Task B-4.5. (20 points) A line through the origin intersects the lines given by the equations $x+y-1=0, x-y-1=0$ at points $A$ and $B$. Determine the geometric locus of the midpoints of the segments $\overline{A B}$.
Solution. The equation of a line passing through the origin is $y=k x$.

The intersection of this line with the line $x+y-1=0$ is obtained by solving the system of equations

$$
y=k x, \quad y=-x+1
$$

The intersection point is $A\left(\frac{1}{1+k}, \frac{k}{1+k}\right)$.

The intersection with the line $x-y-1=0$ is obtained by solving this system of equations

$$
y=k x, \quad y=x-1
$$

The second intersection point is $B\left(\frac{1}{1-k}, \frac{k}{1-k}\right)$.

(5 points)

The midpoint of segment $\overline{A B}$ is

$$
P\left(\frac{\frac{1}{1+k}+\frac{1}{1-k}}{2}, \frac{\frac{k}{1+k}+\frac{k}{1-k}}{2}\right)=P\left(\frac{1}{1-k^{2}}, \frac{k}{1-k^{2}}\right)
$$

The coordinates of point $P$ are $x=\frac{1}{1-k^{2}}, y=\frac{k}{1-k^{2}}$.

![](https://cdn.mathpix.com/cropped/2024_05_30_9c56249806d755c5106ag-16.jpg?height=811&width=865&top_left_y=1308&top_left_x=607)

Now consider the difference $x^{2}-y^{2}$:

$$
x^{2}-y^{2}=(x-y)(x+y)=\frac{1-k}{1-k^{2}} \cdot \frac{1+k}{1-k^{2}}=\frac{1}{1-k^{2}}=x
$$

We obtained the equation $x^{2}-x-y^{2}=0$

which can be rearranged to $\left(x-\frac{1}{2}\right)^{2}-y^{2}=\frac{1}{4}$.