How to find triangle vertices given midpoints? I have a task to find vertices if midpoints are given: $M1(2;1)$, $M2(5;3)$, $M3(3;-4)$. I know one way to solve it through making a system of equations with three variables.
My teacher says there is faster way by using the midline of a triangle, and I can`t find this way of solving it on the Internet. How can I do it?

The matrix equation giving the midpoints from the vertices is
$$
\begin{bmatrix}
\frac12&\frac12&0\\
0&\frac12&\frac12\\
\frac12&0&\frac12
\end{bmatrix}
\begin{bmatrix}
a\vphantom{\frac12}\\b\vphantom{\frac12}\\c\vphantom{\frac12}
\end{bmatrix}
=
\begin{bmatrix}
\frac{a+b}2\\\frac{b+c}2\\\frac{c+a}2
\end{bmatrix}
$$
Inverting the equation above yields
$$
\begin{bmatrix}
1&-1&1\vphantom{\frac12}\\
1&1&-1\vphantom{\frac12}\\
-1&1&1\vphantom{\frac12}
\end{bmatrix}
\begin{bmatrix}
\frac{a+b}2\\\frac{b+c}2\\\frac{c+a}2
\end{bmatrix}
=
\begin{bmatrix}
a\vphantom{\frac12}\\b\vphantom{\frac12}\\c\vphantom{\frac12}
\end{bmatrix}
$$
The last equation simply says
$$
a=\frac{a+b}2+\frac{c+a}2-\frac{b+c}2\\
b=\frac{b+c}2+\frac{a+b}2-\frac{c+a}2\\
c=\frac{b+c}2+\frac{c+a}2-\frac{a+b}2
$$
This amounts to reflecting each midpoint across the midpoint of the line connecting the other two midpoints.
