Given the relative vertices a (0, - 1) and C (2,5) of square ABCD, the coordinates of vertices B and D are obtained

Given the relative vertices a (0, - 1) and C (2,5) of square ABCD, the coordinates of vertices B and D are obtained

Let m (x, y) be the midpoint of AC, then x = 0 + 22 = 1, y = − 1 + 52 = 2, m (x, y) = m (1, 2). Let K be the slope of AC, then k = 3, so the slope of BD is − 1K = − 13. Therefore, the equation of BD is y − 2 = − 13 (x − 1) (1), and the equation of a circle with m point as the center and | Ma | as the radius is (x-1) 2 + (Y-2) 2 = 10 & nbsp; & nbsp; (2) By solving equations (1) and (2), the coordinates of B and D are (4,1) and (- 2,3)