5.2. Given three points $A(2 ; 1), B(3 ;-1), C(-4 ; 0)$, which are vertices of an isosceles trapezoid $A B D C$. Find the coordinates of point $D$, if $\overline{A B}=k \overline{C D}$.
5.2. According to the condition,

$\overline{A B}=k \overline{C D} \Rightarrow \overline{A B} \| \overline{C D}$.

We find the coordinates of vectors ${ }^{-} \overline{A B}$ and $\overline{C D}$ (Fig. 5.2): $\overline{A B}(1 ;-2)$, $\overline{C D}(x+4 ; y)$, where $x$ and $y$ are the coordinates of point $D$. From (1) it follows that

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-598.jpg?height=277&width=479&top_left_y=631&top_left_x=672)

Fig. 5.2

$$
\frac{1}{x+4}=\frac{-2}{y}, \text { or } y=-2 x-8
$$

Since trapezoid $A B D C$ is isosceles, then $|\overline{A C}|=| \overline{B D} \mid$ and $\overline{A C} \perp \overline{B D}$. We find the vectors $\overline{A C}(-6 ;-1)$, $\overline{B D}(x-3 ; y+1)$ and use the fact that $\overline{A C}^{2}=\overline{B D}^{2}$; we have $36+1=(x-3)^{2}+(y+1)^{2}$, or $x^{2}+y^{2}-6 x+2 y=27$.

Solving the system of equations (2) and (3), we get $x_{1}=-1.4, y_{1}=-5.2$ or $x_{2}=-3, y_{2}=-2$. These values correspond to two vectors: $\overline{B D}(-4.4 ;-4.2)$ and $\overline{B D}(-6 ;-1)$. The latter vector is collinear with $\overline{A C}$ and, therefore, does not satisfy the condition. Thus, $D(-1.4 ;-5.2)$.

Answer: $(-1.4 ;-5.2)$.