Question

A variable straight line passes through the points of intersection of the lines $x+2y=1$ and $2x-y=1$ and meets the coordinate axes in $A$ and $B$. Find the locus of the middle point of $AB$..

• A. $x-10xy+3y=0$
• B. $x-xy+3y=0$
• C. $5x-10xy+3y=0$
• D. $x-10xy+y=0$

Answer 1

The correct option is A $x-10xy+3y=0$

Given lines are $x+2y=1$ and $2x-y=1$

The point of intersection of the two given lines is $(\frac{3}{5},\frac{1}{5})$.

Let the midpoint of $AB$ be $(x,y)$. So we have

$A=(2x,0)$

$B=(0,2y)$

Using the intercept form for the variable line and substituting $(3/5,1/5)$ in it we have,

$\frac{\left(3/5\right)}{2x}+\frac{\left(1/5\right)}{2y}=1$

Or in simpler terms, the locus of the midpoint of AB is a hyperbola given by

$x-10xy+3y=0$