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Without using formula, show that points $( -2,\ -1),\ ( 4,\ 0),\ ( 3,\ 3)$ and $( -3,\ 2)$ are the vertices of a parallelogram.
Given: Points $( -2,\ -1),\ ( 4,\ 0),\ ( 3,\ 3)$ and $( -3,\ 2)$
To do: To show that the given points are the vertices of a parallelogram without using the formula.
Solution:
Let points $( -2,\ -1),\ ( 4,\ 0),\ ( 3,\ 3)$ and $( -3,\ 2)$ be respectively denoted by $A,\ B,\ C$ and $D$.
Now, Slope of $AB=\frac{0-( -1)}{4-( -2)}=\frac{0+1}{4+2}=\frac{1}{6}$
Slope 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
Also, Slope of $BC=\frac{3-0}{3-4}=\frac{3}{-1}=-3$
Slope of $AD=\frac{2-( -1)}{-3-( -2)}=\frac{2+1}{-3+2}=-3$
$\Rightarrow$ Slope of $BC =$ Slope of $AD$
$\Rightarrow\ BC$ and $AD$ are parallel to each other.
Therefore, both are pairs of opposite sides of quadrilateral $ABCD$ are parallel.
Hence $ABCD$ is parallelogram.
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