In rhombus ABCD, the co-ordinates of point A and C are (2, -6) and (-4, 8) respectively. Find the equation of the diagonal BD.

In a rhombus,

Diagonals bisect each other and are perpendicular.

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

By formula,

Mid-point = $\Big(\dfrac{x$1 + x2}{2}, \dfrac{y1 + y2}{2}\Big)

Let O be the point of intersection of diagonal AC and BD.

Substituting values we get :

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

= \Big(\dfrac{-2}{2}, \dfrac{2}{2}\Big) \\[1em]

= (-1, 1).O=(22+(−4)​,2(−6)+8​)=(2−2​,22​)=(−1,1).

By formula,

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

Substituting values we get :

Slope of AC = $\dfrac{8 - (-6)}{-4 - 2} = -\dfrac{14}{6} = -\dfrac{7}{3}$.

We know that,

Product of slope of perpendicular lines = -1.

⇒ Slope of AC × Slope of BD = -1

⇒ $-\dfrac{7}{3} \times$ Slope of BD = -1

⇒ Slope of BD = $\dfrac{3}{7}$.

By point-slope form,

Equation of line :

y - y₁ = m(x - x₁)

BD also passes through point O.

So,

Equation of line BD is

$\Rightarrow y - 1 = \dfrac{3}{7}[x - (-1)] \\[1em] \Rightarrow 7(y - 1) = 3[x + 1] \\[1em] \Rightarrow 7y - 7 = 3x + 3 \\[1em] \Rightarrow 7y - 3x = 3 + 7 \\[1em] \Rightarrow 7y - 3x = 10.$

Hence, equation of line BD is 7y - 3x = 10.