Consider the triangle ABC having vertex A(1,1) and its orthocentre is (2,4). Also side AB & BC are members of the family of line, `ax + by + c = 0` where a,b,c are in A.P.
`DeltaABC` is a:

Options

A. obtuse angled triangle
B. Right angled triangle
C. Acute angled triangle
D. Equilateral triangle

Answer

To solve the problem, we need to determine the nature of triangle ABC given the coordinates of its vertex A(1,1) and its orthocenter (2,4). We also know that the sides AB and BC belong to a family of lines represented by the equation \( ax + by + c = 0 \), where \( a, b, c \) are in arithmetic progression (AP).

### Step-by-Step Solution:

1. **Understanding the Orthocenter**:
The orthocenter of a triangle is the point where the three altitudes intersect. In this case, the orthocenter is given as \( O(2,4) \).

2. **Setting Up the Lines**:
Since sides AB and BC belong to the family of lines \( ax + by + c = 0 \) where \( a, b, c \) are in AP, we can express this condition mathematically. If \( a, b, c \) are in AP, we have:
\[ 2b = a + c \]

3. **Finding the Equation of Line AB**:
The line AB passes through point A(1,1). Therefore, substituting \( x = 1 \) and \( y = 1 \) into the line equation gives:
\[ a(1) + b(1) + c = 0 \implies a + b + c = 0 \]

4. **Substituting for c**:
From the AP condition \( 2b = a + c \), we can express \( c \) in terms of \( a \) and \( b \):
\[ c = 2b - a \]
Substituting this into \( a + b + c = 0 \):
\[ a + b + (2b - a) = 0 \implies 3b = 0 \implies b = 0 \]

5. **Finding a and c**:
Since \( b = 0 \), we substitute back to find \( c \):
\[ c = 2(0) - a = -a \]
Thus, the equation of line AB simplifies to:
\[ ax + 0y - a = 0 \implies ax - a = 0 \implies x = 1 \]
Therefore, line AB is vertical at \( x = 1 \).

6. **Finding the Coordinates of Point B**:
Since B lies on line AB, we can denote its coordinates as \( B(1, P) \).

7. **Finding the Equation of Line BC**:
The line BC also belongs to the same family, so we can express it similarly. Let the coordinates of C be \( (Q, 4) \) since the orthocenter's y-coordinate is 4.

8. **Finding the Slope of Line BC**:
The slope of line BC can be expressed as:
\[ \text{slope of BC} = \frac{4 - P}{Q - 1} \]

9. **Finding the Slope of Line AD**:
Since AD is perpendicular to BC, we can use the property of slopes:
\[ \text{slope of AD} = -\frac{Q - 1}{4 - P} \]

10. **Finding the Equation of Line AD**:
The line AD passes through A(1,1) and has the slope we just calculated. Therefore, the equation of line AD is:
\[ y - 1 = -\frac{Q - 1}{4 - P}(x - 1) \]

11. **Substituting the Orthocenter**:
Since the orthocenter (2,4) lies on line AD, we substitute \( (x,y) = (2,4) \) into the equation of line AD to find a relationship between P and Q.

12. **Determining the Nature of the Triangle**:
After solving the equations, we find that the orthocenter lies outside the triangle formed by points A, B, and C. This indicates that triangle ABC is an obtuse triangle.

### Conclusion:
The triangle ABC is an obtuse triangle.