The points A(7, 3) and C(0, -4) are two opposite vertices of a rhombus ABCD. Find the equation of the diagonal BD.

Rhombus ABCD with A(7, 3) and C(0, -4) as the two opposite vertices is shown in the figure below:

Slope of the line AC (m₁),

$= \dfrac{y$2 - y1}{x2 - x1} \\[1em] = \dfrac{-4 - 3}{0 - 7} \\[1em] = \dfrac{-7}{-7} \\[1em] = 1.

Diagonals of rhombus bisect each other at right angles.

∴ BD is perpendicular to AC. Let slope of BD be m₂.

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

Let O be the mid-point of diagonals. It's coordinates are given by,

$= \Big(\dfrac{x$1 + x2}{2}, \dfrac{y1 + y2}{2}\Big) \\[1em] = \Big(\dfrac{7 + 0}{2}, \dfrac{3 + (-4)}{2}\Big) \\[1em] = \Big(\dfrac{7}{2}, -\dfrac{1}{2}\Big).

Equation of BD can be given by point slope form i.e.,

$\Rightarrow y - y$1 = m(x - x1) \\[1em] \Rightarrow y - \Big(-\dfrac{1}{2}\Big) = -1(x - \Big(\dfrac{7}{2}\Big)) \\[1em] \Rightarrow y + \dfrac{1}{2} = -x + \dfrac{7}{2} \\[1em] \Rightarrow \dfrac{2y + 1}{2} = \dfrac{-2x + 7}{2} \\[1em] \Rightarrow 2y + 1 = -2x + 7 \\[1em] \Rightarrow 2y + 2x - 6 = 0 \\[1em] \Rightarrow 2(y + x - 3) = 0 \\[1em] \Rightarrow x + y - 3 = 0.

Hence, the equation of the required line is x + y - 3 = 0.