Prove that the points (-4, -1), (-2, -4), (4, 0) and (2, 3) are the vertices of a quadrilateral.

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Solution

$\begin{matrix} A = (- 4, - 1); B = (- 2, - 4); c = (4, 0); d = (2, 3) D i s tan c e b e t w e e n t w o p o i n t s D = \sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2}} A B = \sqrt{{(- 4 + 2)}^{2} + {(- 1 + 4)}^{2}} = \sqrt{13} S i m i l a r y B C = \sqrt{52} C D = \sqrt{13} A D = \sqrt{52} A B = C D & B C = A D s o, g i v e n v e r t i c e l s a r e v e r t i c e l s o f q u a d r i l a t e r a l a n d i t i s r e c tan g l e . \end{matrix}$

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