1. **State the problem:** We are given the quadratic form equation $$Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0$$ with $$A=1$$, $$B=-2$$, $$C=1$$, and the discriminant condition $$B^2 - 4AC = 0$$. 2. **Identify the conic type:** Since $$B^2 - 4AC = (-2)^2 - 4(1)(1) = 4 - 4 = 0$$, the conic is a parabola or a degenerate conic (pair of lines). 3. **Rewrite the quadratic form:** The equation can be factored as a perfect square: $$Ax^2 + Bxy + Cy^2 = (x - y)^2$$ 4. **Given the factorization:** $$(x - y)^2 = 2ax + 2ay - a^2$$ which can be rewritten as $$(x - y)^2 = a(2x + 2y - a)$$ 5. **Express the pair of lines:** The equation represents two lines: - Line 1: $$x - y = 0$$ or $$y = x$$ - Line 2: $$2x + 2y - a = 0$$ or $$y = \frac{a}{2} - x$$ 6. **Summary:** The original quadratic equation represents two intersecting lines given by: $$y = x$$ and $$y = \frac{a}{2} - x$$ These lines intersect at the point where both equations hold true. **Final answer:** The solution to the given quadratic equation is the pair of lines: $$\boxed{y = x \quad \text{and} \quad y = \frac{a}{2} - x}$$