Question

A line drawn through the origin intersects the lines $2x+y-2=0$ and $x-2y+2=0$ in $A$ and $B$. Show the locus of the mid-point of $AB$ is $2x^2-3xy-2y^2+x+3y=0$

Answer 1

Any line through the origin is $y=mx$ It meets the line $2x+y-2=0$ in $A(\frac{2}{m+2},\frac{2m}{m+2})$
It meets the line $x-2y+2=0$ in point
$B(\frac{2}{2m-1},\frac{2m}{2m-1})$
If $2h=\frac{2}{m+2}+\frac{2}{2m-1}$
or $h=\frac{3m+1}{(m+2)(2m-1)}....\left(1\right)$
$2k=\frac{2}{m+2}+\frac{2m}{2m-1}ork=\frac{m(3m+1)}{(m+2)(2m-1)}....\left(2\right)$
In order to find the locus we have to eliminate the variable $m$
Dividing (1) and (2) we get $\left(k/h\right)=m$. Putting in (1) we get
$h(\frac{k}{h}+2)(\frac{2k}{h}-1)=3.\frac{k}{h}+1$
$2k^2+3hk-2h^2-h=0$
Hence the locus is
$2x^2+3xy-2y^2-3y-x=0$
or $2x^2+3xy-2y^2+x+3y=0$