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Let's verify the statements made by Zackery and Verna regarding the triangle [tex]\( \triangle ABC \)[/tex] with vertices at [tex]\( A(2,3) \)[/tex], [tex]\( B(4,4) \)[/tex], and [tex]\( C(6,3) \)[/tex].
### Step 1: Calculate the lengths of the sides
1. Length of [tex]\( \overline{AB} \)[/tex]:
[tex]\[ \overline{AB} = \sqrt{(4 - 2)^2 + (4 - 3)^2} = \sqrt{2^2 + 1^2} = \sqrt{4 + 1} = \sqrt{5} \][/tex]
2. Length of [tex]\( \overline{BC} \)[/tex]:
[tex]\[ \overline{BC} = \sqrt{(6 - 4)^2 + (3 - 4)^2} = \sqrt{2^2 + (-1)^2} = \sqrt{4 + 1} = \sqrt{5} \][/tex]
3. Length of [tex]\( \overline{AC} \)[/tex]:
[tex]\[ \overline{AC} = \sqrt{(6 - 2)^2 + (3 - 3)^2} = \sqrt{4^2 + 0^2} = \sqrt{16} = 4 \][/tex]
### Step 2: Check for isosceles condition
A triangle is isosceles if at least two of its sides are of equal length. From our calculations:
[tex]\[ \overline{AB} = \sqrt{5}, \quad \overline{BC} = \sqrt{5}, \quad \overline{AC} = 4 \][/tex]
Since [tex]\( \overline{AB} = \overline{BC} \)[/tex], triangle [tex]\( \triangle ABC \)[/tex] is indeed isosceles. Therefore, Zackery is correct.
Reason: [tex]\( \overline{AB} \cong \overline{BC} \)[/tex].
### Step 3: Check for right angles
A triangle is a right triangle if one of its angles is a right angle. To check this, we can use the concept of dot product for the vectors representing the sides.
1. Vector [tex]\( \overline{AB} \)[/tex]:
[tex]\[ \overline{AB} = (4 - 2, 4 - 3) = (2, 1) \][/tex]
2. Vector [tex]\( \overline{BC} \)[/tex]:
[tex]\[ \overline{BC} = (6 - 4, 3 - 4) = (2, -1) \][/tex]
3. Vector [tex]\( \overline{CA} \)[/tex]:
[tex]\[ \overline{CA} = (2 - 6, 3 - 3) = (-4, 0) \][/tex]
For a right angle to exist, the dot product of two vectors should be zero. Compute the dot products:
[tex]\[ \overline{AB} \cdot \overline{BC} = (2, 1) \cdot (2, -1) = 2 \cdot 2 + 1 \cdot (-1) = 4 - 1 = 3 \neq 0 \][/tex]
[tex]\[ \overline{BC} \cdot \overline{CA} = (2, -1) \cdot (-4, 0) = 2 \cdot (-4) + (-1) \cdot 0 = -8 + 0 = -8 \neq 0 \][/tex]
[tex]\[ \overline{CA} \cdot \overline{AB} = (-4, 0) \cdot (2, 1) = -4 \cdot 2 + 0 \cdot 1 = -8 + 0 = -8 \neq 0 \][/tex]
None of the angles in [tex]\( \triangle ABC \)[/tex] is a right angle. Therefore, Verna is incorrect.
### Conclusion
Based on our analysis, the findings are as follows:
- Zackery is correct because [tex]\( \overline{AB} \cong \overline{BC} \)[/tex], making [tex]\( \triangle ABC \)[/tex] an isosceles triangle.
- Verna is incorrect because [tex]\( \triangle ABC \)[/tex] is not a right triangle.
Thus, the correct point is:
Zackery, because [tex]\(\overline{AB} \cong \overline{BC}\)[/tex].
### Step 1: Calculate the lengths of the sides
1. Length of [tex]\( \overline{AB} \)[/tex]:
[tex]\[ \overline{AB} = \sqrt{(4 - 2)^2 + (4 - 3)^2} = \sqrt{2^2 + 1^2} = \sqrt{4 + 1} = \sqrt{5} \][/tex]
2. Length of [tex]\( \overline{BC} \)[/tex]:
[tex]\[ \overline{BC} = \sqrt{(6 - 4)^2 + (3 - 4)^2} = \sqrt{2^2 + (-1)^2} = \sqrt{4 + 1} = \sqrt{5} \][/tex]
3. Length of [tex]\( \overline{AC} \)[/tex]:
[tex]\[ \overline{AC} = \sqrt{(6 - 2)^2 + (3 - 3)^2} = \sqrt{4^2 + 0^2} = \sqrt{16} = 4 \][/tex]
### Step 2: Check for isosceles condition
A triangle is isosceles if at least two of its sides are of equal length. From our calculations:
[tex]\[ \overline{AB} = \sqrt{5}, \quad \overline{BC} = \sqrt{5}, \quad \overline{AC} = 4 \][/tex]
Since [tex]\( \overline{AB} = \overline{BC} \)[/tex], triangle [tex]\( \triangle ABC \)[/tex] is indeed isosceles. Therefore, Zackery is correct.
Reason: [tex]\( \overline{AB} \cong \overline{BC} \)[/tex].
### Step 3: Check for right angles
A triangle is a right triangle if one of its angles is a right angle. To check this, we can use the concept of dot product for the vectors representing the sides.
1. Vector [tex]\( \overline{AB} \)[/tex]:
[tex]\[ \overline{AB} = (4 - 2, 4 - 3) = (2, 1) \][/tex]
2. Vector [tex]\( \overline{BC} \)[/tex]:
[tex]\[ \overline{BC} = (6 - 4, 3 - 4) = (2, -1) \][/tex]
3. Vector [tex]\( \overline{CA} \)[/tex]:
[tex]\[ \overline{CA} = (2 - 6, 3 - 3) = (-4, 0) \][/tex]
For a right angle to exist, the dot product of two vectors should be zero. Compute the dot products:
[tex]\[ \overline{AB} \cdot \overline{BC} = (2, 1) \cdot (2, -1) = 2 \cdot 2 + 1 \cdot (-1) = 4 - 1 = 3 \neq 0 \][/tex]
[tex]\[ \overline{BC} \cdot \overline{CA} = (2, -1) \cdot (-4, 0) = 2 \cdot (-4) + (-1) \cdot 0 = -8 + 0 = -8 \neq 0 \][/tex]
[tex]\[ \overline{CA} \cdot \overline{AB} = (-4, 0) \cdot (2, 1) = -4 \cdot 2 + 0 \cdot 1 = -8 + 0 = -8 \neq 0 \][/tex]
None of the angles in [tex]\( \triangle ABC \)[/tex] is a right angle. Therefore, Verna is incorrect.
### Conclusion
Based on our analysis, the findings are as follows:
- Zackery is correct because [tex]\( \overline{AB} \cong \overline{BC} \)[/tex], making [tex]\( \triangle ABC \)[/tex] an isosceles triangle.
- Verna is incorrect because [tex]\( \triangle ABC \)[/tex] is not a right triangle.
Thus, the correct point is:
Zackery, because [tex]\(\overline{AB} \cong \overline{BC}\)[/tex].
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