Question

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

• A. $\frac{1}{a}+\frac{1}{d}=\frac{1}{b}+\frac{1}{c}$
• B. $\frac{1}{b}+\frac{1}{d}=\frac{1}{c}+\frac{1}{a}$
• C. $\frac{1}{c}+\frac{1}{d}=\frac{1}{a}+\frac{1}{b}$
• D. none of these

Answer 1

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

$\Rightarrow \frac{x}{a}+\frac{y}{b}=1\to 1$

& $\frac{x}{b}+\frac{y}{a}=1\to 2$

are the given lines

The point of intersection

$\Rightarrow 1-2$

$\Rightarrow x(\frac{1}{a}-\frac{1}{b})+y(\frac{1}{b}-\frac{1}{a})=0$

$\Rightarrow x(\frac{1}{a}-\frac{1}{b})=y(\frac{1}{a}-\frac{1}{b})$

$\therefore x=y$ (provided $a\ne b$)

$\therefore$ Point of intersection is:

$\frac{x}{a}+\frac{x}{b}=1$

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

$\Rightarrow x=y=\frac{ab}{a+b}$

Since this point lies on $\frac{x}{c}+\frac{y}{d}=1$

$\Rightarrow \frac{ab}{c(a+b)}+\frac{ab}{d(a+b)}=1$

$\Rightarrow \frac{ab}{a+b}(\frac{1}{c}+\frac{1}{d})=1$

$\Rightarrow \frac{a+b}{ab}=\frac{1}{c}+\frac{1}{d}$

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

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