Prove that the points $(- 7, - 3), (5, 10), (15, 8) a n d (3, - 5)$ taken in order are the corners of a parallelogram.

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Solution

Let A, B, C and D represent the points (-7, -3), (5, 10), (15, 8) and (3, -5) respectively.
Using the distance formula $d = \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}}$ , we find
$A B^{2} = (5 + 7)^{2} + (10 + 3)^{2} = 12^{2} + 13^{2} = 144 + 169 = 313$
$B C^{2} = (15 - 5)^{2} + (8 - 10)^{2} = 10^{2} + (- 2)^{2} = 100 + 4 = 104$
$C D^{2} = (3 - 15)^{2} + (- 5, - 8)^{2} = (- 12)^{2} + (- 13)^{2} = 144 + 169 = 313$
$D A^{2} = (3 + 7)^{2} + (- 5 + 3)^{2} = 10^{2} + (- 2)^{2} = 100 + 4 = 104$
So, $A B = C D = \sqrt{313}$ and $B C = D A = \sqrt{104}$
i.e., The opposite sides are equal. Hence ABCD is a parallelogram.

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