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The problem asks to determine the locus of all points on the plane that are equidistant from the point \( (5, -3) \) and the line given by the equation \( 2x - y = 5 \).

### Step-by-step solution:

1. **Point and Line**: 
   - Point \( A(5, -3) \) is given.
   - Line: \( 2x - y = 5 \) or equivalently, \( y = 2x - 5 \).

2. **Locus of Points Equidistant**:
   - We need to find the locus of points \( P(x, y) \) that are equidistant from the point \( A(5, -3) \) and the line \( 2x - y = 5 \). The distance from a point \( P(x, y) \) to the point \( A(5, -3) \) is given by:
     \[
     d_1 = \sqrt{(x - 5)^2 + (y + 3)^2}
     \]
   - The distance from the point \( P(x, y) \) to the line \( 2x - y = 5 \) is:
     \[
     d_2 = \frac{|2x - y - 5|}{\sqrt{2^2 + (-1)^2}} = \frac{|2x - y - 5|}{\sqrt{5}}
     \]

3. **Equating the distances**:
   The condition for equidistance is \( d_1 = d_2 \). Therefore:
   \[
   \sqrt{(x - 5)^2 + (y + 3)^2} = \frac{|2x - y - 5|}{\sqrt{5}}
   \]
   Squaring both sides, we get:
   \[
   (x - 5)^2 + (y + 3)^2 = \frac{(2x - y - 5)^2}{5}
   \]

4. **Simplifying the equation**:
   Expand both sides:
   - Left side:
     \[
     (x - 5)^2 + (y + 3)^2 = (x^2 - 10x + 25) + (y^2 + 6y + 9) = x^2 + y^2 - 10x + 6y + 34
     \]
   - Right side:
     \[
     \frac{(2x - y - 5)^2}{5} = \frac{(4x^2 - 4xy - 20x + y^2 + 10y + 25)}{5}
     \]
     Simplifying further:
     \[
     = \frac{4x^2 - 4xy - 20x + y^2 + 10y + 25}{5}
     = \frac{4x^2}{5} - \frac{4xy}{5} - \frac{20x}{5} + \frac{y^2}{5} + \frac{10y}{5} + \frac{25}{5}
     = \frac{4x^2}{5} - \frac{4xy}{5} - 4x + \frac{y^2}{5} + 2y + 5
     \]

5. **Final Equation**:
   Multiply the entire equation by 5 to eliminate the fraction:
   \[
   5(x^2 + y^2 - 10x + 6y + 34) = 4x^2 - 4xy - 20x + y^2 + 10y + 25
   \]
   Expand and collect like terms to find the equation of the locus.

This equation will represent the locus of points that are equidistant from the point and the line.

Would you like further steps on simplifying this or have any questions?

### Related Questions:
1. How is the distance from a point to a line derived?
2. What geometric shape is described by points equidistant from a point and a line?
3. How does the general equation of a parabola relate to this problem?
4. Can this locus be classified under conic sections? If yes, which one?
5. What role does symmetry play in determining the shape of the locus?

### Tip:
In problems involving distances from a point to a line, equating distances often leads to quadratic expressions that can result in conic sections, such as parabolas.

Determine the locus of all points on the plane which are equidistant from the point (5, -3) and the line 2x - y = 5.

Learn how to determine the locus of points equidistant from the point (5, -3) and the line 2x - y = 5. This problem involves key concepts of coordinate geometry, including the distance formula and locus theorem. Suitable for high school students in Grades 10-12.

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