Coordinate Geometry-Finding vertices given midpoints If (2,1), (4,5), (1,-3) are the midpoints of the sides of a triangle, find the co-ordinates of its vertices
 A: It can be done by a calculation. 
We will need some notation. Let the triangle be $ABC$, let the midpoint of $BC$ be $(2,1)$, let the midpoint of $AC$ be $(4,5)$, and let the midpoint of $AB$ be $(1,-3)$.
Let the coordinates of $A$ be $(x_1,y_1)$, the coordinates of $B$ be $(x_2,y_2)$, and the coordinates of $C$ be $(x_3,y_3)$.
Draw a labelled picture that has the above information. 
The $x$-coordinate of the midpoint of $BC$ is $\frac{x_2+x_3}{2}$. This we are told is equal to $2$.  We conclude that
$$\frac{x_2+x_3}{2}=2.\tag{1}$$
Similarly, we have
$$\frac{x_1+x_3}{2}=4,\tag{2}$$
and
$$\frac{x_1+x_2}{2}=1.\tag{3}$$
Solve the above system of linear equations for $x_1$, $x_2$, $x_3$.  These are exceptionally simple equations. A nice way to find the solutions is to add the left-hand sides of (1), (2), and (3). This sum is equal to the sum of the right-hand sides. Thus  $x_1+x_2+x_3=7$. 
Now for example to get $x_1$, use the fact that $x_2+x_3=4$. We get that $x_1=3$. Similarly, $x_2=-1$ and $x_3= 5$. 
Finding  $y_1$. $y_2$. and $y_3$ is left to you.
A: Suppose the vertices are given by $(x_1,y_1)$, $(x_2,y_2)$, and $(x_3,y_3)$.  Then the midpoint formula gives us the following six equations:
$$\frac{x_1+x_2}{2}=2\qquad\frac{y_1+y_2}{2}=1\qquad\frac{x_1+x_3}{2}=4\\[.3in]\frac{y_1+y_3}{2}=5\qquad\frac{x_2+x_3}{2}=1\qquad\frac{y_2+y_3}{2}=-3\\$$
Notice these can be separated into two systems of three equations, each with three variables.  See if you can finish off the problem by solving these two systems.
A: Consider a triangle $ABC$ and let $A',B',C'$ be the midpoints of $BC,AC,AB$ respectively. It can be shown that $B'C'$ is parallel to $BC$ by considering the dilation by a factor of $2$ centered at $A$. For the same reason, the other corresponding pairs of lines are parallel as well.
This means that the line containing $A$ and $B$ is the line parallel to $A'B'$ containing $C'$; likewise $BC$ is the line parallel to $B'C'$ containing $A'$; etc.
So in your case, let $A'=(2,1),B'=(4,5),C'=(1,-3)$. The line $AB$ is given by $y=2x-5$ for instance. And $A$ will be the point of intersection of $AB$ and $AC$. Can you finish from here?
