In the plane a line $l$ and a point $A$ outside it are given. Find the locus of the incenters of acute-angled triangles having a vertex $A$ and an opposite side lying on $l$.
1. **Define the problem setup:**
   - Let \( l \) be a line in the plane.
   - Let \( A \) be a point outside \( l \).
   - We need to find the locus of the incenters of acute-angled triangles having a vertex at \( A \) and an opposite side lying on \( l \).

2. **Initial observations:**
   - Let \( D \) be the foot of the perpendicular from \( A \) to \( l \).
   - For \( \triangle ABC \) to be acute-angled, \( B \) and \( C \) must lie on different sides of \( D \) on \( l \).

3. **Special cases and symmetry:**
   - Assume \( B \) lies on the right and \( C \) on the left of \( D \).
   - When \( B \) coincides with \( D \) and \( C \) varies from \( D \) to infinity, the incenters lie on a line segment \( DE \), where \( E \) is a point such that \( \angle AED = 90^\circ \) and \( AE = ED \).
   - The locus is symmetrical about \( AD \), hence we get segment \( DF \), such that \( AEDF \) forms a square.

4. **Coordinate system setup:**
   - Place \( D \) at the origin \((0,0)\) and \( l \) as the \( x \)-axis.
   - Let \( A = (0, a) \).

5. **Incenter calculations:**
   - Let \( I = (x, y) \) be the incenter of \( \triangle ABC \).
   - When \( \angle BAC = 90^\circ \), the distance \( AI = r\sqrt{2} \), where \( r \) is the inradius of \( \triangle ABC \).
   - Since \( r = y \) (the distance from \( I \) to \( BC \)), we have \( AI = y\sqrt{2} \).

6. **Distance formula application:**
   - Using the distance formula, \( (y - a)^2 + x^2 = 2y^2 \).
   - Simplifying, we get \( x^2 - y^2 - 2ya + a^2 = 0 \).
   - Rearranging, \( (y + a)^2 - x^2 = 2a^2 \).

7. **Equation of the hyperbola:**
   - The equation \( \frac{(y + a)^2}{(a\sqrt{2})^2} + \frac{x^2}{(a\sqrt{2})^2} = 1 \) represents a hyperbola.
   - This hyperbola passes through points \( E \) and \( F \).

8. **Region enclosed by the hyperbola:**
   - Let \( \mathcal{H} \) be the region enclosed by the hyperbola and the segments \( DE \) and \( DF \).

9. **Proof of locus:**
   - Fix \( B \) and vary \( C \) from \( D \) to the extent where \( \angle BAC = 90^\circ \).
   - When \( \angle BAC = 90^\circ \), \( I \) lies on the hyperbola.
   - When \( C = D \), \( I \) lies on the segment \( DF \).
   - As \( C \) varies uniformly on \( l \), \( I \) also varies uniformly on a straight line within \( \mathcal{H} \).

10. **Every point in \( \mathcal{H} \) as an incenter:**
    - Let \( A' \) be the foot of the perpendicular from \( I \) to \( l \).
    - Since \( I \in \mathcal{H} \), \( AI \geq IA' \sqrt{2} \).
    - Drawing a circle with radius \( IA' \) centered at \( I \) and drawing tangents from \( A \) to this circle intersecting \( l \) gives points \( B \) and \( C \).
    - Since \( \angle A \leq 90^\circ \), \( B \) and \( C \) lie on different sides of \( D \) on \( l \), ensuring \( \angle B \) and \( \angle C \) are non-obtuse.

11. **Conclusion:**
    - The required locus is the interior of \( \mathcal{H} \), excluding the boundary points where one angle is a right angle.

The final answer is the interior of the region \( \boxed{ \mathcal{H} } \) enclosed by the hyperbola and the segments \( DE \) and \( DF \).