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  • Show that the following points taken in order form the vertices of a parallelogram. (9, 5), (6, 0), ( - 2, - 3) and (1, 2)

Show that the following points taken in order form the vertices of a parallelogram. (9, 5), (6, 0), ( - 2, - 3) and (1, 2)

Answers (1)

Let A(9, 5), B(6, 0), C( - 2, - 3) and D(1, 2) be the points

The distance formula:

D = \sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}

AB= \sqrt{(6-9)^2+(0-5)^2} =\sqrt{9+25} = \sqrt{34}

BC= \sqrt{(-2-6)^2+(-0-3)^2} =\sqrt{64+9} = \sqrt{73}

CD= \sqrt{(1-(-2))^2+(2-(-3))^2} =\sqrt{9+25} = \sqrt{34}

AD= \sqrt{(1-9)^2+(2-5)^2} =\sqrt{64+9} = \sqrt{73}

And the diagonals:

AC =\sqrt{(-2-9)^2+(-3-5)^2} = \sqrt{121+64} = \sqrt{185}

BD =\sqrt{(1-6)^2+(2-0)^2} = \sqrt{25+4} = \sqrt{29}

Here we can observe that the opposite sides of this quadrilateral are of the same length.

However, the diagonals are of different lengths.

Therefore, the given points are the vertices of a parallelogram.

Posted by

Safeer PP

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