Let $ A$ and $ B$ be fixed distinct points on the $ X$ axis, none of which coincides with the origin $ O(0, 0),$ and let $ C$ be a point on the $ Y$ axis of an orthogonal Cartesian coordinate system. Let $ g$ be a line through the origin $ O(0, 0)$ and perpendicular to the line $ AC.$ Find the locus of the point of intersection of the lines $ g$ and $ BC$ if $ C$ varies along the $ Y$ axis. Give an equation and a description of the locus.
1. **Define the points and lines:**
   - Let \( A = (a, 0) \) and \( B = (b, 0) \) be fixed points on the \( X \)-axis.
   - Let \( C = (0, k) \) be a point on the \( Y \)-axis, where \( k \) is a variable.
   - The line \( g \) passes through the origin \( O(0, 0) \) and is perpendicular to the line \( AC \).

2. **Find the slope of line \( AC \):**
   - The slope of \( AC \) is given by:
     \[
     \text{slope of } AC = \frac{k - 0}{0 - a} = -\frac{k}{a}
     \]

3. **Find the slope of line \( g \):**
   - Since \( g \) is perpendicular to \( AC \), the slope of \( g \) is the negative reciprocal of the slope of \( AC \):
     \[
     \text{slope of } g = \frac{a}{k}
     \]
   - Therefore, the equation of line \( g \) is:
     \[
     y = \frac{a}{k} x
     \]

4. **Find the equation of line \( BC \):**
   - The slope of \( BC \) is:
     \[
     \text{slope of } BC = \frac{k - 0}{0 - b} = -\frac{k}{b}
     \]
   - Therefore, the equation of line \( BC \) is:
     \[
     y = -\frac{k}{b} x + k
     \]

5. **Find the intersection of lines \( g \) and \( BC \):**
   - Set the equations of \( g \) and \( BC \) equal to each other:
     \[
     \frac{a}{k} x = -\frac{k}{b} x + k
     \]
   - Multiply through by \( k \) to clear the fraction:
     \[
     a x = -\frac{k^2}{b} x + k^2
     \]
   - Rearrange to solve for \( x \):
     \[
     a x + \frac{k^2}{b} x = k^2
     \]
     \[
     x \left( a + \frac{k^2}{b} \right) = k^2
     \]
     \[
     x = \frac{k^2}{a + \frac{k^2}{b}}
     \]
     \[
     x = \frac{k^2 b}{a b + k^2}
     \]

6. **Find the corresponding \( y \)-coordinate:**
   - Substitute \( x \) back into the equation of line \( g \):
     \[
     y = \frac{a}{k} x = \frac{a}{k} \cdot \frac{k^2 b}{a b + k^2} = \frac{a k b}{a b + k^2}
     \]

7. **Simplify the coordinates:**
   - The coordinates of the intersection point are:
     \[
     \left( \frac{k^2 b}{a b + k^2}, \frac{a k b}{a b + k^2} \right)
     \]

8. **Find the locus as \( k \) varies:**
   - Let \( k \to \infty \):
     \[
     \left( \frac{k^2 b}{a b + k^2}, \frac{a k b}{a b + k^2} \right) \to (b, 0)
     \]
   - Let \( k \to 0 \):
     \[
     \left( \frac{k^2 b}{a b + k^2}, \frac{a k b}{a b + k^2} \right) \to (0, 0)
     \]
   - The locus of the point of intersection is a line segment from \( (0, 0) \) to \( (b, 0) \).

The final answer is \( \boxed{ (0, 0) } \) to \((b, 0)\).