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

Asked by Abhisek | 1 year ago |  139

##### Solution :-

Let the given point be A (-2, -1) , B (4, 0) , C ( 3, 3) and D ( -3, 2)

So now, The slope of AB = $$\dfrac{(0+1)}{(4+2)}$$

= $$\dfrac{1}{6}$$

The slope of CD

$$\dfrac{(3-2)}{(3+3)}$$$$\dfrac{1}{6}$$

Hence, slope of AB = Slope of CD

AB ∥ CD

Now,

The slope of BC = $$\dfrac{(3-0)}{(3-4)}$$

$$\dfrac{3}{-1}$$ = -3

The slope of AD =$$\dfrac{(2+1)}{(-3+2)}$$

$$\dfrac{3}{-1}$$ = -3

Hence, slope of BC = Slope of AD

Thus the pair of opposite sides are quadrilateral are parallel, so we can say that ABCD is a parallelogram.

Hence the given vertices, A (-2, -1), B (4, 0), C(3, 3) and D(-3, 2) are vertices of a parallelogram

Answered by Pragya Singh | 1 year ago

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