Question

The point of intersection of the lines $\frac{x}{a}+\frac{y}{b}=1$ and $\frac{x}{b}+\frac{y}{a}=1$ lines on :

• A. $x-y=0$
• B. $x+y=0$
• C. $x+y=1$
• D. $x-2y=0$

Answer 1

The correct option is A $x-y=0$

$\text{Given equations of lines}$

$\frac{x}{a}+\frac{y}{b}=1\text{and}\frac{x}{b}+\frac{y}{d}=1$

On simplifying we get

$\begin{array}{c}bx+ay=ab\times a⟶\left(1\right)\\ \text{and}ax+by=ab\times b-\left(2\right)\end{array}$

$\text{Put}y\text{in}\left(1\right)\text{we get}$

$bx+a\left(\frac{ab}{a+b}\right)=ab$

$\Rightarrow bx=\frac{ab(a+b)-a^2}{a+b}$

$\Rightarrow x=\frac{a^2+ab-ab}{a+b}=\frac{a^2}{a+b}$

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

$x=a\left(\frac{y}{b}\right)\Rightarrow x=y\left(\frac{a}{b}\right)$