The points $$A(5$$, $$-$$ $$1$$, $$1)$$ $$B(7$$, $$-$$ $$4$$,$$7)$$,$$C(1$$, $$-$$ $$6$$, $$10)$$ and $$D(1$$ $$-$$ $$3$$,$$4)$$ are vertices of a

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Infinity Learn · Accepted Answer

Given that, $$A(5$$, $$-$$ $$1$$, $$1)$$ $$B(7$$, $$-$$ $$4$$,$$7)$$,$$C(1$$, $$-$$ $$6$$, $$10)$$ and $$D(1$$ $$-$$ $$3$$,$$4)now$$,$$AB=(7$$−$$5)2+($$−$$4+1)2+(7$$−$$1)2=4+9+36=7BC=(1$$−$$7)2+($$−$$6+4)2+(10$$−$$7)2=36+4+9=7CD=($$−$$1$$−$$1)2+($$−$$3+6)2+(4$$−$$10)2=4+9+36=7DA=(5+1)2+($$−$$1+3)2+(1$$−$$4)2=36+4+9=7$$∴ $$AB=BC=CD=DA=7$$,Also, $$AC=(1$$−$$5)2+($$−$$6+1)2+(10$$−$$1)2=16+25+81=122BD=($$−$$1$$−$$7)2+($$−$$4+3)2+(7$$−$$4)2=64+1+9=74Diagonal$$ AC≠BDHence, ABCD is not a square, so it is a rhombus.

The points A(5, - 1, 1) B(7, - 4,7),C(1, - 6, 10) and D(1 - 3,4) are vertices of a

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