Question

If three consecutive vertices of a parallelogram ABCD areA(1, -2), B(3,6) and C(5,10), find its fourth vertex D.

Answer 1

Let A (1, -2), B (3, 6), C (5, 10) be the three vertices of a parallelogram ABCD and the fourth vertex be D (a, b)

Join AC and BD, intersecting at O

We know that the diagonals of a parallelogram bisect each other

Therefore, O is the midpoint of AC as well as BD

Midpoint of AC = $(\frac{1+5}{2},\frac{-2+10}{2})$

= $(\frac{6}{2},\frac{8}{2})$

= (3, 4)

Midpoint of BD = $(\frac{3+a}{2},\frac{6+b}{2})$

Therefore, $\frac{3+a}{2}=3,\frac{6+b}{2}=4$

⇒ 3 + a = 6 , 6 + b = 8

⇒ a = 6 - 3, b = 8 - 6

⇒ a = 3 and b = 2

Therefore, the fourth vertex is D (3,2)