Locus of midpoint of the line cutting the other two lines

Question

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.

Answer 1

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 $$\color{blue}{4h^2-k^2-k+2=0}.$$

Answer 2

Here is 1) a different way to establish the equation 2) geometrical remarks pertaining to the fact that the obtained curve is a hyperbola.

With the parametric coordinates of $A$ and $B$ you have given, one finds without difficulty that midpoint $M$ has coordinates $(h,k)$ (I use the same notations than in the other solution given by @Anurag A) with :

$$h=\dfrac{3m}{4-m^2}, \ \ k=\dfrac{4+2m^2}{4-m^2}\tag{1}$$

One can transform the last expression into :

$$k=\dfrac{4-m^2+3m^2}{4-m^2}=1+\dfrac{3m^2}{4-m^2}=1+mh\tag{1}$$

From (1), one can extract

$$m=\dfrac{k-1}{h}\tag{2}$$

Plugging the expression of $m$ given by (2) into the first equation of (1), we obtain,

$$h=\dfrac{3\tfrac{k-1}{h}}{4-\left(\tfrac{k-1}{h}\right)^2}$$

Simplifying, one obtains :

$$3\dfrac{k-1}{h}=\dfrac{4h^2-(k-1)^2}{h}$$

yielding the desired equation :

$$4h^2-k^2-k+2=0$$

It should be said that this equation is that of a hyperbola, because it can be written under the form :

$$(k+1/2)^2-4h^2=9/4$$

(red curve of the figure below). Moreover, the asymptotic lines (green lines) of this hyperbola are parallel to the given lines, a fact that could have been foreseen by studying limit cases, for example, if the slope of line passing through $M$ is $2-\varepsilon$ with $\varepsilon \to 0$, the point of intersection $B$ tends to a fixed point $B_2(-3/4,-1/2)$, and the point of intersection $A$ is tending to infinity on the first line ; therefore, the midpoint $M$ of $[AB]$ is close to the mid point of $[AM]$ which tens to be to a line halfway from the asymptotic line when seen from $M$ (line with equation $y=2x-1/2$). See figure below.