Locus problem for straight lines

Question

$y=mx$ intersects $2x+y-2=0$ at $A$ and $x-2y+2=0$ at $B$. Find locus of mid-point of $AB$.

I tried to find the points in terms of m and then applying midpoint formula but I could not eliminate m.

Answer 1

Solving for A you have $$A\left(\frac{2}{2+m},\frac{2m}{2+m}\right)$$

Similarly for B you have $$B\left(\frac{2}{2m-1},\frac{2m}{2m-1}\right)$$

Then the $x$ coordinate of the midpoint is given by $$x=\frac{1}{2+m}+\frac{1}{2m-1}$$

Now you can replace $m$ with $$m=\frac yx$$ and obtain a Cartesian equation for the locus...

Can you finish this?

Answer 2

with line 1, $2x+mx=2$ therefore $x={2\over2+m}$ and $y={2m\over2+m}$
with line 2, $x-2mx+2=0$ therefore $x={2\over2m-1}$ and $y={2m\over2m-1}$

Hence the midpoint is $$\left({1\over2+m}+{1\over2m-1},{m\over2+m}+{m\over2m-1}\right)$$ and thus $$x={3m+1\over2m^2+3m-2}\text{ and }y={3m^2+m\over2m^2+3m-2}$$

Now in both of these equations solve for m using quadratic formulae and then set the two results equal. You'll get the eq of the curve. It's a hyperbola