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Question
Prove that the points (3, -2), (4, 0), (6, -3) and (5, -5) are the vertices of a parallelogram.
Solution
Let A (3,-2); B (4, 0); C (6,-3) and D (5,-5) be the vertices of a quadrilateral. We have to prove that the quadrilateral ABCD is the parallelogram.
We should proceed with the fact that if the diagonals of a quadrilateral bisect each other than the quadrilateral is a parallelogram.
Now to find the mid-point P(x,y) of two points `A(x_1,y_1)` and `B(x_2,y_2)` we use section formula as,
`P(x,y) = ((x_1 + x_2)/2, (y_1+y_2)/2)`
So the mid-point of the diagonal AC is,
`Q(x,y) = ((3 + 6)/2, (-2-3)/2)`
`=(9/2, - 5/2)`
Similarly mid-point of diagonal BD is,
`R(x,y) = ((4 + 5)/2, (-5+0)/2)`
`= (9/2, -5/2)`
Therefore the mid-points of the diagonals are coinciding and thus diagonal bisects each other.
Hence ABCD is a parallelogram.
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