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प्रश्न
Show that the points A (1, −2), B (3, 6), C (5, 10) and D (3, 2) are the vertices of a parallelogram.
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उत्तर
The distance d between two points `(x_1,y_1) and (x_2, y_2)` is given by the formula
`d = sqrt((x_1 - x_2)^2 +(y_1 - y_2)^2)`
In a parallelogram the opposite sides are equal in length.
Here the four points are A(1, −2), B(3, 6), C(5, 10) and D(3, 2).
Let us check the length of the opposite sides of the quadrilateral that is formed by these points.
`AB = sqrt((1 - 3)^2 + (2 - 6))`
`= sqrt((-2)^2 + (-8)^2)`
`= sqrt(4 + 64)`
`AB = sqrt(68)`
`CD = sqrt((5 - 3)^2 + (10 - 2)^2)`
`= sqrt((2)^2 + (8)^2)`
`= sqrt(4 + 64)`
`CD = sqrt(68)`
We have one pair of opposite sides equal.
Now, let us check the other pair of opposite sides.
`BC = sqrt((3 - 5)^2 + (6 - 10)^2)`
`= sqrt((-2)^2 + (-4)^2)`
`=sqrt(4 + 16)`
`AD = sqrt20`
The other pair of opposite sides is also equal. So, the quadrilateral formed by these four points is definitely a parallelogram.
Hence we have proved that the quadrilateral formed by the given four points is a parallelogram
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