ABCD is a parallelogram in which vertices B = (5, 8), C = (4, 7) and D = (2, -4). Find :

(i) the coordinates of vertex A.

(ii) the equation of diagonal BD.

(i) We know that,

Diagonals of parallelogram bisect each other.

Let E be the point of intersection of diagonals.

So, E will be the mid-point of BD as well as of AC.

By mid-point formula,

Mid-point (M) = $\Big(\dfrac{x$1 + x2}{2}, \dfrac{y1 + y2}{2}\Big)

So, mid-point of BD = $\Big(\dfrac{5 + 2}{2}, \dfrac{8 + (-4)}{2}\Big) = \Big(\dfrac{7}{2}, \dfrac{4}{2}\Big) = (3.5, 2)$.

E is mid-point of AC as well. Let coordinates of A be (a, b).

$\therefore (3.5, 2) = \Big(\dfrac{a + 4}{2}, \dfrac{b + 7}{2}\Big) \\[1em] \Rightarrow \dfrac{a + 4}{2} = 3.5 \text{ and } \dfrac{b + 7}{2} = 2 \\[1em] \Rightarrow a + 4 = 7 \text{ and } b + 7 = 4 \\[1em] \Rightarrow a = 7 - 4 = 3 \text{ and } b = 4 - 7 = -3.$

A = (a, b) = (3, -3).

Hence, coordinates of A are (3, -3).

(ii) By two point form,

Equation of a line : y - y₁ = $\dfrac{y$2 - y1}{x2 - x1}(x - x_1)

Substituting values we get :

Equation of BD :

$\Rightarrow y - 8 = \dfrac{-4 - 8}{2 - 5}(x - 5) \\[1em] \Rightarrow y - 8 = \dfrac{-12}{-3}(x - 5) \\[1em] \Rightarrow y - 8 = 4(x - 5) \\[1em] \Rightarrow y - 8 = 4x - 20 \\[1em] \Rightarrow 4x - y - 20 + 8 = 0 \\[1em] \Rightarrow 4x - y - 12 = 0 \\[1em] \Rightarrow y = 4x - 12.$

Hence, equation of BD is y = 4x - 12.