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Points A (4, 3), B (6, 4), C (5, –6) and D (–3, 5) are the vertices of a parallelogram.

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Answer.     [False]
Solution.        
The given points area A(4, 3), B(6, 4), C(5, –6), D(–3, 5)
Distance \: between\: AB = \sqrt{\left ( 6-4 \right )^{2}+\left ( 4-3 \right )^{2}}
AB = \sqrt{\left ( 2 \right )^{2}+\left ( 1 \right )^{2}}= \sqrt{5}
Distance \: between\: BC = \sqrt{\left ( 5-6 \right )^{2}+\left ( -6-4 \right )^{2}}
BC= \sqrt{1+100}= \sqrt{101}
Distance \: between\: CD = \sqrt{\left ( -3-5 \right )^{2}+\left ( 5+6 \right )^{2}}
CD = \sqrt{64+121}= \sqrt{185}
Distance \: between\: DA = \sqrt{\left ( 4+3 \right )^{2}+\left ( 3-5 \right )^{2}}
DA =\sqrt{49+4}= \sqrt{53}
Here, opposite sides are not equal i.e.
AB\neq CD,BC\neq DA
Hence, it is not a parallelogram

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