If the parabola `y=(a-b)x^2+(b-c)x+(c-a)` touches x- axis then the line `ax+by+c=0` (a) always passes through a fixed point (b) represents the family of parallel lines (c) is always perpendicular to x-axis (d) always has negative slope

Options

A. always passes through a fixed point
B. represents the family of parallel lines
C. is always perpendicular to x-axis
D. always has negative slope

Explanation

To solve the problem, we need to analyze the given parabola and the line equation.

### Step-by-Step Solution:

1. **Identify the Parabola Equation**:
The equation of the parabola is given as:
\[ y = (a-b)x^2 + (b-c)x + (c-a) \]

2. **Condition for Tangency with the x-axis**:
For the parabola to touch the x-axis, the quadratic equation must have exactly one solution. This occurs when the discriminant is zero. The general form of a quadratic equation is \(Ax^2 + Bx + C = 0\), where:
- \(A = a - b\)
- \(B = b - c\)
- \(C = c - a\)

The discriminant (\(D\)) is given by:
\[ D = B^2 - 4AC \]
Setting the discriminant to zero for tangency:
\[ (b-c)^2 - 4(a-b)(c-a) = 0 \]

3. **Expand and Simplify the Discriminant**:
Expanding the equation:
\[ (b-c)^2 - 4[(a-b)(c-a)] = 0 \]
Expanding each term:
\[ (b^2 - 2bc + c^2) - 4(ac - ab - bc + ba) = 0 \]
Rearranging gives:
\[ b^2 + c^2 - 2bc - 4ac + 4ab + 4bc = 0 \]
Simplifying further, we can combine like terms.

4. **Rearranging the Equation**:
Rearranging the terms leads to:
\[ 2a^2 - (b+c)^2 + 4bc = 0 \]
This can be factored or rearranged to find relationships between \(a\), \(b\), and \(c\).

5. **Finding the Fixed Point**:
From the manipulation, we find that:
\[ 2a - b - c = 0 \implies b + c = 2a \]
Now, substituting \(b\) and \(c\) into the line equation \(ax + by + c = 0\):
\[ ax + (2a - c)y + c = 0 \]
This can be solved for specific values of \(x\) and \(y\).

6. **Conclusion**:
After solving, we find that the line \(ax + by + c = 0\) always passes through the point \((2, -1)\). Therefore, the correct answer is:
\[ \text{(a) always passes through a fixed point} \]