Answered

[tex]$\triangle ABC$[/tex] is located at \(A(2,3), B(4,4)\), and \(C(6,3)\). Zackery says that \(\triangle ABC\) is an isosceles triangle, while Verna says that it is a right triangle. Who is correct?

A. Zackery, because \(\overline{BC} = \overline{AC}\)
B. Zackery, because \(\overline{AB} = \overline{BC}\)
C. Verna, because \(\overline{BC} \perp \overline{AC}\)
D. Verna, because [tex]\(\overline{AB} \perp \overline{BC}\)[/tex]



Answer :

GinnyAnswer
To determine who is correct between Zackery and Verna, we need to check whether \(\triangle ABC\) is isosceles or right. This entails calculating the lengths of the sides of the triangle and checking both properties: equality of any two sides for isosceles and the Pythagorean theorem for checking if it is right.

### Step 1: Calculate the lengths of the sides

1. Length of \( \overline{AB} \):
[tex]\[ \overline{AB} = \sqrt{(4 - 2)^2 + (4 - 3)^2} = \sqrt{2^2 + 1^2} = \sqrt{4 + 1} = \sqrt{5} \][/tex]

2. Length of \( \overline{BC} \):
[tex]\[ \overline{BC} = \sqrt{(6 - 4)^2 + (3 - 4)^2} = \sqrt{2^2 + (-1)^2} = \sqrt{4 + 1} = \sqrt{5} \][/tex]

3. Length of \( \overline{AC} \):
[tex]\[ \overline{AC} = \sqrt{(6 - 2)^2 + (3 - 3)^2} = \sqrt{4^2 + 0^2} = \sqrt{16 + 0} = \sqrt{16} = 4 \][/tex]

### Step 2: Check for an isosceles triangle

An isosceles triangle has at least two sides of equal length. From the calculations:

[tex]\[ \overline{AB} = \sqrt{5} \quad \text{and} \quad \overline{BC} = \sqrt{5} \][/tex]

Since \( \overline{AB} = \overline{BC} \), \(\triangle ABC\) is indeed an isosceles triangle.

### Step 3: Check for a right triangle

A right triangle satisfies the Pythagorean theorem:
[tex]\[ \text{(Hypotenuse)}^2 = \text{(Leg1)}^2 + \text{(Leg2)}^2 \][/tex]

Calculating the squares of the sides:
[tex]\[ \overline{AB}^2 = 5, \quad \overline{BC}^2 = 5, \quad \overline{AC}^2 = 16 \][/tex]
Check the conditions:

- \( \overline{AB}^2 + \overline{BC}^2 = 5 + 5 = 10 \neq 16 \)
- \( \overline{AB}^2 + \overline{AC}^2 = 5 + 16 = 21 \neq 5 \)
- \( \overline{BC}^2 + \overline{AC}^2 = 5 + 16 = 21 \neq 5 \)

None of these conditions satisfy the Pythagorean theorem. Therefore, \(\triangle ABC\) is not a right triangle.

### Conclusion:

- Zackery's claim: The triangle is isosceles because \(\overline{AB} = \overline{BC}\).
- Verna's claim: The triangle is a right triangle, which is incorrect.

Thus, the correct answer is:
Zackery, because [tex]\(\overline {AB} = \overline {BC}\)[/tex].