Question

Using the concept of slope, show that the points $(-2,-1),(4,0),(3,3)$ and $(-3,2)$ taken in order form a parallelogram.

Answer 1

Let $A(-2,-1,B(4,0),C(3,3)$ and $D(-3,2)$ be the given points taken in order. Now the slope of $ABAB=\frac{0+1}{4+2}=\frac{1}{6}$ Slope of $CD=\frac{2-3}{-3-3}=\frac{1}{6}$ $\therefore$ Slope of $AB=$ slope of $CD$ Hence, $AB$ is parallel to $CD$ Now the slope of $BC=\frac{3-0}{3-4}=-3$ Slope of $AD=\frac{2+1}{-3+2}=-3$ $\therefore$ Slope of $BC=$ Slope of $AD$ Hence, $BC$ is parallel of $AD$. From $\left(1\right)$ an d$\left(2\right)$ we see that opposite sides of quadrilateral $ABCD$ are parallel $\therefore$ $ABCD$ is a parallelogram.