Question

Show that four points (0,-1),(6,7),(-2,3) & (8,3) are vertices of a rectangle

Answer 1

$<b> Answer: Refer the attachment$

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Answer 2

AnswEr:

Let A(0,-1) , B(6,7) , C(-2,3) and D(8,3) be the given points. Then,

$\bf \: AD = \sf \sqrt{(8 - 0) {}^{2} + {(3 + 1)}^{2} } \\ \\ \sf = \sqrt{64 + 16} = \sf4 \sqrt{5} \\ \\ \bf \: BC =\sf \sqrt{( {6 +2 )}^{2} + {(7 - 3)}^{2} } \\ \\ \bf = \sf\sqrt{64 + 16} = \sf4 \sqrt{5} \\ \\ \bf \: AC = \sf \sqrt{ {( - 2 - 0)}^{2} + {(3 + 1)}^{2} } \\ \\ \sf = \sqrt{4 + 16} = \sf2 \sqrt{5} \qquad \: \: \: and, \\ \\ \bf \: BD = \sf\sqrt{ {(8 - 6)}^{2} + {(3 - 7)}^{2} } \\ \\ \sf = \sqrt{4 + 16} = \sf 2 \sqrt{5}$

$\therefore$ $\sf\underline{AD=BC\:and\:AC=BD}$

So, ADBC is a parallelogram.

$\huge\underline\bold{Now,}$

$\sf \: AB = \sqrt{ {(6 - 0)}^{2} + {(7 + 1)}^{2} } \\ = \sf \sqrt{36 + 64} = 10 \\ \\ \rm \: and \: \\ \\ \sf \: CD = \sqrt{ {(8 + 2)}^{2} + {(3 - 3)}^{2} } \\ \sf = 10$

Clearly , AB² = AD² + DB² and CD² = CB² + BD². Hence, ADBC is a rectangle.