Question 13 Marks
Show that points $P(1,-2)$ $Q(5,2)$, $R(3,-1)$, $S(-1,-5)$ are the vertices of a parallelogram.
Answer
View full question & answer→Slope of PQ = $\frac{2 - (-2)}{5 - 1} = \frac{4}{4} = 1$
Slope of RS = $\frac{-5 - (-1)}{-1 - 3} = \frac{-4}{-4} = 1$
Slope of QR = $\frac{-1 - 2}{3 - 5} = \frac{-3}{-2} = \frac{3}{2}$
Slope of PS = $\frac{-5 - (-2)}{-1 - 1} = \frac{-3}{-2} = \frac{3}{2}$
Since slope of PQ = slope of RS (PQ || RS) and slope of QR = slope of PS (QR || PS), PQRS is a parallelogram.
Slope of RS = $\frac{-5 - (-1)}{-1 - 3} = \frac{-4}{-4} = 1$
Slope of QR = $\frac{-1 - 2}{3 - 5} = \frac{-3}{-2} = \frac{3}{2}$
Slope of PS = $\frac{-5 - (-2)}{-1 - 1} = \frac{-3}{-2} = \frac{3}{2}$
Since slope of PQ = slope of RS (PQ || RS) and slope of QR = slope of PS (QR || PS), PQRS is a parallelogram.