Question

A straight line through the point of intersection of the lines $x+2y=4$ and $2x+y=4$ meets the coordinates axes at $A$ and $B$. The locus of the mid-point of $AB$ is

• A. $3(x+y)=2xy$
• B. $2(x+y)=3xy$
• C. $2(x+y)=xy$
• D. $x+y=3xy$

Answer 1

The correct option is B $2(x+y)=3xy$

Given lines are,
$x+2y=4$ ..... (i)
and $2x+y=4$ ..... (ii)
Solving Eqs. (i) and (ii), we get
$x=\frac{4}{3},y=\frac{4}{3}$
Let $AB:\frac{x}{a}+\frac{y}{b}=1$ ..... (iii)
Meets $x$-axis at $A$ and $y$-axis at $B$.
If the straight line (iii) passes through the point $(\frac{4}{3},\frac{4}{3})$, then $\frac{4}{3a}+\frac{4}{3b}=1$
$\Rightarrow \frac{1}{a}+\frac{1}{b}=\frac{3}{4}$ ...... (iv)
Let the mid-point of $AB$ be $(h,k)$.
$\therefore h=\frac{a+0}{2},k=\frac{0+b}{2}$
$\Rightarrow a=2h,b=2k$
$\therefore$ From Eq. (iv), $\frac{1}{2h}+\frac{1}{2k}=\frac{3}{4}$
$\Rightarrow 2h+2k=3hk$
Replace $x,y$ from $h,k$ respectively, we get required locus
i.e., $2(x+y)=3xy$