Question

A variable straight line is drawn through the point of intersection of the straight lines $\frac{x}{2}+\frac{y}{3}=1$ and $\frac{x}{3}+\frac{y}{2}=1$ and meets the coordinate axes at $A$ and $B$. Then the locus of the mid-point of $AB$ is

• A. $3x+3y-5xy=0$
• B. $3x-3y-5xy=0$
• C. $5x+5y-3xy=0$
• D. $5x-5y-3xy=0$

Answer 1

The correct option is A $3x+3y-5xy=0$

Given, $\frac{x}{2}+\frac{y}{3}=1...\left(1\right)$
$\frac{x}{3}+\frac{y}{2}=1...\left(2\right)$
Solving $\left(1\right)$ and $\left(2\right)$, we get
$x=\frac{6}{5}$ and $y=\frac{6}{5}$
$\therefore$ Intersection point $P$ is $(\frac{6}{5},\frac{6}{5})$
Now, variable line passing through $P$ meets the axes at points $A$ and $B$.


Let the mid-point of $AB$ be $M(h,k)$ whose locus is to be found.
Then coordinates of $A$ and $B$ are $(2h,0)$ and $(0,2k)$ respectively.
Clearly, points $A,B$ and $P$ are collinear.
Then $\Delta =\left|\begin{array}{ccc}2h& 0& 1\\ 0& 2k& 1\\ \frac{6}{5}& \frac{6}{5}& 1\end{array}\right|=0$
$\Rightarrow 2h(2k-\frac{6}{5})+1(0-2k\cdot \frac{6}{5})=0$
$\Rightarrow 4hk-\frac{12}{5}h-\frac{12}{5}k=0$
$\Rightarrow 3h+3k-5hk=0$
Hence, the locus of $M$ is $3x+3y-5xy=0$