Problems involving LOCUS Okay so I'm really having a hard time finding the locus of curves or point of intersection of curves.
The problem : I am able to get the geometrical condition in most of the questions but then I have no idea what to do to get the locus.
The question :A variable straight line drawn through the point of intersection of the straight line $x+2y-4=0$ and $2x+y-4=0$ meet the coordinate axes in $A$ and $B$.Then what is the locus of midpoint of $AB$ ?
My try : I found the equation of the variable straight line as $3x+3y-8=0$,and the required mid-point in the question to be $(4/3,4/3)$.How do I proceed from here ? 
 A: Th equation of any line passing through the intersection of $$x+2y-4=0,2x+y-4=0$$ is 
$$0=x+2y-4+m(2x+y-4)=x(1+2m)+y(2+m)-(4m+4)$$ where $m$ is an arbitrary constant
$$\implies\dfrac x{\dfrac{4m+4}{2m+1}}+\dfrac y{\dfrac{4m+4}{m+2}}=1$$
So, $A:\left(\dfrac{4m+4}{2m+1},0\right)$ and $B;\left(0,\dfrac{4m+4}{m+2}\right)$
Now if $P(h,k)$ is the midpoint,
we have $$h=\dfrac{\dfrac{4m+4}{2m+1}+0}2\iff m=?$$
$$k=\dfrac{0+\dfrac{4m+4}{m+2}}2\iff m=?$$
Equate the values of $m$ to eliminate $m,$ the variable we have introduced.
Can you take it from here?   
A: Let $M(a,b)$ be a point on our locus and $C\left(\frac{4}{3},\frac{4}{3}\right)$.
Thus, $$m_{MC}=\frac{b-\frac{4}{3}}{a-\frac{4}{3}}$$ and we have an equation of $AM$:
$$y-\frac{4}{3}=\frac{b-\frac{4}{3}}{a-\frac{4}{3}}\left(x-\frac{4}{3}\right),$$
which for $x=0$ gives
$$y=\frac{4}{3}-\frac{4}{3}\cdot\frac{b-\frac{4}{3}}{a-\frac{4}{3}}$$ and it gives an equation of the locus:
$$\frac{\frac{4}{3}-\frac{4}{3}\cdot\frac{b-\frac{4}{3}}{a-\frac{4}{3}}+0}{2}=b$$ or
$$b=\frac{2a}{3a-2}$$ or
$$y=\frac{2x}{3x-2}.$$
A: if the variable straight goes through $$P(4/3;4/3)$$ then we get
$$y=mx-\frac{4}{3}m+\frac{4}{3}$$ now we can calculate the intersection Points with the axis:
$$x=0$$ then $$y=\frac{4}{3}(1-m)$$
$$y=0$$ then $$x=\frac{-\frac{4}{3}+\frac{4}{3}m}{m}$$
can you finish?
the coordintaes of the midpoint is given by $$M\left(\frac{-\frac{4}{3}+\frac{4}{3}m}{2m};\frac{\frac{4}{3}-\frac{4}{3}m}{2}\right)$$
eliminating we get $$-3xy+2x+2y=0$$
