Question 490711
First, you can show that AB, BC, CD, and DA are all of equal length. For example,


*[tex \LARGE AB = \sqrt{(-2 - 6)^2 + (2 - (-13))^2}]


However this only proves that the quadrilateral is a rhombus. To prove it is a square, you could perhaps show that one of the angles is 90, then it will follow that the quadrilateral is a square.


Take the slope of two segments. For example, the slope of line AB is


*[tex \LARGE m_1 = \frac{2 - (-13)}{-2 - 6} = \frac{15}{-8}]


and the slope of BC is


*[tex \LARGE m_2 = \frac{10 - 2}{13 - (-2)} = \frac{8}{15}]


Hence, AB and BC are perpendicular, and it follows that ABCD is a square.