A variable line $\frac{x}{a}+\frac{y}{b}=1$ is such that $a+b=4$. The locus of the mid-point of the portion of the line intercepted between the axes is

• A

  $x+y=4$
• B

  $x+y=8$
• C

  $x+y=1$
• Correct answer
  D

  $x+y=2$

Solution

We have, equation of line $\frac{x}{a}+\frac{y}{b}=1$. Let $P(h, k)$ be the required point. Now, coordinates $A$ and $B$ are $(a, 0)$ and $(0, b)$ respectively. Since, $P$ is mid-point of $A B$. Therefore, $(h, k)=\left(\frac{a+0}{2}, \frac{0+b}{2}\right)$ $\Rightarrow \quad h=\frac{a}{2}$ and $k=\frac{b}{2}$ $\Rightarrow \quad a=2 h$ and $b=2 k$ Now, it is given that $$ \begin{aligned} & a+b=4 \\ \Rightarrow & 2 h+2 k=4 \\ \Rightarrow & h+k=2 \end{aligned} $$ So, locus of $P$ is $x+y=2$

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