Question

The equation to the line passing through the intersection of $\frac{x}{a}+\frac{y}{b}=1,\frac{x}{b}+\frac{y}{a}=1$, where $ab=a+b$ and $(1,2)$ is

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

Answer 1

Answer: (A)

$x=1$

The equations can be written as

$bx+ay=ab$ ........(i) and

$ax+by=ab$ .........(ii)

Solving them simultaneously and eliminating $x$, we get

$y=\frac{ab(a-b)}{a^2-b^2}=\frac{(a+b)(a-b)}{(a+b)(a-b)}$ ....$(\because ab=a+b)$

$=1$

Substituting $y=1$ in equation (i), we get

$bx+a=ab$

$\Rightarrow x=1$

Therefore, the point of intersection is

$(x_1,y_1)$ is $(1,1)$

The line passes through $(x_2,y_2)$ which is $(1,2)$ ....$\because x_1=x_2$

The line is parallel to $y$-axis.

So, the equation of the line is $(x-x_1)=0$

$\Rightarrow x-1=0$

$\Rightarrow x=1$