Answer
$C=(2\sqrt{3},2)$ or $(-2\sqrt{3},2)$
(two such triangles are possible)
Work Step by Step
As seen in previous exercises, the median is a line segment connecting a vertex and the midpoint of the opposite side.
The median of an equilateral triangle is also its altitude.
The altitude is perpendicular to the side.
The given side has length $4$ and is on the y-axis, so the median is parallel to the x-axis.
This means the third point has the same y-coordinate as the midpoint of the side.
$A=(0,0), B=(0,4)$
$M_{AB}=\displaystyle \left(\frac{0+0}{2},\frac{4+0}{2}\right)=(0,2).$
$C=(x,2)$
The x-coordinate of the third point will be one altitude to the left, or one altitude to the right.
The altitude (height) of an equilateral triangle is $h=\displaystyle \frac{a\sqrt{3}}{2}=\frac{4\sqrt{3}}{2}=2\sqrt{3}.$
The x-coordinate can be $+2\sqrt{3}$ or $-2\sqrt{3}.$
$C=(2\sqrt{3},2)$ or $(-2\sqrt{3},2)$
(two such triangles are possible)