Question

If $\frac{x}{c}+\frac{y}{d}=1$ be any line passing through point of intersection of lines $\frac{x}{a}+\frac{y}{b}=1\text{and}\frac{x}{b}+\frac{y}{a}=1$ then

• A. $\frac{1}{c}-\frac{1}{d}=\frac{1}{a}-\frac{1}{b}$
• B. $\frac{1}{c}+\frac{1}{d}=\frac{1}{a}+\frac{1}{b}$
• C. $\frac{1}{c}-\frac{1}{b}=\frac{1}{a}-\frac{1}{d}$
• D. $\frac{1}{d}-\frac{1}{c}=\frac{1}{a}-\frac{1}{b}$

Answer 1

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

$\frac{x}{c}+\frac{y}{d}=1,\frac{x}{a}+\frac{y}{b}=1$ and $\frac{x}{b}+\frac{y}{a}=1$ are concurrent,
$\Rightarrow \left|\begin{array}{ccc}1& \frac{1}{c}& \frac{1}{d}\\ 1& \frac{1}{a}& \frac{1}{b}\\ 1& \frac{1}{b}& \frac{1}{a}\end{array}\right|=0\Rightarrow \frac{1}{c}+\frac{1}{d}=\frac{1}{a}+\frac{1}{b}$