\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 \( \triangle \mathrm{ABC} \) 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}

To determine who is correct about triangle \( \Delta \mathrm{ABC} \), we can calculate the lengths of the sides using the distance formula. The lengths are as follows: \( \overline{AB} \) is approximately \( 2.24 \), \( \overline{BC} \) is approximately \( 2.24 \), and \( \overline{AC} \) is \( 4 \). This means that \( \overline{AB} \cong \overline{BC} \), confirming that Zackery is correct in identifying \( \Delta \mathrm{ABC} \) as an isosceles triangle. Next, to check if it's a right triangle, we can examine the slopes of the lines. The slope of \( \overline{AB} \) is \( \frac{1}{2} \) and the slope of \( \overline{BC} \) is \( \frac{-1}{2} \). Since the product of the slopes \( \left(\frac{1}{2} \times -\frac{1}{2} = -1\right) \) does not equal \( -1 \), this triangle does not contain a right angle. Therefore, while Zackery is right about it being isosceles, Verna's claim that it is a right triangle is incorrect.