<s>[INST] Locus of midpoint of the line cutting the other two lines Given two lines $L_1: y = 2x-2$ and $L_2: y = -2x-2$. The variable line $L$ passes through the point $R(0,1)$ and meets $L_1$ and $L_2$ at $A$ and $B$, respectively. The slope of variable line is $m$. Find the equation of the locus of the midpoint $M$ of $AB$.
So far I've got to the point that coordinates of $A$ and $B$ are:
$\left(x_A, y_A\right) = \left( \dfrac{3}{2-m}, \dfrac{2+2m}{2-m} \right) $
$\left(x_B, y_B\right) = \left( - \dfrac{3}{2+m}, \dfrac{2-2m}{2+m} \right) $
I can also calculate the coordinates of the midpoint, but don't know how to eliminate $m$. Just give me a hint how to proceed from here. I should get the equation $4x^2-y^2-y+2=0$ as final solution.
 [/INST] Let $(h,k)$ be the midpoint $M$ of $AB$.
\begin{align*}
h&=\frac{x_A+x_B}{2}=\frac{3m}{4-m^2}
\end{align*}
But $R(0,1)$ also lies on the line $L$, so $m=\frac{k-1}{h-0}=\frac{k-1}{h}$. Thus
$$h=\frac{3m}{4-m^2}=\frac{3\left(\frac{k-1}{h}\right)}{\left(4-\left(\frac{k-1}{h}\right)^2\right)}.$$
Simplify this and you get
$$\color{blue}{4h^2-k^2-k+2=0}.$$
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