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?

---

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!