Question 4 (1 point) (04.01 MC) \( \triangle A B C \) is located at \( A(3,1), B(4,4) \), and \( C(5,1) \). Zackery says that \( \triangle A B C \) is an equilateral triangle, while Verna says that it is an isosceles triangle. Who is correct? a Zackery, because all of the sides are congruent b Zackery, Decause there are no congruent sides c Verna, because \( \overline{\mathrm{BC}} \cong \overline{\mathrm{AC}} \) d Verna, because \( \overline{A B} \cong \overline{\mathrm{BC}} \)

To determine who is correct, we need to calculate the lengths of each side of \( \triangle ABC \). Using the distance formula, we find: - \( AB = \sqrt{(4 - 3)^2 + (4 - 1)^2} = \sqrt{1^2 + 3^2} = \sqrt{10} \) - \( BC = \sqrt{(5 - 4)^2 + (1 - 4)^2} = \sqrt{1^2 + (-3)^2} = \sqrt{10} \) - \( AC = \sqrt{(5 - 3)^2 + (1 - 1)^2} = \sqrt{2^2 + 0^2} = 2 \) We can see \( AB \cong BC \) (both equal \( \sqrt{10} \)), while \( AC \neq AB \). Therefore, Verna is correct, and the triangle is isosceles since it has at least two congruent sides. Answer: c Verna, because \( \overline{\mathrm{BC}} \cong \overline{\mathrm{AC}} \)