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 

Hint:
$$2k=\dfrac{6+3t}{3+2t}+\dfrac{6+t}{3+t}$$
$$2k-\dfrac32-1=\dfrac{6+3t}{3+2t}-\dfrac32+\dfrac{6+t}{3+t}-1=\dfrac{3}{2(3+2t)}+\dfrac3{3+t}$$
$$2h=\dfrac3{3+t}+\dfrac3{3+2t}$$
Solve the simultaneous equations for $\dfrac1{3+2t},\dfrac1{3+t}$
Finally use $$2(3+t)-(3+2t)=3$$
Optionally we can simplify the result 
