Question

Show that P($-$2,4 ), Q(4, 8), R(10, 5) and S(4, 1) are vertices of parallelogram.

Answer 1

Given: P($-$2,4 ), Q(4, 8), R(10, 5) and S(4, 1)



We have:
Slope of line = $\frac{y_2-y_1}{x_2-x_1}=\frac{y_1-y_2}{x_1-x_2}$

Slope of line PQ =$\frac{8-4}{4-(-2)}=\frac{4}{6}=\frac{2}{3}$ ... (i)

Slope of line QR = $\frac{5-8}{10-4}=\frac{-3}{6}=\frac{-1}{2}$ … (ii)

Slope of line PS = $\frac{1-4}{4-(-2)}=-\frac{3}{6}=\frac{-1}{2}$ … (iii)

Slope of line RS = $\frac{1-5}{4-10}=\frac{-4}{-6}=\frac{2}{3}$ … (iv)

From (i) and (iv), slope of line PQ = slope of line RS
i.e., line PQ is parallel to RS.
Also, from (ii) and (iii), slope of line QR = slope of line PS
i.e., line QR is parallel to PS.
Here, both pairs of the opposite sides are parallel.
∴ PQRS is a parallelogram.