## Midpoints of Triangle Sides

Points $X(-1,1),Y(3,5)$, and $Z(17,1)$ are the midpoints of the sides of triangle $ABC$. Find the coordinates of the points $A,B$, and $C$. (Labels $A,B$, and $C$ are arbitrarily assigned to coordinates).
Source: NCTM Mathematics Teacher 2008

SOLUTION

Draw a first line passing through $Y$ and parallel to $\overleftrightarrow{XZ}$. Equation of the first line
$y=5\qquad\qquad (1)$
Slope of $\overleftrightarrow{XY}$
$\dfrac{1-5}{-1-3}=1$
Draw a second line passing through point Z and parallel to $\overleftrightarrow{XY}$. Equation of the second line
$y=1x+b$
Substitute the values of $Z(17,1)$ into the equation
$1=17+b$
$b=-16$
Equation of the second line
$y=x-16\qquad (2)$
The two lines intersect at point $A(x,y)$ the coordinates of which satisfy both Eq. $(1)$ and $(2)$.
Solving for $x$ yields
$5=x-16$
$x=21$
Coordinates of vertex $A=(21,5)$.
Suppose $Z(17,1)$ is the midpoint of side $\overline{AB}$. The midpoint formula yields the coordinates of vertex $B(x,y)$
$\dfrac{x+21}{2}=17$
$x=13$
$\dfrac{y+5}{2}=1$
$y=-3$
Coordinates of vertex $B=(13,-3)$.
Similarly $X(-1,1)$ is the midpoint of side $\overline{BC}$. In this case the midpoint formula yields the coordinates of $B(x,y)$
$\dfrac{x+13}{2}=-1$
$x=-15$
$\dfrac{y-3}{2}=1$
$y=5$
Coordinates of vertex $C=(-15,5)$.

Answer: $(21,5),(13,-3),(-15,5)$