Prove that the diagonals of a rectangle ABCD with vertices A(2, -1), B(5, -1), C(5, 6) and D(2, 6) are equal and bisect each other.

Explanation

Diagonal `AC = sqrt((5-2)^(2) + (6+1)^(2))`
` = sqrt(3^(2) + 7^(2)) = sqrt(9 + 49) = sqrt(58)`
Diagonal `BD = sqrt((5-2)^(2) + (-1-6)^(2))`
` = sqrt(3^(2) + (-7)^(2)) = sqrt(9+49) = sqrt(58)`
`therefore` Diag. AC = Diag. BD.
Midpoint of `AC = ((5+2)/(2), (6-1)/(2)) = ((7)/(2), (5)/(2))`
Midpoint of `BD = ((5+2)/(2), (-1+6)/(2)) = ((7)/(2), (5)/(2))`