In square ABCD, the vertex A = (2, 8) and vertex C = (8, 3). Find the equation of the diagonal BD.

In a square,

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 :

$O = \Big(\dfrac{2 + 8}{2}, \dfrac{8 + 3}{2}\Big) \\[1em]$

= \Big(\dfrac{10}{2}, \dfrac{11}{2}\Big) \\[1em]

= \Big(5, \dfrac{11}{2}\Big).O=(22+8​,28+3​)=(210​,211​)=(5,211​).

By formula,

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

Substituting values we get :

Slope of AC = $\dfrac{3 - 8}{8 - 2} = -\dfrac{5}{6}$.

We know that,

Product of slope of perpendicular lines = -1.

⇒ Slope of AC × Slope of BD = -1

⇒ $-\dfrac{5}{6} \times$ Slope of BD = -1

⇒ Slope of BD = $\dfrac{6}{5}$.

By point-slope form,

Equation of line :

y - y₁ = m(x - x₁)

BD also passes through point O.

So,

Equation of line BD is

$\Rightarrow y - \dfrac{11}{2} = \dfrac{6}{5}(x - 5) \\[1em] \Rightarrow \dfrac{2y - 11}{2} = \dfrac{6x - 30}{5} \\[1em] \Rightarrow 5(2y - 11) = 2(6x - 30) \\[1em] \Rightarrow 10y - 55 = 12x - 60 \\[1em] \Rightarrow 12x - 10y - 60 + 55 = 0 \\[1em] \Rightarrow 10y = 12x - 5.$

Hence, equation of line BD is 10y = 12x - 5.