\begin{tabular}{l} Question 4 (Multiple Choice Worth 1 points) \ (04.01 MC) \ $$ \Delta \mathrm{ABC} $$ is located at $$ \mathrm{A}(2,3), \mathrm{B}(4,4) $$, and $$ \mathrm{C}(6,3) $$. Zackery says that $$ riangle \mathrm{ABC} $$ is an isosceles triangle, while Verna says that it is a right triangle. Who is correct? \ O Zackery, because $$ \overline{\mathrm{BC}} \cong \overline{\mathrm{AC}} $$ \ Zackery, because $$ \overline{\mathrm{AB}} \cong \overline{\mathrm{BC}} $$ \ verna, because $$ \overline{\mathrm{BC}} \perp \overline{\mathrm{AC}} $$ \ Verna, because $$ \overline{\mathrm{AB}} \perp \overline{\mathrm{BC}} $$ \ \hline\end{tabular}

Question

\begin{tabular}{l} Question 4 (Multiple Choice Worth 1 points) \ (04.01 MC) \ $$ \Delta \mathrm{ABC} $$ is located at $$ \mathrm{A}(2,3), \mathrm{B}(4,4) $$, and $$ \mathrm{C}(6,3) $$. Zackery says that $$ 	riangle \mathrm{ABC} $$ is an isosceles triangle, while Verna says that it is a right triangle. Who is correct? \ O Zackery, because $$ \overline{\mathrm{BC}} \cong \overline{\mathrm{AC}} $$ \ Zackery, because $$ \overline{\mathrm{AB}} \cong \overline{\mathrm{BC}} $$ \ verna, because $$ \overline{\mathrm{BC}} \perp \overline{\mathrm{AC}} $$ \ Verna, because $$ \overline{\mathrm{AB}} \perp \overline{\mathrm{BC}} $$ \ \hline\end{tabular}

UpStudy AI · Accepted Answer

**Step 1. Compute the lengths of the sides of $$ \triangle ABC $$.**

- For side $$ \overline{AB} $$ (from $$ A(2,3) $$ to $$ B(4,4) $$):
  $$
  AB = \sqrt{(4-2)^2+(4-3)^2}=\sqrt{2^2+1^2}=\sqrt{4+1}=\sqrt{5}
  $$

- For side $$ \overline{BC} $$ (from $$ B(4,4) $$ to $$ C(6,3) $$):
  $$
  BC = \sqrt{(6-4)^2+(3-4)^2}=\sqrt{2^2+(-1)^2}=\sqrt{4+1}=\sqrt{5}
  $$

- For side $$ \overline{AC} $$ (from $$ A(2,3) $$ to $$ C(6,3) $$):
  $$
  AC = \sqrt{(6-2)^2+(3-3)^2}=\sqrt{4^2+0^2}=\sqrt{16} = 4
  $$

**Step 2. Identify the Type of Triangle.**

Since $$ AB = BC = \sqrt{5} $$ and these two sides are congruent, the triangle is **isosceles**.

**Step 3. Verify if $$ \triangle ABC $$ is a right triangle.**

- A triangle is right if one pair of sides is perpendicular. The slopes of sides are:
  
  - Slope of $$ \overline{AB} $$:
    $$
    m_{AB} = \frac{4-3}{4-2}=\frac{1}{2}
    $$
  
  - Slope of $$ \overline{BC} $$:
    $$
    m_{BC} = \frac{3-4}{6-4}=\frac{-1}{2}
    $$

- Check if $$ m_{AB} \times m_{BC} = -1 $$:
  $$
  \frac{1}{2} \times \left(-\frac{1}{2}\right)=-\frac{1}{4} \neq -1
  $$

Thus, no pair of sides is perpendicular, and the triangle is **not** a right triangle.

**Step 4. Conclusion.**

Zackery is correct because the triangle is isosceles with $$ \overline{AB} \cong \overline{BC} $$.

The correct answer is:

**Zackery, because $$ \overline{AB} \cong \overline{BC} $$.**
Zackery is correct because $$ \overline{AB} \cong \overline{BC} $$.

Upstudy AI Solution

Answer

Solution

The Deep Dive

Related Questions

Latest Geometry Questions