B(-5, 6) and D(1, 4) are the vertices of rhombus ABCD. Find the equations of diagonals BD and AC.

The rhombus ABCD is shown in the figure below:

By formula,

Slope = $\dfrac{y$2 - y1}{x2 - x1}

Substituting values we get,

$\text{Slope of BD } = \dfrac{4 - 6}{1 - (-5)} \\[1em] = \dfrac{-2}{6} \\[1em] = -\dfrac{1}{3}.$

Equation of BD by point-slope form,

⇒ y - y₁ = m(x - x₁)

⇒ y - 6 = $-\dfrac{1}{3}$[x - (-5)]

⇒ 3(y - 6) = -1(x + 5)

⇒ 3y - 18 = -x - 5

⇒ 3y + x - 18 + 5 = 0

⇒ x + 3y - 13 = 0

⇒ x + 3y = 13.

Since, diagonals of rhombus are perpendicular to each other.

So, product of their slopes will be -1.

Slope of AC × Slope of BD = -1

$\text{Slope of AC} \times -\dfrac{1}{3}$ = -1

Slope of AC = -1 × -3 = 3.

We know that,

Diagonals of rhombus bisect each other. Let diagonals meet at point O.

∴ Mid-point of AC = Mid-point of BD.

Co-ordinates of mid-point of BD (i.e. O)

= $\Big(\dfrac{x$1 + x2}{2}, \dfrac{y1 + y2}{2}\Big) = \Big(\dfrac{-5 + 1}{2}, \dfrac{6 + 4}{2}\Big) = \Big(\dfrac{-4}{2}, \dfrac{10}{2}\Big) = (-2, 5).

∴ Co-ordinates of mid-point of AC = (-2, 5).

Equation of AC by point-slope form,

⇒ y - y₁ = m(x - x₁)

⇒ y - 5 = 3[x - (-2)]

⇒ y - 5 = 3[x + 2]

⇒ y - 5 = 3x + 6

⇒ 3x - y + 6 + 5 = 0

⇒ 3x - y + 11 = 0

⇒ y = 3x + 11.

Hence, equation of BD is x + 3y = 13 and equation of AC is y = 3x + 11.