Question

The three vertices of a parallelogram ABCD, taken in order, are A(1, −2), B(3, 6) and C(5, 10). Find the coordinates of the fourth vertex D.

Answer 1

Let A(1, −2), B(3, 6) and C(5, 10) be the three vertices of a parallelogram ABCD and let its fourth vertex be D(a, b).
Join AC and BD.
Let AC and BD intersect at point O.



We know that the diagonals of a parallelogram bisect each other.
So, O is the mid-point of AC as well as of BD.
Mid-point of AC = $\left(\frac{1+5}{2},\frac{-2+10}{2}\right)$ ⇒ (3, 4)
Mid-point of BD = $\left(\frac{3+a}{2},\frac{6+b}{2}\right)$
∴ $\frac{3+a}{2}=3$ and $\frac{6+b}{2}=4$
⇒ 3 + a = 6 and 6 + b = 8
⇒ a = 3 and b = 2
Hence, the fourth vertex is D(3, 2).