Question

If $\frac{x}{c}+\frac{y}{d}=1$ is a line through the intersection of $\frac{x}{a}+\frac{y}{b}=1$ and $\frac{x}{b}+\frac{y}{a}=1$ and the lengths of the perpendiculars drawn from the origin to these lines are equal in lengths then

• A. $\frac{1}{a^2}+\frac{1}{b^2}=\frac{1}{c^2}+\frac{1}{d^2}$
• B. $\frac{1}{a^2}-\frac{1}{b^2}=\frac{1}{c^2}-\frac{1}{d^2}$
• C. $\frac{1}{a}+\frac{1}{b}=\frac{1}{c}+\frac{1}{d}$
• D. none

Answer 1

The correct option is A $\frac{1}{a^2}+\frac{1}{b^2}=\frac{1}{c^2}+\frac{1}{d^2}$

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

these lines have equal in length perpendiculars drawn from origin.

We know, length of perpendicular from origin of a line ax+bt+c=0 is given as

$d=|\frac{c}{\sqrt{a^2+b^2}}|$. Hence, Here

$\frac{1}{\sqrt{\frac{1}{c^2}+\frac{1}{d^2}}}=\frac{1}{\sqrt{\frac{1}{a^2}+\frac{1}{b^2}}}\Rightarrow \frac{1}{a^2}+\frac{1}{b^2}=\frac{1}{c^2}+\frac{1}{d^2}$

Option A