A) square
B) parallelogram
C) rectangle
D) rhombus
Correct Answer: B
Solution :
Let\[A=(-1,1),B=(0,-3),C=(5,2)\]and D=(4,6) \[AB=\sqrt{{{(0+1)}^{2}}+{{(-3-1)}^{2}}}=\sqrt{1+{{(-4)}^{2}}}=\sqrt{17}\] \[BC=\sqrt{{{(5-0)}^{2}}+{{(2+3)}^{2}}}=\sqrt{25+25}=5\sqrt{2}\] \[CD=\sqrt{{{(4-5)}^{2}}+{{(6-2)}^{2}}}=\sqrt{1+16}=\sqrt{17}\] \[DA=\sqrt{{{(-1-4)}^{2}}+{{(1-6)}^{2}}}=\sqrt{25+25}=5\sqrt{2}\] Slope of \[AB=\frac{-3-1}{0+1}=-4\] Slope of \[BC=\frac{2+3}{5-0}=1\] Slope of \[CD=\frac{6-2}{4-5}=-4\] Slope of \[DA=\frac{1-6}{-1-4}=1\] \[\because \] \[AB=CD\]and\[BC=DA\] and slope of AB = slope of CD and slope of BC = slope of DA Hence, given points are the vertices of a parallelogram.
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