2}+\frac{1}{b^2}=\frac{1}{p^2}+\frac{1}{q^2} \end{gathered}$$

Hence, option (B) is the correct answer.