Task B-4.2. (20 points) In the Cartesian coordinate system, the points $A\left(-\frac{17}{2}, 0\right), B(2,0), C(-1,0)$ are given. Find all points on the line $y=x-3$ from which the segments $\overline{A C}$ and $\overline{B C}$ are seen at the same angle different from zero.
First solution. Let $T$ be the desired point. Its coordinates are $T(x, x-3)$. Since the angles $\angle C T A$ and $\angle B T C$ are equal, the segment $\overline{T C}$ lies on the angle bisector of $\angle B T A$. From the angle bisector theorem, we have the ratio $|A T|:|B T|=|A C|:|B C|$. Substituting the coordinates of the points, we get

$$
\sqrt{\left(x+\frac{17}{2}\right)^{2}+(x-3)^{2}}: \sqrt{(x-2)^{2}+(x-3)^{2}}=\frac{15}{2}: 3
$$

![](https://cdn.mathpix.com/cropped/2024_05_30_978576e245d98ac92b08g-19.jpg?height=800&width=1360&top_left_y=454&top_left_x=225)

By squaring and simplifying, we obtain the equation $x^{2}-7 x=0$, whose solutions are $x_{1}=0$, $x_{2}=7$. The corresponding ordinates are $y_{1}=-3, y_{2}=4$.

Thus, the desired points are $T_{1}(0,-3)$ and $T_{2}(7,4)$.