In rhombus ABCD, A(7, 3) and C(0, -4) are two opposite vertices. Find :

(i) mid-point of diagonal AC

(ii) mid-point of diagonal BD

(iii) slope of diagonal AC

(iv) the equation of diagonal BD.

Rhombus ABCD is shown in the figure below:

(i) By formula,

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

Let M be the mid-point of AC.

Substituting values we get :

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

Hence, mid-point of AC = $\Big(\dfrac{7}{2}, -\dfrac{1}{2}\Big).$

(ii) We know that,

Diagonals of rhombus bisect each other at point of intersection.

⇒ Mid-point of BD = Mid-point of AC = $\Big(\dfrac{7}{2}, -\dfrac{1}{2}\Big).$

Hence, mid-point of BD = $\Big(\dfrac{7}{2}, -\dfrac{1}{2}\Big).$

(iii) By formula,

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

Let slope of AC be s.

Substituting values we get :

$s = \dfrac{-4 - 3}{0 - 7} \\[1em] = \dfrac{-7}{-7} \\[1em] = 1.$

Hence, slope of AC = 1.

(iv) We know that,

Diagonals of Rhombus are perpendicular to each other,

Let slope of BD be z.

We know that,

Product of slope of perpendicular lines is equal to -1.

∴ z × s = -1

⇒ z × 1 = -1

⇒ z = -1.

∴ Diagonal BD has slope = -1 and passes through point $\Big(\dfrac{7}{2}, -\dfrac{1}{2}\Big).$

By point-slope form,

Equation of line :

⇒ y - y₁ = m(x - x₁)

⇒ $y - \Big(-\dfrac{1}{2}\Big) = -1\Big(x - \dfrac{7}{2}\Big)$

⇒ $y + \dfrac{1}{2} = -x + \dfrac{7}{2}$

⇒ x + y = $\dfrac{7}{2} - \dfrac{1}{2}$

⇒ x + y = $\dfrac{7 - 1}{2}$

⇒ x + y = $\dfrac{6}{2}$

⇒ x + y = 3.

Hence, equation of diagonal BD is x + y = 3.