Prove that $A (1, 7)$,$B (2, 4)$,and $C (5, 5)$ are the vertices of an isosceles right triangle.

(N/A) To prove that the points $A(1, 7)$,$B(2, 4)$,and $C(5, 5)$ form an isosceles right triangle,we calculate the lengths of the sides using the distance formula: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.
$1$. Length of $AB$: $AB = \sqrt{(2 - 1)^2 + (4 - 7)^2} = \sqrt{1^2 + (-3)^2} = \sqrt{1 + 9} = \sqrt{10}$.
$2$. Length of $BC$: $BC = \sqrt{(5 - 2)^2 + (5 - 4)^2} = \sqrt{3^2 + 1^2} = \sqrt{9 + 1} = \sqrt{10}$.
$3$. Length of $AC$: $AC = \sqrt{(5 - 1)^2 + (5 - 7)^2} = \sqrt{4^2 + (-2)^2} = \sqrt{16 + 4} = \sqrt{20}$.
Since $AB = BC = \sqrt{10}$,the triangle is isosceles.
Now,check for the right-angled condition using the Pythagorean theorem $(AB^2 + BC^2 = AC^2)$:
$AB^2 + BC^2 = 10 + 10 = 20$.
$AC^2 = 20$.
Since $AB^2 + BC^2 = AC^2$,the triangle is a right-angled triangle.
Thus,$A, B,$ and $C$ are the vertices of an isosceles right triangle.