If the base of a triangle and the ratio of the lengths of the other two unequal sides are given, then the vertex lies on

Options

A. straight line
B. circle
C. ellipse
D. parabola

Explanation

To solve the problem, we need to determine the locus of the vertex of a triangle given the base and the ratio of the lengths of the other two sides. Let's break this down step by step.

### Step-by-Step Solution

1. **Define the Base of the Triangle**:
Let the base of the triangle be the line segment \( BC \) where the coordinates of point \( B \) are \( (0, 0) \) and the coordinates of point \( C \) are \( (a, 0) \).

2. **Define the Vertex**:
Let the vertex \( A \) have coordinates \( (h, k) \), where \( h \) and \( k \) are variables we will determine.

3. **Set Up the Ratio of the Sides**:
We know the ratio of the lengths of the sides \( AB \) and \( AC \) is given as \( \frac{AB}{AC} = \lambda \) where \( \lambda \neq 1 \) (since the sides are unequal).

4. **Use the Distance Formula**:
Using the distance formula, we can express the lengths of the sides:
- \( AB = \sqrt{(h - 0)^2 + (k - 0)^2} = \sqrt{h^2 + k^2} \)
- \( AC = \sqrt{(h - a)^2 + (k - 0)^2} = \sqrt{(h - a)^2 + k^2} \)

5. **Set Up the Equation**:
From the ratio \( \frac{AB}{AC} = \lambda \), we can square both sides to eliminate the square roots:
\[ \frac{AB^2}{AC^2} = \lambda^2 \implies \frac{h^2 + k^2}{(h - a)^2 + k^2} = \lambda^2 \]

6. **Cross-Multiply**:
Cross-multiplying gives us:
\[ h^2 + k^2 = \lambda^2 \left( (h - a)^2 + k^2 \right) \]

7. **Expand the Equation**:
Expanding the right-hand side:
\[ h^2 + k^2 = \lambda^2 (h^2 - 2ah + a^2 + k^2) \]

8. **Rearrange the Equation**:
Rearranging gives:
\[ h^2 + k^2 - \lambda^2 h^2 + \lambda^2 (2ah - a^2) + \lambda^2 k^2 = 0 \]
This simplifies to:
\[ (1 - \lambda^2)h^2 + (1 - \lambda^2)k^2 + 2a\lambda^2 h - a^2\lambda^2 = 0 \]

9. **Factor Out Common Terms**:
Factoring out \( (1 - \lambda^2) \):
\[ h^2 + k^2 + \frac{2a\lambda^2}{1 - \lambda^2} h - \frac{a^2\lambda^2}{1 - \lambda^2} = 0 \]

10. **Identify the Locus**:
This equation represents a circle in the \( (h, k) \) plane, indicating that the vertex \( A \) lies on a circle.

### Final Conclusion
Thus, the vertex \( A \) of the triangle lies on a circle.