Question

what is true about △abc? select three options. ab ⊥ ac. the triangle is a right triangle. the triangle is an isosceles triangle. the triangle is an equilateral triangle. bc ∥ ac.

Explanation:

Step1: Calculate slopes of lines

The slope formula is $m = \frac{y_2 - y_1}{x_2 - x_1}$. For line $AB$ with $A(-1,3)$ and $B(-5,-1)$, $m_{AB}=\frac{3 + 1}{-1+5}=\frac{4}{4} = 1$. For line $AC$ with $A(-1,3)$ and $C(3,-1)$, $m_{AC}=\frac{3 + 1}{-1 - 3}=\frac{4}{-4}=-1$. Since $m_{AB}\times m_{AC}=1\times(-1)= - 1$, $\overline{AB}\perp\overline{AC}$, so the triangle is a right - triangle.

Step2: Calculate lengths of sides

The distance formula is $d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$.
$AB=\sqrt{(-1 + 5)^2+(3 + 1)^2}=\sqrt{16 + 16}=\sqrt{32}=4\sqrt{2}$.
$AC=\sqrt{(-1 - 3)^2+(3 + 1)^2}=\sqrt{16 + 16}=\sqrt{32}=4\sqrt{2}$.
$BC=\sqrt{(3 + 5)^2+(-1 + 1)^2}=\sqrt{64}=8$. Since $AB = AC=4\sqrt{2}$, the triangle is isosceles.

Step3: Check parallel condition

The $y$ - coordinate of $B$ and $C$ is $-1$, so the slope of $BC$ is $m_{BC}=\frac{-1+1}{3 + 5}=0$. The slope of $AC$ is $-1$. Since $m_{BC}
eq m_{AC}$, $BC$ is not parallel to $AC$.

Answer:

A. $\overline{AB}\perp\overline{AC}$
B. The triangle is a right triangle.
C. The triangle is an isosceles triangle.