Three vertices of a parallelogram ABCD taken in order are A(3, 6), B(5, 10) and C(3, 2). Find :

(i) the co-ordinates of the fourth vertex D

(ii) length of diagonal BD

(iii) equation of side AB of the parallelogram ABCD

The parallelogram ABCD is shown in the figure below:

(i) We know that the diagonals of a parallelogram bisect each other. Let (x, y) be the coordinates of D.

Mid-point of diagonal AC = $\dfrac{x$1 + x2}{2}, \dfrac{y1 + y2}{2} = \dfrac{3 + 3}{2}, \dfrac{6 + 2}{2} = (3, 4)

Mid-point of diagonal BD = $\dfrac{5 + x}{2}, \dfrac{10 + y}{2}$

These two should be same. On equating we get,

$\dfrac{5 + x}{2} = 3, \dfrac{10 + y}{2} = 4$

⇒ 5 + x = 6 and 10 + y = 8

⇒ x = 6 − 5 and y = 8 − 10

⇒ x = 1 and y = −2.

Hence, coordinates of D are (1, -2).

(ii) By distance formula the distance between B(5, 10) and D(1, -2) is given by $\sqrt{(x$2 - x1)^2 + (y2 - y1)^2}

Substituting values we get BD,

$= \sqrt{(1 - 5)^2 + (-2 - 10)^2} \\[1em] = \sqrt{(-4)^2 + (-12)^2} \\[1em] = \sqrt{16 + 144} \\[1em] = \sqrt{160} \\[1em] = 4\sqrt{10}$

Hence, the length of diagonal BD is $4\sqrt{10}$ units.

(iii) Equation of side AB can be given by two point formula i.e.,

$y - y$1 = \dfrac{y2 - y1}{x2 - x1} (x - x1)

Substituting values we get,

⇒ y − 6 = $\dfrac{10 - 6}{5 - 3}$ (x - 3)

⇒ y − 6 = $\dfrac{4}{2}$ (x - 3)

⇒ (y − 6) = 2(x - 3)

⇒ (y − 6) = 2x - 6

⇒ 2x - y -6 + 6 = 0

⇒ 2x - y = 0

Hence, equation of the line is 2x - y = 0.