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 the lengths of sides

Use the distance formula \(d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\)
For \(AB\): \(A(-1,3)\), \(B(-5,-1)\)
\(AB=\sqrt{(-5 + 1)^2+(-1 - 3)^2}=\sqrt{(-4)^2+(-4)^2}=\sqrt{16 + 16}=\sqrt{32}=4\sqrt{2}\)
For \(AC\): \(A(-1,3)\), \(C(3,-1)\)
\(AC=\sqrt{(3 + 1)^2+(-1 - 3)^2}=\sqrt{(4)^2+(-4)^2}=\sqrt{16+16}=\sqrt{32}=4\sqrt{2}\)
For \(BC\): \(B(-5,-1)\), \(C(3,-1)\)
\(BC=\sqrt{(3 + 5)^2+(-1+1)^2}=\sqrt{(8)^2+0^2}=8\)

Step2: Check the properties

Since \(AB = AC = 4\sqrt{2}\), the triangle is isosceles.
Check if it's a right - triangle: \(AB^{2}+AC^{2}=(4\sqrt{2})^{2}+(4\sqrt{2})^{2}=32 + 32 = 64\), \(BC^{2}=8^{2}=64\). So \(AB^{2}+AC^{2}=BC^{2}\), by Pythagorean theorem, \(\angle A = 90^{\circ}\), so \(\overline{AB}\perp\overline{AC}\) and the triangle is a right - triangle.
For parallelism: The slope of \(\overline{BC}\): \(m_{BC}=\frac{-1+1}{3 + 5}=0\) (horizontal line). The slope of \(\overline{AC}\): \(m_{AC}=\frac{-1 - 3}{3+1}=\frac{-4}{4}=-1\). Since \(m_{BC}
eq m_{AC}\), \(\overline{BC}\) is not parallel to \(\overline{AC}\).
Since \(AB = AC
eq BC\), it's not equilateral.

Answer:

\(\overline{AB}\perp\overline{AC}\), The triangle is a right triangle, The triangle is an isosceles triangle.