Find the perimeter of triangle $ABC$, if the coordinates of its vertices $A(-3,5), B(3,-3)$ and point $M(6$, 1), which is the midpoint of side $BC$, are known.

#
Use the formulas for the coordinates of the midpoint of a segment and the formulas for the distance between two points.

## Solution

Let $(x, y)$ be the coordinates of vertex $C$. According to the condition, $1 / 2(x+3)=6, 1 / 2(y-3)=1$, from which $x=9, y=5$.

Using the distance formula between two points, we find the sides of triangle $ABC: AB=$

$\sqrt{(3-(-3))^{2}+(-3-5)^{2}}=\sqrt{36+64}=10$,

$\left.BC=\sqrt{(9-3)^{2}+\left(5-(-3)^{2}\right.}=\sqrt{36+64}=10, AC=9-(-3)\right)=12$. Therefore, $P_{ABC}=10+10+12=32$.

![](https://cdn.mathpix.com/cropped/2024_05_06_243bf411927c48073e50g-20.jpg?height=397&width=514&top_left_y=1005&top_left_x=772)

Answer

32.