Question

Without using distance formula, show that the points (-2,-1),(4,0),(3,3) and (-3,2) are the vertices of a parallelogram.

Answer 1

Let A(-2,-1), B(4,0), C(3,3) and D(-3,2) are the vertices of a parallelogram.

Mid-point of AC = $(\frac{3-2}{2},\frac{3-1}{2})=(\frac{1}{2},1)$ $[\because \text{mid-point of two points}(x_1,y_1\left)\text{and}\right(x_2,y_2)=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})]$

Mid-point of BD = $(\frac{4-3}{2},\frac{0+2}{2})=(\frac{1}{2},1)$

$\because$ Mid-point of AC = Mid-point of BD

$\therefore$ ABCD is a parallelogram. Hence proved.