Q79 / 180 · COMEDK 2015 · Mathematics

Let $a$ and $b$ be non-zero real such that $a \neq b$. 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
    $a x+b y=0$
  • B
    $b x+a y=0$
  • Correct answer
    C
    $y-x=0$
  • D
    $x+y=0$

Solution

Given, equation of lines are $\begin{aligned} & \frac{x}{a}+\frac{y}{b} &=1 & \ldots \text { (i) } \\ \text { and } & \frac{x}{b}+\frac{y}{a} &=1 & \ldots \text { (ii) } \\ \Rightarrow & b x+a y &=a b & \ldots \text { (iii) } \\ \text { and } & a x+b y &=a b & \ldots \text { (iv) } \end{aligned}$ Solving Eqs. (iii) and (iv), we get $$ \left(a^{2}-b^{2}\right) y=a^{2} b-a b^{2}=a b(a-b) \Rightarrow y=\frac{a b}{a+b} $$ Substituting the value of $y$ in Eq. (iii), we get $$ \begin{aligned} & b x+a\left(\frac{a b}{a+b}\right)=a b \Rightarrow b x=a b-\frac{a^{2} b}{a+b} \\ \Rightarrow & b x=\frac{a b^{2}}{a+b} \Rightarrow x=\frac{a b}{a+b} \end{aligned} $$ $\therefore$ Point of intersection is $\left(\frac{a b}{a+b}, \frac{a b}{a+b}\right)$. Since, equation of the line passing through origin is $y=m x$ $\therefore$ When it passes through $\left(\frac{a b}{a+b}, \frac{a b}{a+b}\right)$, then we get $$ m=1 $$ Hence, requiret equation of line is $y-x=0$.

Hence, the correct answer is option: 'C'.

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