Show that $A(1, 4)$,$B(7, -2)$,and $C(9, 6)$ are the vertices of an isosceles triangle.

(N/A) To show that the points $A(1, 4)$,$B(7, -2)$,and $C(9, 6)$ form an isosceles 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 $AB$:
$AB = \sqrt{(7 - 1)^2 + (-2 - 4)^2} = \sqrt{6^2 + (-6)^2} = \sqrt{36 + 36} = \sqrt{72} = 6\sqrt{2}$ units.
$2$. Length of $BC$:
$BC = \sqrt{(9 - 7)^2 + (6 - (-2))^2} = \sqrt{2^2 + 8^2} = \sqrt{4 + 64} = \sqrt{68} = 2\sqrt{17}$ units.
$3$. Length of $AC$:
$AC = \sqrt{(9 - 1)^2 + (6 - 4)^2} = \sqrt{8^2 + 2^2} = \sqrt{64 + 4} = \sqrt{68} = 2\sqrt{17}$ units.
Since $BC = AC$,two sides of the triangle are equal in length. Therefore,$\triangle ABC$ is an isosceles triangle.