The points A(1, 3) and C(6, 8) are two opposite vertices of a square ABCD. Find the equation of the diagonal BD.

Given,

A(1, 3) and C(6, 8)

$m${AC}= \dfrac{y2 - y1}{x2 - x_1} = \dfrac{8 - 3}{6 - 1} = \dfrac{5}{5} = 1.

We know that diagonal AC is a perpendicular bisector of diagonal BD.

⇒ m_AC × m_BD = -1

⇒ 1 × m_BD = -1

⇒ m_BD = -1

Let O be the point of intersection of diagonals, which is the mid-point of both the diagonals.

$O = \Big(\dfrac{x$1 + x2}{2}, \dfrac{y1 + y2}{2}\Big) \\[1em] = \Big(\dfrac{1 + 6}{2}, \dfrac{3 + 8}{2}\Big) \\[1em] = \Big(\dfrac{7}{2}, \dfrac{11}{2}\Big).

By point-slope formula equation of AC is

$\Rightarrow y - y$1 = m(x - x1) \\[1em] \Rightarrow y - \Big(\dfrac{11}{2}\Big) = -1 \Big[x - \Big(\dfrac{7}{2}\Big)\Big] \\[1em] \Rightarrow y - \dfrac{11}{2} = -x + \dfrac{7}{2} \\[1em] \Rightarrow y + x - \dfrac{11}{2} - \dfrac{7}{2} = 0 \\[1em] \Rightarrow y + x - \dfrac{18}{2} = 0 \\[1em] \Rightarrow x + y - 9 = 0.

Hence, the equation of the required line is x + y - 9 = 0.