(1, 5) and (-3, -1) are the co-ordinates of vertices A and C respectively of rhombus ABCD. Find the equations of the diagonals AC and BD.

Given, A = (1, 5) and C = (-3, -1) of rhombus ABCD.

We know that in a rhombus, diagonals bisect each other at right angle.

So, mid-point of AC = mid-point of BD.

Let’s take O to be the point of intersection of the diagonals AC and BD.

Then, the co-ordinates of O

= $\Big[\dfrac{1 + (-3)}{2}, \dfrac{5 + (-1)}{2}\Big] = \Big(\dfrac{-2}{2}, \dfrac{4}{2}\Big) = (-1, 2).$

$\text{Slope of AC }= \dfrac{y$2 - y1}{x2 - x1} \\[1em] = \dfrac{-1 - 5}{-3 - 1} \\[1em] = \dfrac{-6}{-4} = \dfrac{3}{2}.

Then, the equation of the line AC is

⇒ y - y₁ = m(x - x₁)

⇒ y - 5 = $\dfrac{3}{2}$(x - 1)

⇒ 2y - 10 = 3x - 3

⇒ 3x - 2y -3 + 10 = 0

⇒ 3x - 2y + 7 = 0.

Now, the line BD is perpendicular to AC.

∴ Slope of BD × Slope of AC = -1

⇒ Slope of BD × $\dfrac{3}{2}$ = -1

⇒ Slope of BD = $-\dfrac{2}{3}$.

BD will also pass through O.

By point-slope form,

Equation of the line BD,

⇒ y – y₁ = m(x – x₁)

⇒ y – 2 = $-\dfrac{2}{3}$[x - (-1)]

⇒ 3(y – 2) = -2[x + 1]

⇒ 3y - 6 = -2x - 2

⇒ 2x + 3y = -2 + 6

⇒ 2x + 3y = 4.

Hence, equation of line AC is 3x - 2y + 7 = 0 and BD is 2x + 3y = 4.