To find the number of values of \( \alpha \) such that the points \( ( \alpha, 6), (-5, 0) \), and \( (5, 0) \) form an isosceles triangle, we will analyze the distances between the points and set them equal to find the possible values of \( \alpha \).
### Step 1: Define the points
Let:
- Point A: \( A(\alpha, 6) \)
- Point B: \( B(-5, 0) \)
- Point C: \( C(5, 0) \)
### Step 2: Calculate the distances
We will calculate the distances between the points \( A \), \( B \), and \( C \).
1. Distance \( AB \):
\[
AB = \sqrt{(\alpha - (-5))^2 + (6 - 0)^2} = \sqrt{(\alpha + 5)^2 + 6^2} = \sqrt{(\alpha + 5)^2 + 36}
\]
2. Distance \( BC \):
\[
BC = \sqrt{((-5) - 5)^2 + (0 - 0)^2} = \sqrt{(-10)^2} = 10
\]
3. Distance \( AC \):
\[
AC = \sqrt{(\alpha - 5)^2 + (6 - 0)^2} = \sqrt{(\alpha - 5)^2 + 6^2} = \sqrt{(\alpha - 5)^2 + 36}
\]
### Step 3: Set up equations for isosceles triangle
An isosceles triangle has at least two equal sides. We will consider three cases:
#### Case 1: \( AB = AC \)
Set the distances equal:
\[
\sqrt{(\alpha + 5)^2 + 36} = \sqrt{(\alpha - 5)^2 + 36}
\]
Squaring both sides:
\[
(\alpha + 5)^2 + 36 = (\alpha - 5)^2 + 36
\]
Canceling \( 36 \) and expanding:
\[
(\alpha + 5)^2 = (\alpha - 5)^2
\]
Expanding both sides:
\[
\alpha^2 + 10\alpha + 25 = \alpha^2 - 10\alpha + 25
\]
Simplifying:
\[
20\alpha = 0 \implies \alpha = 0
\]
#### Case 2: \( AB = BC \)
Set the distances equal:
\[
\sqrt{(\alpha + 5)^2 + 36} = 10
\]
Squaring both sides:
\[
(\alpha + 5)^2 + 36 = 100
\]
Solving:
\[
(\alpha + 5)^2 = 64
\]
Taking square roots:
\[
\alpha + 5 = 8 \quad \text{or} \quad \alpha + 5 = -8
\]
This gives:
\[
\alpha = 3 \quad \text{or} \quad \alpha = -13
\]
#### Case 3: \( AC = BC \)
Set the distances equal:
\[
\sqrt{(\alpha - 5)^2 + 36} = 10
\]
Squaring both sides:
\[
(\alpha - 5)^2 + 36 = 100
\]
Solving:
\[
(\alpha - 5)^2 = 64
\]
Taking square roots:
\[
\alpha - 5 = 8 \quad \text{or} \quad \alpha - 5 = -8
\]
This gives:
\[
\alpha = 13 \quad \text{or} \quad \alpha = -3
\]
### Step 4: Collect all values of \( \alpha \)
From the three cases, we have the following values:
- From Case 1: \( \alpha = 0 \)
- From Case 2: \( \alpha = 3, -13 \)
- From Case 3: \( \alpha = 13, -3 \)
### Final values of \( \alpha \)
The distinct values of \( \alpha \) are:
- \( 0, 3, -13, 13, -3 \)
### Conclusion
Thus, the total number of distinct values of \( \alpha \) such that the points form an isosceles triangle is **5**.
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