Prove that the points A(1, 7), B (4, 2),

Solution:

The distance $d$ between two points $\left(x_{1}, y_{\mathrm{I}}\right)$ 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 square all the sides are equal in length. Also, the diagonals are equal in length in a square.

Here the four points are A(1, 7), B(4, 2), C(−1, −1) and D(−4, 4).

First let us check if all the four sides are equal.

$A B=\sqrt{(1-4)^{2}+(7-2)^{2}}$

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

 

$=\sqrt{9+25}$

$A B=\sqrt{34}$

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

$=\sqrt{(5)^{2}+(3)^{2}}$

 

$=\sqrt{25+9}$

$B C=\sqrt{34}$

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

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

 

$=\sqrt{9+25}$

$C D=\sqrt{34}$

$A D=\sqrt{(1+4)^{2}+(7-4)^{2}}$

$=\sqrt{(5)^{2}+(3)^{2}}$

 

$=\sqrt{25+9}$

$A D=\sqrt{34}$

Since all the sides of the quadrilateral are the same it is a rhombus.

For the rhombus to be a square the diagonals also have to be equal to each other.

$A C=\sqrt{(1+1)^{2}+(7+1)^{2}}$

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

$=\sqrt{4+64}$

$A C=\sqrt{68}$

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

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

$=\sqrt{64+4}$

$B D=\sqrt{68}$

Since the diagonals of the rhombus are also equal to each other the rhombus is a square.

Hence the quadrilateral formed by the given points is a.