A(7,-8) and C(1,4) are vertices of a square ABCD. Find equation of diagonal BD?

Diagonals of a square bisect each other perpendicularly.

=> BD perpendicularly bisects diagonal joining  A(7,-8) and C(1,4) at O, thus O is the mid point of AC and BD.

=> Coordinates of O = $$(\frac{7 + 1}{2} , \frac{-8 + 4}{2})$$

= $$(\frac{8}{2} , \frac{-4}{2}) = (4,-2)$$

Now, slope of AC = $$\frac{y_2 - y_1}{x_2 - x_1} = \frac{(4 + 8)}{(1 - 7)}$$

= $$\frac{12}{-6} = -2$$

Let slope of line BD = $$m$$

Product of slopes of two perpendicular lines = -1

=> $$m \times -2 = -1$$

=> $$m = \frac{1}{2}$$

Equation of a line passing through point $$(x_1,y_1)$$ and having slope $$m$$ is $$(y - y_1) = m(x - x_1)$$

$$\therefore$$ Equation of line BD passing thorugh O(4,-2)

=> $$(y + 2) = \frac{1}{2}(x - 4)$$

=> $$2y + 4 = x - 4$$

=> $$x - 2y = 4 + 4 = 8$$

=> Ans - (B)