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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.

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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}

The slope of AD

\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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