Locus of mid point of $AB$ 
If the family of lines $tx+3y-6=0.$ where $t$ is variable intersect the lines $x-2y+3=0$ and $x-y+1=0$ at point $A$  and $B.$ Then locus of mid point of $AB$ is

what i try 
Intersection of line $tx+3y-6=0$ and $x-2y+3=0$ is 
$\displaystyle A\bigg(\frac{3}{3+2t},\frac{6+3t}{3+2t}\bigg)$
and Intersection of $tx+3y-6=0$ and $x-y+1=0$ is 
$\displaystyle B\bigg(\frac{3}{3+t},\frac{6+t}{3+t}\bigg)$
Let locus of mid point of $AB$ is at  $M(h,k)$
Then 
$\displaystyle h=\frac{1}{2}\bigg(\frac{3}{3+2t}+\frac{3}{3+t}\bigg)$ and $\displaystyle k = \frac{1}{2}\bigg(\frac{6+3t}{3+2t}+\frac{6+t}{3+t}\bigg)$
How do i eliminate $t$ in an easy way
or please help me is any other easier ways to solve it thanks 

Let $M(x,y).$
Thus, $$y-x=\frac{1}{2}\left(\frac{3t+3}{2t+3}+1\right)$$ or
$$t=\frac{6x-6y+6}{4y-4x-5}.$$
Can you end it now?
