Show that the points $(1,7), (4,2), (-1,-1)$ and $(-4,4)$ are the vertices of a square.

(N/A) Let $A(1,7), B(4,2), C(-1,-1)$ and $D(-4,4)$ be the given points.
To prove that $ABCD$ is a square,we must show that all its sides are equal and its diagonals are equal.
Using the distance formula $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$:
$AB = \sqrt{(4-1)^2 + (2-7)^2} = \sqrt{3^2 + (-5)^2} = \sqrt{9+25} = \sqrt{34}$
$BC = \sqrt{(-1-4)^2 + (-1-2)^2} = \sqrt{(-5)^2 + (-3)^2} = \sqrt{25+9} = \sqrt{34}$
$CD = \sqrt{(-4 - (-1))^2 + (4 - (-1))^2} = \sqrt{(-3)^2 + 5^2} = \sqrt{9+25} = \sqrt{34}$
$DA = \sqrt{(1 - (-4))^2 + (7-4)^2} = \sqrt{5^2 + 3^2} = \sqrt{25+9} = \sqrt{34}$
Now,calculating the diagonals:
$AC = \sqrt{(-1-1)^2 + (-1-7)^2} = \sqrt{(-2)^2 + (-8)^2} = \sqrt{4+64} = \sqrt{68}$
$BD = \sqrt{(-4-4)^2 + (4-2)^2} = \sqrt{(-8)^2 + 2^2} = \sqrt{64+4} = \sqrt{68}$
Since $AB = BC = CD = DA$ and $AC = BD$,all four sides are equal and the diagonals are equal. Therefore,$ABCD$ is a square.