The points P(2, 1), Q(4, 5), and R(7, 4) are the midpoints of the sides of a triangle. Graph the three midsegments to form the midsegment triangle. Then use your graph and the properties of midsegments to draw the original triangle.

Each mid-segment is parallel to the side of the triangle whose midpoint is the vertex opposite that segment. So, we used a graphing tool to draw, for example, a line parallel to PQ through point R. The vertices of the original triangle are at the intersection points of these lines (or 1 mid-segment length from the midpoint).

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Analytically, we can make use of the fact that each mid-segment is effectively a diagonal of a parallelogram. For example, PQ is a diagonal of APRQ. The other diagonal is AR. The two diagonals cross at the midpoint of each. This means we can find point A using the midpoint relation:

midpoint AR = midpoint PQ

(A +R)/2 = (P +Q)/2 . . . use the midpoint formula

A +R = P +Q . . . . . . . . multiply by 2

A = P +Q -R . . . . . . . . subtract R to get an expression for A

A = (2, 1) +(4, 5) -(7, 4) = (2+4-7, 1+5-4) = (-1, 2)

The other vertices of ΔABC can be found in similar fashion:

B = R +P -Q = (7, 4) +(2, 1) -(4, 5) = (7+2-4, 4+1-5) = (5, 0)

C = Q +R -P = (4, 5) +(7, 4) -(2, 1) = (4+7-2, 5+4-1) = (9, 8)

So, the original triangle's vertices are ...

A(-1, 2), B(5, 0), C(9, 8)