Question

Let $a$ and $b$ be non-zero reals such that $a\ne b$. Then the equation of the line passing through the origin and the point of intersection of $\frac{x}{a}+\frac{y}{b}=1$ and $\frac{x}{b}+\frac{y}{a}=1$ is

• A. $ax+by=0$
• B. $bx+ay=0$
• C. $y-x=0$
• D. $x+y=0$

Answer 1

Answer: (C)

Given equation of lines are
$\frac{x}{a}+\frac{y}{b}=1$ ...(i)
and $\frac{x}{b}+\frac{y}{a}=1$...(ii)
$\Rightarrow bx+ay=ab$ ...(iii)
and $ax+by=ab$ ...(iv)
Solving (iii) and (iv), we get
$(a^2-b^2)y=a^{2}b-a{b}^2=ab(a-b)$
$\Rightarrow y=\frac{ab}{a+b}$
Substituting the value of y in (iii}, we get
$bx+a\cdot \frac{ab}{a+b}=ab\Rightarrow bx=ab-\frac{a^{2}b}{a+b}$
$\Rightarrow bx=\frac{a{b}^2}{a+b}\Rightarrow x=\frac{ab}{a+b}$
$\therefore$ Point of intersection is $(\frac{ab}{a+b},\frac{ab}{a+b})$
Since equation of the line passing through origin is $y=mx$
$\therefore$ When it pass through $(\frac{ab}{a+b},\frac{ab}{a+b})$ then, we get $m=1$
Hence, required equation of line is,
$y-x=0$