Let P, Q, R and S be the points on the plane | vedclass

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(A) The coordinates of the points are $P(-2, -1)$,$Q(4, 0)$,$R(3, 3)$,and $S(-3, 2)$.Calculate the slopes of the sides:$m_{PQ} = \frac{0 - (-1)}{4 - (

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parallelogram,which is neither a rhombus nor a rectangle.

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(A) The coordinates of the points are $P(-2, -1)$,$Q(4, 0)$,$R(3, 3)$,and $S(-3, 2)$.Calculate the slopes of the sides:$m_{PQ} = \frac{0 - (-1)}{4 - (-2)} = \frac{1}{6}$$m_{SR} = \frac{3 - 2}{3 - (-3)} = \frac{1}{6}$$m_{QR} = \frac{3 - 0}{3 - 4} = \frac{3}{-1} = -3$$m_{PS} = \frac{2 - (-1)}{-3 - (-2)} = \frac{3}{-1} = -3$Since $m_{PQ} = m_{SR}$ and $m_{QR} = m_{PS}$,the opposite sides are parallel,so $PQRS$ is a parallelogram.Now,check the lengths of the sides:$PQ = \sqrt{(4 - (-2))^2 + (0 - (-1))^2} = \sqrt{6^2 + 1^2} = \sqrt{37}$$QR = \sqrt{(3 - 4)^2 + (3 - 0)^2} = \sqrt{(-1)^2 + 3^2} = \sqrt{10}$Since $PQ 
eq QR$,it is not a rhombus.Check the product of slopes of adjacent sides:$m_{PQ} 	imes m_{QR} = \frac{1}{6} 	imes (-3) = -0.5 
eq -1$.Since the product of slopes is not $-1$,the adjacent sides are not perpendicular,so it is not a rectangle.Therefore,$PQRS$ is a parallelogram,which is neither a rhombus nor a rectangle.

Let $P, Q, R$ and $S$ be the points on the plane with position vectors $-2 \hat{i}-\hat{j}$,$4 \hat{i}$,$3 \hat{i}+3 \hat{j}$ and $-3 \hat{i}+2 \hat{j}$ respectively. Then the quadrilateral $PQRS$ must be a

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