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

Q: 9         Without using the distance formula, show that points  $(-2,-1),(4,0),(3,3)$  and $(-3,2)$  are  the vertices of a parallelogram.

Views

Given points are  $A(-2,-1),B(4,0),C(3,3)$  and $D(-3,2)$
We know the pair of the opposite side are parallel to each other in a parallelogram
Which means their slopes are also equal
$Slope = m = \frac{y_2-y_1}{x_2-x_1}$
The slope of AB =

$\frac{0+1}{4+2} = \frac{1}{6}$

The slope of BC =

$\frac{3-0}{3-4} = \frac{3}{-1} = -3$

The slope of CD =

$\frac{2-3}{-3-3} = \frac{-1}{-6} = \frac{1}{6}$

$\frac{2+1}{-3+2} = \frac{3}{-1} = -3$
We can clearly see that
The slope of AB = Slope of CD               (which means they are parallel)
and
The slope of BC = Slope of AD               (which means they are parallel)
Hence pair of opposite sides are parallel to each other
Therefore, we can say that points  $(-2,-1),(4,0),(3,3)$  and $(-3,2)$  are  the vertices of a parallelogram

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