Three vertices of a parallelogram, taken in order, are A(1, 2), B(4, 8) and C(9, 6). What are the coordinates of the fourth vertex D?

In parallelogram ABCD the diagonals $AC$ and $BD$ bisect each other, so the midpoint of $AC$ must equal the midpoint of $BD$. The midpoint of $AC$ is ((1 + 9)/2, (2 + 6)/2) = (5, 4), so $D=2\times (5,4)-B$ = (10 − 4, 8 − 8) = (6, 0). Equivalently $D=A+C-B$. A check on the sides: $AB$ runs (3, 6) and $DC$ runs (9 − 6, 6 − 0) = (3, 6), so they are parallel and equal. The trap (12, 12) computes $B+C-A$, which makes the fourth vertex of the parallelogram ABDC with the letters out of order; (−4, 4) computes $A+B-C$ in the same careless way; and (6, 12) keeps the right $x$ but reflects the $y$ in the wrong direction.