Show that the points A (1, −2), B (3, 6), C (5, 10)
Question:

Show that the points A (1, −2), B (3, 6), C (5, 10) and D (3, 2) are the vertices of a parallelogram.

Solution:

The distance $d$ between two points $\left(x_{1}, y_{1}\right)$ and $\left(x_{2}, y_{2}\right)$ is given by the formula

$d=\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}}$

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(32).

Let us check the length of the opposite sides of the quadrilateral that is formed by these points.

$A B=\sqrt{(1-3)^{2}+(-2-6)^{2}}$

$=\sqrt{(-2)^{2}+(-8)^{2}}$

 

$=\sqrt{4+64}$

$A B=\sqrt{68}$

$C D=\sqrt{(5-3)^{2}+(10-2)^{2}}$

$=\sqrt{(2)^{2}+(8)^{2}}$

 

$=\sqrt{4+64}$

$C D=\sqrt{68}$

We have one pair of opposite sides equal.

Now, let us check the other pair of opposite sides.

$B C=\sqrt{(3-5)^{2}+(6-10)^{2}}$

$=\sqrt{(-2)^{2}+(-4)^{2}}$

$=\sqrt{4+16}$

$B C=\sqrt{20}$

$A D=\sqrt{(1-3)^{2}+(-2-2)^{2}}$

$=\sqrt{(-2)^{2}+(-4)^{2}}$

$=\sqrt{4+16}$

$A D=\sqrt{20}$

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 .

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