Question

Point A
𝐴
lies at (1,2)
(
1
,
2
)
and point B
𝐵
lies at (4,5)
(
4
,
5
)
. Which point in needed so that △ABC
△
𝐴
𝐵
𝐶
is isosceles? (1 point)
Responses

C (1,5)
𝐶

(
1
,
5
)
C (1,5)
𝐶

(
1
,
5
)

C (4,2)
𝐶

(
4
,
2
)
C (4,2)
𝐶

(
4
,
2
)

C (7,8)
𝐶

(
7
,
8
)
C (7,8)
𝐶

(
7
,
8
)

C (3,2)
𝐶

(
3
,
2
)
C (3,2)

Answer 1

To determine which point \( C \) would make triangle \( \triangle ABC \) isosceles, we need to find a point \( C \) that makes at least two sides of the triangle equal in length.

1. **Calculate the distance from point A to B**:
\[
AB = \sqrt{(4 - 1)^2 + (5 - 2)^2} = \sqrt{3^2 + 3^2} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2}
\]

2. **Now, we will evaluate each of the given points for \( C \)**:

- **For \( C(1, 5) \)**:
\[
AC = \sqrt{(1 - 1)^2 + (5 - 2)^2} = \sqrt{0 + 3^2} = 3
\]
\[
BC = \sqrt{(4 - 1)^2 + (5 - 5)^2} = \sqrt{3^2 + 0} = 3
\]
\( AC = BC \) => This would form an isosceles triangle.

- **For \( C(4, 2) \)**:
\[
AC = \sqrt{(4 - 1)^2 + (2 - 2)^2} = \sqrt{3^2 + 0} = 3
\]
\[
BC = \sqrt{(4 - 4)^2 + (5 - 2)^2} = \sqrt{0 + 3^2} = 3
\]
\( AC = BC \) => This also forms an isosceles triangle.

- **For \( C(7, 8) \)**:
\[
AC = \sqrt{(7 - 1)^2 + (8 - 2)^2} = \sqrt{6^2 + 6^2} = \sqrt{72} = 6\sqrt{2}
\]
\[
BC = \sqrt{(7 - 4)^2 + (8 - 5)^2} = \sqrt{3^2 + 3^2} = 3\sqrt{2}
\]
\( AC \neq BC \) => This does not form an isosceles triangle.

- **For \( C(3, 2) \)**:
\[
AC = \sqrt{(3 - 1)^2 + (2 - 2)^2} = \sqrt{2^2 + 0} = 2
\]
\[
BC = \sqrt{(4 - 3)^2 + (5 - 2)^2} = \sqrt{1^2 + 3^2} = \sqrt{10}
\]
\( AC \neq BC \) => This does not form an isosceles triangle.

3. **Conclusion**:
The points \( C(1, 5) \) and \( C(4, 2) \) both result in isosceles triangles with \( AB \). The correct responses that allow \( \triangle ABC \) to be isosceles are \( C(1, 5) \) and \( C(4, 2) \).

If you are looking for one answer, choose either \( C(1, 5) \) or \( C(4, 2) \).