Question

Point A%0D%0A𝐴%0D%0A lies at (2,4)%0D%0A(%0D%0A2%0D%0A,%0D%0A4%0D%0A)%0D%0A and point B%0D%0A𝐵%0D%0A lies at (3,6)%0D%0A(%0D%0A3%0D%0A,%0D%0A6%0D%0A)%0D%0A. Which point is needed so that △ABC%0D%0A△%0D%0A𝐴%0D%0A𝐵%0D%0A𝐶%0D%0A is isosceles?(1 point)%0D%0AResponses%0D%0A%0D%0AC(3,4)%0D%0A𝐶%0D%0A(%0D%0A3%0D%0A,%0D%0A4%0D%0A)%0D%0Aupper C left parenthesis 3 comma 4 right parenthesis%0D%0A%0D%0AC(5,5)%0D%0A𝐶%0D%0A(%0D%0A5%0D%0A,%0D%0A5%0D%0A)%0D%0Aupper C left parenthesis 5 comma 5 right parenthesis%0D%0A%0D%0AC(5,4)%0D%0A𝐶%0D%0A(%0D%0A5%0D%0A,%0D%0A4%0D%0A)%0D%0Aupper C left parenthesis 5 comma 4 right parenthesis%0D%0A%0D%0AC(0,4)

Answer 1

To find the point \( C \) such that triangle \( \triangle ABC \) is isosceles, we'll check each option by calculating the distances from \( C \) to points \( A \) and \( B \) and then comparing those distances.

1. **Distance Formula**: The distance between two points \( (x_1, y_1) \) and \( (x_2, y_2) \) is given by:
\[
d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
\]

### Given points:
- \( A(2, 4) \)
- \( B(3, 6) \)

### Option A: \( C(3, 4) \)
- Distance \( AC \):
\[
AC = \sqrt{(3 - 2)^2 + (4 - 4)^2} = \sqrt{1^2 + 0^2} = 1
\]
- Distance \( BC \):
\[
BC = \sqrt{(3 - 3)^2 + (6 - 4)^2} = \sqrt{0^2 + 2^2} = 2
\]

Since \( AC \neq BC \), this does not form an isosceles triangle.

### Option B: \( C(5, 5) \)
- Distance \( AC \):
\[
AC = \sqrt{(5 - 2)^2 + (5 - 4)^2} = \sqrt{3^2 + 1^2} = \sqrt{9 + 1} = \sqrt{10}
\]
- Distance \( BC \):
\[
BC = \sqrt{(5 - 3)^2 + (5 - 6)^2} = \sqrt{2^2 + (-1)^2} = \sqrt{4 + 1} = \sqrt{5}
\]

Since \( AC \neq BC \), this does not form an isosceles triangle.

### Option C: \( C(5, 4) \)
- Distance \( AC \):
\[
AC = \sqrt{(5 - 2)^2 + (4 - 4)^2} = \sqrt{3^2 + 0^2} = 3
\]
- Distance \( BC \):
\[
BC = \sqrt{(5 - 3)^2 + (4 - 6)^2} = \sqrt{2^2 + (-2)^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2}
\]

Since \( AC \neq BC \), this does not form an isosceles triangle.

### Option D: \( C(0, 4) \)
- Distance \( AC \):
\[
AC = \sqrt{(0 - 2)^2 + (4 - 4)^2} = \sqrt{(-2)^2 + 0^2} = 2
\]
- Distance \( BC \):
\[
BC = \sqrt{(0 - 3)^2 + (4 - 6)^2} = \sqrt{(-3)^2 + (-2)^2} = \sqrt{9 + 4} = \sqrt{13}
\]

Since \( AC \neq BC \), this does not form an isosceles triangle.

### Conclusion:
None of the points \( C(3, 4) \), \( C(5, 5) \), \( C(5, 4) \), or \( C(0, 4) \) forms an isosceles triangle with the points \( A \) and \( B \). Please double-check the given coordinates or criteria for forming an isosceles triangle.