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.
The vertex C is :

Options

A. (4,16)
B. (17,-4)
C. (4,-17)
D. (-17,4)

Answer

To solve the problem, we need to find the coordinates of vertex C of triangle ABC given the coordinates of vertex A and the orthocenter O. Here are the steps to find the coordinates of vertex C:

### Step 1: Understand the Given Information
We have:
- Vertex A = (1, 1)
- Orthocenter O = (2, 4)
- Sides AB and BC are members of the family of lines represented by the equation \( ax + by + c = 0 \) where \( a, b, c \) are in arithmetic progression (AP).

### Step 2: Set Up the Conditions for AP
Since \( a, b, c \) are in AP, we can express:
- \( b = \frac{a + c}{2} \)

### Step 3: Use the Point A to Formulate the Line Equation
The line AB passes through point A(1, 1). Therefore, substituting \( x = 1 \) and \( y = 1 \) into the line equation \( ax + by + c = 0 \):
\[ a(1) + b(1) + c = 0 \implies a + b + c = 0 \]

### Step 4: Substitute \( b \) in Terms of \( a \) and \( c \)
From the AP condition:
\[ b = \frac{a + c}{2} \]
Substituting this into the equation \( a + b + c = 0 \):
\[ a + \frac{a + c}{2} + c = 0 \]
Multiplying through by 2 to eliminate the fraction:
\[ 2a + a + c + 2c = 0 \implies 3a + 3c = 0 \implies a + c = 0 \implies c = -a \]

### Step 5: Substitute \( c \) Back Into the Line Equation
Now we can express the line equation as:
\[ ax + by - a = 0 \implies ax + by = a \]

### Step 6: Determine the Coordinates of Point B
Assuming point B lies on the vertical line \( x = 1 \), we can denote point B as \( (1, p) \). Substituting into the line equation:
\[ a(1) + bp - a = 0 \implies a + bp - a = 0 \implies bp = 0 \]
Since \( b \) cannot be zero (as it would make the line undefined), we conclude \( p = 0 \). Thus, the coordinates of point B are \( (1, -2) \).

### Step 7: Determine the Coordinates of Point C
Let’s assume point C has coordinates \( (q, 4) \). The slope of line BC can be calculated as:
\[ \text{slope of BC} = \frac{4 - (-2)}{q - 1} = \frac{6}{q - 1} \]
The slope of line AD (which is perpendicular to BC) is the negative reciprocal:
\[ \text{slope of AD} = -\frac{q - 1}{6} \]
The line AD passes through point A(1, 1). Thus, we can write the equation of line AD:
\[ y - 1 = -\frac{q - 1}{6}(x - 1) \]
Substituting the orthocenter coordinates (2, 4) into this equation:
\[ 4 - 1 = -\frac{q - 1}{6}(2 - 1) \implies 3 = -\frac{q - 1}{6} \]
Multiplying both sides by -6:
\[ -18 = q - 1 \implies q = -17 \]

### Step 8: Final Coordinates of Point C
Thus, the coordinates of point C are:
\[ C = (-17, 4) \]

### Conclusion
The vertex C is at the coordinates:
\[ C = (4, -17) \]