Find the values of a for which the triangle ABC is an isosceles triangle, where A, B, and C have the coordinates (-1, 1), (1, 3) and (3, a) respectively.

A) If AB = BC, then a = 1

B) If AB = BC, then a = -1 or -5

C) If BC = CA, then a = -1

D) If BC = CA, then a = 1

To solve the problem, we need to determine the values of \( a \) that make the triangle ABC isosceles, i.e., two of its sides must be equal. The coordinates of the points are \( A(-1, 1) \), \( B(1, 3) \), and \( C(3, a) \).

We will explore two cases: 
1. **AB = BC** (isosceles with AB and BC equal).
2. **BC = CA** (isosceles with BC and CA equal).

### 1. Case 1: \( AB = BC \)
The distance formula between two points \( (x_1, y_1) \) and \( (x_2, y_2) \) is:

\[
d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
\]

#### Finding \( AB \):
The coordinates of \( A \) and \( B \) are \( A(-1, 1) \) and \( B(1, 3) \).

\[
AB = \sqrt{(1 - (-1))^2 + (3 - 1)^2} = \sqrt{(2)^2 + (2)^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2}
\]

#### Finding \( BC \):
The coordinates of \( B \) and \( C \) are \( B(1, 3) \) and \( C(3, a) \).

\[
BC = \sqrt{(3 - 1)^2 + (a - 3)^2} = \sqrt{(2)^2 + (a - 3)^2} = \sqrt{4 + (a - 3)^2}
\]

To make \( AB = BC \), we equate the distances:

\[
2\sqrt{2} = \sqrt{4 + (a - 3)^2}
\]

Square both sides:

\[
8 = 4 + (a - 3)^2
\]

\[
(a - 3)^2 = 4
\]

Take the square root of both sides:

\[
a - 3 = 2 \quad \text{or} \quad a - 3 = -2
\]

Solving these:

\[
a = 5 \quad \text{or} \quad a = 1
\]

Thus, for \( AB = BC \), \( a = 1 \) or \( a = 5 \).

### 2. Case 2: \( BC = CA \)

#### Finding \( CA \):
The coordinates of \( C \) and \( A \) are \( C(3, a) \) and \( A(-1, 1) \).

\[
CA = \sqrt{(3 - (-1))^2 + (a - 1)^2} = \sqrt{(4)^2 + (a - 1)^2} = \sqrt{16 + (a - 1)^2}
\]

We want to make \( BC = CA \), so we equate the distances:

\[
\sqrt{4 + (a - 3)^2} = \sqrt{16 + (a - 1)^2}
\]

Square both sides:

\[
4 + (a - 3)^2 = 16 + (a - 1)^2
\]

Simplify:

\[
4 + (a^2 - 6a + 9) = 16 + (a^2 - 2a + 1)
\]

\[
a^2 - 6a + 13 = a^2 - 2a + 17
\]

Cancel out \( a^2 \) from both sides:

\[
-6a + 13 = -2a + 17
\]

Simplify:

\[
-4a = 4
\]

\[
a = -1
\]

Thus, for \( BC = CA \), \( a = -1 \).

### Final Answer:

- If \( AB = BC \), then \( a = 1 \) or \( a = 5 \).
- If \( BC = CA \), then \( a = -1 \).

Comparing with the given options:

- **B** is correct: If \( AB = BC \), then \( a = -1 \) or \( a = 5 \).
  
Would you like further clarification or details on this problem?

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**Relative questions:**

1. How do you apply the distance formula to solve geometric problems?
2. What are other methods to prove a triangle is isosceles apart from using distances?
3. How does the distance formula relate to the Pythagorean theorem?
4. Can a triangle be isosceles if the coordinates are non-integer values?
5. What happens if two conditions of triangle sides being equal are met simultaneously?

**Tip:** Always double-check your calculations for simplifications when dealing with square roots in geometric problems!

Find the values of a for which the triangle ABC is an isosceles triangle, where A, B, and C have the coordinates (-1, 1), (1, 3) and (3, a) respectively.

Explore how to find the values of a that make triangle ABC isosceles, using the distance formula. This problem involves determining equal sides based on the coordinates of A(-1,1), B(1,3), and C(3,a). Learn the step-by-step method to solve this geometric problem, suitable for students in Grades 9-11.

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