What is the equation of a locus of a moving point which is equidistant from (0, 0) and (4, -2)?

The problem asks for the equation of the locus of a moving point that is equidistant from two fixed points, \((0, 0)\) and \((4, -2)\). The locus of points equidistant from two fixed points is the **perpendicular bisector** of the line segment joining those two points.

### Step 1: Find the midpoint of the line segment
The midpoint \(M\) of the segment connecting \((0, 0)\) and \((4, -2)\) can be found using the midpoint formula:
\[
M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)
\]
Substituting the values:
\[
M = \left(\frac{0 + 4}{2}, \frac{0 + (-2)}{2}\right) = \left(2, -1\right)
\]

### Step 2: Find the slope of the line segment
The slope \(m\) of the line segment connecting \((0, 0)\) and \((4, -2)\) is:
\[
m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-2 - 0}{4 - 0} = \frac{-2}{4} = -\frac{1}{2}
\]

### Step 3: Find the slope of the perpendicular bisector
The slope of the perpendicular bisector will be the negative reciprocal of \(-\frac{1}{2}\):
\[
\text{Slope of perpendicular bisector} = \frac{2}{1} = 2
\]

### Step 4: Write the equation of the perpendicular bisector
The equation of the line with slope 2 that passes through the midpoint \((2, -1)\) can be written in point-slope form:
\[
y - y_1 = m(x - x_1)
\]
Substituting the values:
\[
y - (-1) = 2(x - 2)
\]
Simplifying:
\[
y + 1 = 2x - 4
\]
\[
y = 2x - 5
\]

### Conclusion:
The equation of the locus of the moving point equidistant from \((0, 0)\) and \((4, -2)\) is:
\[
\boxed{y = 2x - 5}
\]

Would you like to explore any more related concepts or have any questions?

### Related Questions:
1. What is the geometric interpretation of the locus of points equidistant from two fixed points?
2. How do you find the perpendicular bisector of a line segment in 3D space?
3. What is the equation of the locus of a point equidistant from a point and a line?
4. How would the equation change if the two fixed points were \((a, b)\) and \((c, d)\)?
5. Can the perpendicular bisector concept be applied to non-Euclidean geometries?

**Tip:** To find the locus of points equidistant from two points, always consider the perpendicular bisector of the segment connecting those points.

Explore how to find the equation of the locus of a moving point equidistant from the points (0, 0) and (4, -2) using coordinate geometry. Understand the application of the midpoint formula, distance formula, and perpendicular bisector theorem. Suitable for high school students.

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