Question

what is true about \\(\delta abc\\)? select three options

\\(\overline{ab} \perp \overline{ac}\\)

the triangle is a right triangle.

the triangle is an isosceles triangle.

the triangle is an equilateral triangle.

\\(\overline{bc} \parallel \overline{ac}\\)

Explanation:

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<reasoning>

Calculate slopes of the triangle sides

\[

$$\begin{aligned} &A(-1, 3),\quad B(-5, -1),\quad C(3, -1)\\ &\text{Slope of } \overline{AB} = \frac{-1 - 3}{-5 - (-1)} = \frac{-4}{-4} = 1\\ &\text{Slope of } \overline{AC} = \frac{-1 - 3}{3 - (-1)} = \frac{-4}{4} = -1\\ &\text{Slope of } \overline{BC} = \frac{-1 - (-1)}{3 - (-5)} = 0 \end{aligned}$$

\]

Verify perpendicularity and right triangle properties

\[

$$\begin{aligned} &(\text{Slope of } \overline{AB}) \cdot (\text{Slope of } \overline{AC}) = 1 \cdot (-1) = -1 \implies \overline{AB} \perp \overline{AC}\\ &\angle A = 90^\circ \implies \triangle ABC \text{ is a right triangle} \end{aligned}$$

\]

Calculate side lengths to determine triangle type

\[

$$\begin{aligned} &AB = \sqrt{(-5 - (-1))^2 + (-1 - 3)^2} = \sqrt{(-4)^2 + (-4)^2} = \sqrt{32} = 4\sqrt{2}\\ &AC = \sqrt{(3 - (-1))^2 + (-1 - 3)^2} = \sqrt{4^2 + (-4)^2} = \sqrt{32} = 4\sqrt{2}\\ &BC = \sqrt{(3 - (-5))^2 + (-1 - (-1))^2} = \sqrt{8^2 + 0^2} = 8\\ &AB = AC eq BC \implies \triangle ABC \text{ is an isosceles triangle} \end{aligned}$$

\]
</reasoning>

<answer>
<mcq-correct>\(\overline{AB} \perp \overline{AC}\)</mcq-correct>
<mcq-correct>The triangle is a right triangle.</mcq-correct>
<mcq-correct>The triangle is an isosceles triangle.</mcq-correct>
<mcq-option>The triangle is an equilateral triangle.</mcq-option>
<mcq-option>\(\overline{BC} \parallel \overline{AC}\)</mcq-option>
</answer>

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Answer:

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<reasoning>

Calculate slopes of the triangle sides

\[

$$\begin{aligned} &A(-1, 3),\quad B(-5, -1),\quad C(3, -1)\\ &\text{Slope of } \overline{AB} = \frac{-1 - 3}{-5 - (-1)} = \frac{-4}{-4} = 1\\ &\text{Slope of } \overline{AC} = \frac{-1 - 3}{3 - (-1)} = \frac{-4}{4} = -1\\ &\text{Slope of } \overline{BC} = \frac{-1 - (-1)}{3 - (-5)} = 0 \end{aligned}$$

\]

Verify perpendicularity and right triangle properties

\[

$$\begin{aligned} &(\text{Slope of } \overline{AB}) \cdot (\text{Slope of } \overline{AC}) = 1 \cdot (-1) = -1 \implies \overline{AB} \perp \overline{AC}\\ &\angle A = 90^\circ \implies \triangle ABC \text{ is a right triangle} \end{aligned}$$

\]

Calculate side lengths to determine triangle type

\[

$$\begin{aligned} &AB = \sqrt{(-5 - (-1))^2 + (-1 - 3)^2} = \sqrt{(-4)^2 + (-4)^2} = \sqrt{32} = 4\sqrt{2}\\ &AC = \sqrt{(3 - (-1))^2 + (-1 - 3)^2} = \sqrt{4^2 + (-4)^2} = \sqrt{32} = 4\sqrt{2}\\ &BC = \sqrt{(3 - (-5))^2 + (-1 - (-1))^2} = \sqrt{8^2 + 0^2} = 8\\ &AB = AC eq BC \implies \triangle ABC \text{ is an isosceles triangle} \end{aligned}$$

\]
</reasoning>

<answer>
<mcq-correct>\(\overline{AB} \perp \overline{AC}\)</mcq-correct>
<mcq-correct>The triangle is a right triangle.</mcq-correct>
<mcq-correct>The triangle is an isosceles triangle.</mcq-correct>
<mcq-option>The triangle is an equilateral triangle.</mcq-option>
<mcq-option>\(\overline{BC} \parallel \overline{AC}\)</mcq-option>
</answer>

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