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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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Question

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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Solution

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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Chapter 6: Co-ordinate Geometry - Exercise 6.2 [Page 15]

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R.D. Sharma Mathematics [English] Class 10
Chapter 6 Co-ordinate Geometry
Exercise 6.2 | Q 7 | Page 15

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