Question

Three vertices of a parallelogram taken in order are (−1, −6), (2, −5) and (7, 2). The fourth vertex is

(a) (1, 4)
(b) (4, 1)
(c) (1, 1)
(d) (4, 4)
(e) (0, 0)

Answer 1

(b) (4,1)

Let A(−1, −6), B(2, −5) and C(7, 2) be the given vertex. Let D(h, k) be the fourth vertex.

The midpoints of AC and BD are $\left(3,-2\right)\mathrm{and}\left(\frac{2+h}{2},\frac{-5+k}{2}\right)$ respectively.

We know that the diagonals of a parallelogram bisect each other.

$$\begin{gathered} \therefore 3=\frac{2+h}{2}\mathrm{and}-2=\frac{-5+k}{2} \\ \Rightarrow h=4\mathrm{and}k=1 \end{gathered}$$