State whether the following statement is true or false. Justify your answer.
Points $A(-6, 10)$,$B(-4, 6)$,and $C(3, -8)$ are collinear such that $AB = \frac{2}{9} AC$.

(A) True.
First,we check if the points are collinear by calculating the area of the triangle formed by them.
Area of triangle $= \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$.
Here,$x_1 = -6, x_2 = -4, x_3 = 3$ and $y_1 = 10, y_2 = 6, y_3 = -8$.
Area $= \frac{1}{2} | -6(6 - (-8)) + (-4)(-8 - 10) + 3(10 - 6) |$.
Area $= \frac{1}{2} | -6(14) + (-4)(-18) + 3(4) |$.
Area $= \frac{1}{2} | -84 + 72 + 12 | = \frac{1}{2} | 0 | = 0$.
Since the area is $0$,the points are collinear.
Next,we calculate the distances $AB$ and $AC$ using the distance formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.
$AB = \sqrt{(-4 - (-6))^2 + (6 - 10)^2} = \sqrt{2^2 + (-4)^2} = \sqrt{4 + 16} = \sqrt{20} = 2\sqrt{5}$.
$AC = \sqrt{(3 - (-6))^2 + (-8 - 10)^2} = \sqrt{9^2 + (-18)^2} = \sqrt{81 + 324} = \sqrt{405} = 9\sqrt{5}$.
Now,check the relation $AB = \frac{2}{9} AC$:
$\frac{2}{9} AC = \frac{2}{9} \times 9\sqrt{5} = 2\sqrt{5}$.
Since $AB = 2\sqrt{5}$,the relation holds true.