ABCD is a rhombus. The coordinates of A and C are (3, 6) and (-1, 2) respectively. Write down the equation of BD.

Slope of AC is given by,

$m_1 = \dfrac{2 - 6}{-1 - 3} = \dfrac{-4}{-4}$ = 1.

We know that diagonal of a rhombus bisect each other at right angles. So, the diagonal BD is perpendicular to diagonal AC.

Let the slope of BD be m₂. Then,

$\Rightarrow m$1 \times m2 = -1 \\[1em] \Rightarrow 1 \times m2 = -1 \\[1em] \Rightarrow m 2 = -1.

Coordinates of mid-point of AC and BD are same which are

$= \Big(\dfrac{3 + (-1)}{2}, \dfrac{6 + 2}{2}\Big) \\[1em] = \Big(\dfrac{2}{2}, \dfrac{8}{2}\Big) \\[1em] = (1, 4).$

Equation of the line having the slope = -1 and passing through (1, 4) can be given by point-slope formula i.e.,

$\Rightarrow y - y$1 = m(x - x1) \\[1em] \Rightarrow y - 4 = -1(x - 1) \\[1em] \Rightarrow y - 4 = -x + 1 \\[1em] \Rightarrow x + y - 4 - 1 = 0 \\[1em] \Rightarrow x + y - 5 = 0.

Hence, the equation of BD is x + y - 5 = 0.