The points B(1, 3) and D(6, 8) are two opposite vertices of a square ABCD. Find the equation of the diagonal AC.

Slope of BD is given by

m₁ = $\dfrac{8 - 3}{6 - 1} = \dfrac{5}{5}$ = 1.

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

So, the slope of AC (m₂) will be,

$\Rightarrow m$1 \times m2 = -1 \\[1em] \Rightarrow 1 \times m2 = -1 \\[1em] \Rightarrow m2 = -1.

Coordinates of mid-point of BD and AC will be same as diagonals of a square meet at their mid-point.

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

By point-slope formula equation of AC is

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