If `A(3,4)` and `B(-5,-2)` are the extremities of the base of an isosceles triangle ABC with tan `C = 2`, then point C can be

Options

A. `((3sqrt(5)-1)/(2),-(1+2sqrt(5)))`
B. `(-((3sqrt(5)+5))/(2),3+2sqrt(5))`
C. `((3sqrt(5)-1)/(2),3-2sqrt(5))`
D. `(-((3sqrt(5)-5))/(2),-(1-2sqrt(5)))`

Explanation

To solve the problem, we need to find the coordinates of point C in triangle ABC, where A(3, 4) and B(-5, -2) are the extremities of the base, and the angle at C (angle ACB) has a tangent of 2.

### Step-by-Step Solution:

1. **Find the coordinates of points A and B:**
- A = (3, 4)
- B = (-5, -2)

2. **Calculate the slope of line AB:**
The slope \( m \) of line AB can be calculated using the formula:
\[ m = \frac{y_2 - y_1}{x_2 - x_1} \]
Substituting the coordinates of points A and B:
\[ m = \frac{-2 - 4}{-5 - 3} = \frac{-6}{-8} = \frac{3}{4} \]

3. **Find the slope of line AC and BC:**
Since tan C = 2, we can find the slope of line AC (or BC) using the definition of tangent:
\[ \tan C = \frac{\text{opposite}}{\text{adjacent}} = 2 \]
This means the slope of line AC can be \( m_1 = 2 \) or \( m_2 = -2 \) (for the two possible positions of point C).

4. **Use the point-slope form to find the equations of lines AC and BC:**
- For line AC with slope \( m_1 = 2 \):
\[ y - 4 = 2(x - 3) \implies y = 2x - 6 + 4 \implies y = 2x - 2 \]
- For line BC with slope \( m_2 = -2 \):
\[ y + 2 = -2(x + 5) \implies y = -2x - 10 - 2 \implies y = -2x - 12 \]

5. **Find the intersection of lines AC and BC:**
Set the equations equal to each other to find the coordinates of point C:
\[ 2x - 2 = -2x - 12 \]
Solving for x:
\[ 2x + 2x = -12 + 2 \implies 4x = -10 \implies x = -\frac{5}{2} \]
Now substitute \( x = -\frac{5}{2} \) back into one of the equations to find y:
\[ y = 2\left(-\frac{5}{2}\right) - 2 = -5 - 2 = -7 \]
Thus, point C can be at \( C\left(-\frac{5}{2}, -7\right) \).

6. **Check the other possible slope:**
If we consider the other slope \( m_2 = -2 \) for line AC:
\[ y - 4 = -2(x - 3) \implies y = -2x + 6 + 4 \implies y = -2x + 10 \]
Set this equal to the equation of line BC:
\[ -2x + 10 = -2x - 12 \]
This leads to a contradiction, indicating that the other configuration does not yield a valid point C.

### Final Answer:
The coordinates of point C can be \( C\left(-\frac{5}{2}, -7\right) \).