Question

Point A lies at (2, 4) and point B lies at (3, 6). Which point is needed so that AABC is isosceles? (1 point)
О C (5,5)
O C (3,4)
OC (0,4)
OC (5,4)

Answer 1

To determine which point \( C \) can form an isosceles triangle \( \triangle ABC \) with points \( A(2, 4) \) and \( B(3, 6) \), we need to find a point \( C \) such that at least two sides of the triangle are of equal length.

1. **Calculate the distance \( AB \):**
\[
AB = \sqrt{(3 - 2)^2 + (6 - 4)^2} = \sqrt{1^2 + 2^2} = \sqrt{1 + 4} = \sqrt{5}
\]

2. **Check the given options for point \( C \):**
- **Option 1: \( C(5, 5) \)**
\[
AC = \sqrt{(5 - 2)^2 + (5 - 4)^2} = \sqrt{3^2 + 1^2} = \sqrt{9 + 1} = \sqrt{10}
\]
\[
BC = \sqrt{(5 - 3)^2 + (5 - 6)^2} = \sqrt{2^2 + (-1)^2} = \sqrt{4 + 1} = \sqrt{5}
\]
(AB ≠ AC, but AB = BC)

- **Option 2: \( C(3, 4) \)**
\[
AC = \sqrt{(3 - 2)^2 + (4 - 4)^2} = \sqrt{1^2 + 0^2} = 1
\]
\[
BC = \sqrt{(3 - 3)^2 + (6 - 4)^2} = \sqrt{0^2 + 2^2} = 2
\]
(No equal lengths)

- **Option 3: \( C(0, 4) \)**
\[
AC = \sqrt{(0 - 2)^2 + (4 - 4)^2} = \sqrt{(-2)^2 + 0^2} = 2
\]
\[
BC = \sqrt{(0 - 3)^2 + (4 - 6)^2} = \sqrt{(-3)^2 + (-2)^2} = \sqrt{9 + 4} = \sqrt{13}
\]
(No equal lengths)

- **Option 4: \( C(5, 4) \)**
\[
AC = \sqrt{(5 - 2)^2 + (4 - 4)^2} = \sqrt{3^2 + 0^2} = 3
\]
\[
BC = \sqrt{(5 - 3)^2 + (4 - 6)^2} = \sqrt{2^2 + (-2)^2} = \sqrt{4 + 4} = \sqrt{8}
\]
(No equal lengths)

3. **Conclusion:**
The only case where two sides are equal is with \( C(5, 5) \), where \( AB = BC \).

Therefore, the correct answer is:

**Option 1: \( C(5, 5) \)**.