Question

The three vertices of a parallelogram ABCD are A(−2, 3) B(6, 7) and C(8, 3). The fourth vertex D is

(a) (1, 0)
(b) (0, 1)
(c) (−1, 0)
(d) (0, −1)

Answer 1

(d) (0, −1)
Let the fourth vertex be D(a, b).
Join AC and BD.
Let AC and BD intersect at the point O.
We know that the diagonals of a parallelogram bisect each other.
$$\begin{gathered} \text{Midpoint of}AC\text{is}\left(\frac{-2+8}{2},\frac{3+3}{2}\right)\Rightarrow \left(3,3\right) \\ \text{Midpoint of}BD\text{is}\left(\frac{6+a}{2},\frac{7+b}{2}\right) \end{gathered}$$
Therefore,
$$\begin{gathered} \frac{6+a}{2}=3\text{and}\frac{7+b}{2}=3 \\ \Rightarrow 6+a=6\text{and}7+b=6 \end{gathered}$$
$\Rightarrow a=0\text{and}b=-1$
Hence, the fourth vertex is D(0,−1).