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Prove that A(3, 5), B(6, 0), C(1, -3) and D(-2, 2) are the vertices of a square. |
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Answer» Step-by-step EXPLANATION: midpoint of AC(3+1/2,5-3/2)=(2,1) midpoint of BD(6-2/2,0+2/2)=(2,1) diagonal bisect each other ABCD is parallelogram AB=√(6-3)^2+(0-5)^2=√9+25=√34 BC=√(6-1)^2++(0+3)^2=√25+9=√34 since in parallelogram opposite side are EQUAL THEREFORE in ABCD all sides are equal AC=√(1-3)^2+(-3-5)^2=√4+64=√68 BD=√(-2-6)^2+(2-0)^2=√68 AC=BD AB^2+BC^2=AC^2 34+34=68 similarly AD^2+AB^2=BD^2 34+34=68 all angles are 90° therefore ABCD is a SQUARE |
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