To find the equation for the set of all points in the plane that are equidistant from a point and a line, you can use the definition of an equidistant locus. In this case, the locus of points is equidistant from the point $P = (3, \frac{23}{4})$ and the line $y = -\frac{3}{4}$.

Step-by-step process:

1. Distance from a point to a line: The formula for the distance from a point $(x, y)$ to a horizontal line $y = c$ is: $\text{Distance to line} = |y - c|$ Here, $c = -\frac{3}{4}$, so the distance from any point $(x, y)$ to the line $y = -\frac{3}{4}$ is: $\text{Distance to line} = |y + \frac{3}{4}|$

2. Distance from a point to another point: The distance from the point $P = (3, \frac{23}{4})$ to any point $(x, y)$ in the plane is given by the Euclidean distance formula: $\text{Distance to } P = \sqrt{(x - 3)^2 + \left(y - \frac{23}{4}\right)^2}$

3. Equidistant condition: For the point $(x, y)$ to be equidistant from both $P$ and the line, we set the distances equal: $\sqrt{(x - 3)^2 + \left(y - \frac{23}{4}\right)^2} = |y + \frac{3}{4}|$

4. Square both sides to eliminate the square root: $(x - 3)^2 + \left(y - \frac{23}{4}\right)^2 = \left(y + \frac{3}{4}\right)^2$

5. Simplify the equation: First, expand both sides: $(x - 3)^2 + \left(y - \frac{23}{4}\right)^2 = \left(y + \frac{3}{4}\right)^2$ Expand each term: $(x^2 - 6x + 9) + \left(y^2 - 2y \cdot \frac{23}{4} + \left(\frac{23}{4}\right)^2 \right) = \left( y^2 + 2y \cdot \frac{3}{4} + \left(\frac{3}{4}\right)^2 \right)$

   After simplifying: $(x - 3)^2 + \left( y - \frac{23}{4} \right)^2 = \left( y + \frac{3}{4} \right)^2$

This is the final result for the equation of the set of points equidistant from both the point and the line.

Do you need further details or have any questions about this process?

Here are 5 related questions:

1. How do you interpret the geometric meaning of this equation?
2. What would happen if the point $P$ or the line were moved to different positions?
3. How can you simplify this equation further?
4. How does the distance formula change for non-horizontal lines?
5. What other shapes can you generate by changing the equidistant conditions (e.g., between two points or two lines)?

Tip: When squaring both sides of an equation, be cautious of potential extraneous solutions, especially if the context involves absolute values or other constraints.