3. Let triangle $ABC$ have vertices $A(-1,2), B(-2,-3)$, and $C(2,-1)$. From point $A$, we construct a perpendicular to side $BC$. The perpendicular intersects $BC$ at point $E$. Accurately calculate the length of segment $AE$.
3. We find that in the given triangle $v_{a}=d(A, E)$. We calculate the area of the triangle $\mathrm{ABC}: S=9$. The length of side $a$ is $d(B, C)=2 \sqrt{5}$. Then we express $v_{a}$ from the area $v_{a}=\frac{2 \cdot S}{a}=\frac{9 \sqrt{5}}{5}$. The length $v_{a}=\frac{9 \sqrt{5}}{5}$.

The finding that $d(A, E)=v_{a} \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots$. point

Using the appropriate formula for the area point

![](https://cdn.mathpix.com/cropped/2024_06_07_59669f975316a35884bfg-07.jpg?height=623&width=565&top_left_y=228&top_left_x=1291)

![](https://cdn.mathpix.com/cropped/2024_06_07_59669f975316a35884bfg-07.jpg?height=49&width=1065&top_left_y=678&top_left_x=210)
Calculated $a=d(B, C)=2 \sqrt{5} \ldots \ldots \ldots \ldots \ldots \ldots .1$ point Correctly substituted data into the formula $S=\frac{a v_{a}}{2} \ldots .1$ point $B(-2,-3)$

Result $v_{a}=\frac{9 \sqrt{5}}{5}$

1 point