Prove that the points $A(1, 1)$,$B(4, 4)$,and $C(6, 2)$ are the vertices of a right-angled triangle.

(N/A) Let the given points be $A(1, 1)$,$B(4, 4)$,and $C(6, 2)$.
Using the distance formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$:
$1$. Length of $AB = \sqrt{(4 - 1)^2 + (4 - 1)^2} = \sqrt{3^2 + 3^2} = \sqrt{9 + 9} = \sqrt{18}$.
$2$. Length of $BC = \sqrt{(6 - 4)^2 + (2 - 4)^2} = \sqrt{2^2 + (-2)^2} = \sqrt{4 + 4} = \sqrt{8}$.
$3$. Length of $AC = \sqrt{(6 - 1)^2 + (2 - 1)^2} = \sqrt{5^2 + 1^2} = \sqrt{25 + 1} = \sqrt{26}$.
Now,check for the Pythagorean theorem $(AB^2 + BC^2 = AC^2)$:
$AB^2 = 18$,$BC^2 = 8$,and $AC^2 = 26$.
Since $18 + 8 = 26$,we have $AB^2 + BC^2 = AC^2$.
Therefore,the triangle is a right-angled triangle with the right angle at vertex $B$.