Question

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

• A. $x+3y-10xy=0$
• B. $x-3y-10xy=0$
• C. $x+3y+10xy=0$
• D. $None$

Answer 1

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

Let the equation of any line passing through the point of intersection of the given line be

$(x+2y-1)+a(2x-y-1)=0$

Reducing the equation to its intercept form

$\frac{x(1+2a)}{(1+a)}+\frac{y(2-1)}{(1+a)}=1$

Therefore coordinates of A and B, where this line meets the coordinate axie respectively.

$A=(\frac{1+a}{1+2a},0)$ on x-axis

$B=(0,\frac{1+a}{2-a})$ on y-axis

Mid point of $AB=(\frac{1+a}{2+4a},\frac{1+a}{4-2a})$

Now we find the locus of this point by eliminating $a$ between the two expressions

$x=\frac{1+a}{2+4a}$

$y=\frac{1+a}{4-2a}$

$y=\frac{x}{10x-3}$

$x=10xy-3y$

$x+3y-10xy=0$