Without using distance formula, show that points $(-2,-1),(4,0),(3,3)$ and $(-3,2)$ are vertices of a parallelogram.
Without using distance formula, show that points $(-2,-1),(4,0),(3,3)$ and $(-3,2)$ are vertices of a parallelogram.
Let points $(-2,-1),(4,0),(3,3)$ and $(-3,2)$ be respectively denoted by $A , B , C ,$ and $D$.
Slopes of $AB =\frac{0+1}{4+2}=\frac{1}{6}$
Slopes of $CD =\frac{2-3}{-3-3}=\frac{-1}{-6}=\frac{1}{6}$
$\Rightarrow$ Slope of $AB =$ Slope of $CD$
$\Rightarrow AB$ and $CD$ are parallel to each other.
Now, slope of $BC =\frac{3-0}{3-4}=\frac{3}{-1}=-3$
Slope of $AD =\frac{2+1}{-3+2}=\frac{3}{-1}=-3$
$\Rightarrow$ Slope of $BC =$ Slope of $AD$
$\Rightarrow BC$ and $AD$ are parallel to each other.
Therefore, both pairs of opposite side of quadrilateral $ABCD$ are parallel. Hence, $ABCD$ is a parallelogram.
Thus, points $(-2,-1),(4,0),(3,3)$ and $(-3,2)$ are the vertices of a parallelogram.
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