suppose I have an equilateral triangle ABC whose base is AB and apex is C. given co-ordinates A(ax,ay), B(bx, by) and C(x,y). Given a fixed point V which is on side AB and whose co-ordinates is(p,q). Let's assume A is moving on a straight line ay=ma*ax+ca and B is moving on a straight line by=mb*bx+cb, but point V is on a fixed co-ordinate on AB. How to caluculate equation of locus of point C using all the given points, but keep in mind the final equation will contain only x,y,p and q and no other variable will be in the final equation?

To determine the equation of the locus of point \( C \) of the equilateral triangle \( ABC \) with the given conditions, we can follow these steps:

1. **Equation of Line AB:**

   Since \( A \) and \( B \) move along given straight lines, let's write their parametric forms:
   \[
   A: (ax, ay) = (t, ma \cdot t + ca)
   \]
   \[
   B: (bx, by) = (u, mb \cdot u + cb)
   \]

2. **Fixed Point V:**

   The point \( V \) with coordinates \( (p, q) \) lies on the line segment \( AB \). The line \( AB \) can be parameterized as follows:
   \[
   (1 - \lambda) A + \lambda B = (p, q)
   \]
   where \( 0 \leq \lambda \leq 1 \). This gives us the system of equations:
   \[
   (1 - \lambda) t + \lambda u = p
   \]
   \[
   (1 - \lambda) (ma \cdot t + ca) + \lambda (mb \cdot u + cb) = q
   \]

3. **Solve for \(\lambda\):**

   From the first equation:
   \[
   t = \frac{p - \lambda u}{1 - \lambda}
   \]
   
   Substitute \( t \) in the second equation:
   \[
   (1 - \lambda) \left(ma \cdot \frac{p - \lambda u}{1 - \lambda} + ca\right) + \lambda (mb \cdot u + cb) = q
   \]

4. **Equation of Point C:**

   Given \( C \) is the apex of an equilateral triangle, we can find its coordinates using the midpoint \( M \) of \( AB \) and a perpendicular line to \( AB \):
   \[
   M = \left( \frac{t + u}{2}, \frac{ma \cdot t + ca + mb \cdot u + cb}{2} \right)
   \]
   Since \( AB \) is the base, and the height is perpendicular, the coordinates of \( C \) can be found using rotation of \( \frac{\pi}{3} \) radians (or 60 degrees) from \( M \).

   The height \( h \) from \( C \) to \( AB \):
   \[
   h = \frac{\sqrt{3}}{2} \cdot AB = \frac{\sqrt{3}}{2} \sqrt{(u - t)^2 + (mb \cdot u + cb - ma \cdot t - ca)^2}
   \]

5. **Eliminate Parameters:**

   Substitute \( t \) and \( u \) into the coordinates of \( C \), and express \( x \) and \( y \) in terms of \( p \) and \( q \) only.

Since we have a system involving parametric lines and fixed points, solving it explicitly can be quite complex. However, the key idea is to substitute the parametric points \( A \) and \( B \), express \( V \) and then derive the final expression for \( C \). Using symbolic computation tools could assist in simplifying these equations further.

For a comprehensive solution, more specific steps and possibly numerical methods or algebraic simplifications would be required. If the algebra becomes too cumbersome, numerical techniques or software tools (like MATLAB, Python's SymPy, or similar) might be necessary to handle the algebraic manipulation.

Would you like to go into further detail on a specific part of the process or explore this algebraically with specific values?

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Here are 5 questions for you to consider next:
1. How do you find the coordinates of the midpoint of a line segment?
2. What is the formula for the distance between two points in a coordinate plane?
3. How do you derive the equation of a line given two points on it?
4. How can you express the parametric equations for a point moving along a line?
5. What are the properties of an equilateral triangle that could simplify calculations?

**Tip:** Using algebraic software tools can greatly assist in simplifying complex systems of equations involving multiple parameters.

Learn how to derive the equation of the locus of point C in equilateral triangle ABC using coordinate geometry and parametric equations. This problem involves finding a fixed point V on side AB and deriving the locus equation involving only coordinates x, y, p, and q.

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