Show that the points (−4, −1), (−2, −4) (4, 0) and

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 rectangle, the opposite sides are equal in length. The diagonals of a rectangle are also equal in length.

Here the four points are A(−4,−1), B(−2,−4), C(4,0) and D(2,3).

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

$A B=\sqrt{(-4+2)^{2}+(-1+4)^{2}}$

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

$=\sqrt{4+9}$

$A B=\sqrt{13}$

$C D=\sqrt{(4-2)^{2}+(0-3)^{2}}$

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

$=\sqrt{4+9}$

$C D=\sqrt{13}$

We have one pair of opposite sides equal.

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

$B C=\sqrt{(-2-4)^{2}+(-4-0)^{2}}$

$=\sqrt{(-6)^{2}+(-4)^{2}}$

$=\sqrt{36+16}$

$B C=\sqrt{52}$

$A D=\sqrt{(-4-2)^{2}+(-1-3)^{2}}$

$=\sqrt{(-6)^{2}+(-4)^{2}}$

$=\sqrt{36+16}$

$A D=\sqrt{52}$

The other pair of opposite sides are also equal. So, the quadrilateral formed by these four points is definitely a parallelogram.

For a parallelogram to be a rectangle we need to check if the diagonals are also equal in length.

$A C=\sqrt{(-4-4)^{2}+(-1-0)^{2}}$

$=\sqrt{(-8)^{2}+(-1)^{2}}$

$=\sqrt{64+1}$

$A C=\sqrt{65}$

$B D=\sqrt{(-2-2)^{2}+(-4-3)^{2}}$

$=\sqrt{(-4)^{2}+(-7)^{2}}$

$=\sqrt{16+49}$

$B D=\sqrt{65}$

Now since the diagonals are also equal we can say that the parallelogram is definitely a rectangle.

Hence we have proved that the quadrilateral formed by the four given points is a.