Question

If A (1, 0), B (5, 3), C (2, 7) and D (x, y) are vertices of a parallelogram ABCD, the co-ordinates of D are

(a) (–2, –3)

(b) (–2, 4)

(c) (2, –3)

(d) (3, 5)

Answer 1

Correct option is (b) (–2, 4)

We know that the diagonals of a parallelogram bisect each other. So, the
the midpoint of AC is the same as the midpoint of BD.

Mid point of two points (x₁​, y₁​) and (x₂​, y₂​) is  calculated by the formula \(\left(\frac{x _1 + x_2}2, \frac{y_1 + y_2}2\right)\)

So, midpoint of AC = Mid point of BD

⇒ \(\left(\frac{1 + 2}2, \frac{0+ 7}2\right) = \left(\frac{5+x}2, \frac{3+y}2\right)\)

⇒ \(\left(\frac{3}2, \frac{7}2\right) = \left(\frac{5+x}2, \frac{3+y}2\right)\)

⇒ 5 + x = 3; 3 + y = 7

⇒ x = −2; y = 4

Hence, D = (−2, 4).