The points B(7, 3) and D(0, –4) are two opposite vertices of a rhombus ABCD. Find the equation of diagonal AC.

Slope of the line BD (m₁),

$m$1 = \dfrac{y2 - y1}{x2 - x_1} \\[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 AC be m₂.

⇒ m₁ × m₂ = -1

⇒ 1 × m₂ = -1

⇒ m₂ = -1

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

$O = \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 AC 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 \Big[x - \Big(\dfrac{7}{2}\Big)\Big] \\[1em] \Rightarrow y + \Big(\dfrac{1}{2}\Big) = \Big[-x + \Big(\dfrac{7}{2}\Big)\Big] \\[1em] \Rightarrow \Big(\dfrac{2y + 1}{2}\Big) = \Big(\dfrac{-2x + 7}{2}\Big) \\[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 \\[1em] \Rightarrow x + y = 3.

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