If P(3,7) is a point on the line joining A(1,1) and B(6,16), then the harmonic conjugate Q of point P has the coordinates

Options

A. (9,29)
B. (-9,29)
C. (9,-29)
D. (-9,-29)

Answer

To find the harmonic conjugate \( Q \) of the point \( P(3, 7) \) which lies on the line segment joining points \( A(1, 1) \) and \( B(6, 16) \), we can follow these steps:

### Step 1: Identify the coordinates of points A and B
We have:
- \( A(1, 1) \)
- \( B(6, 16) \)

### Step 2: Use the section formula to find the ratio in which point P divides AB
The coordinates of point \( P \) are given as \( (3, 7) \). According to the section formula, if a point \( P \) divides the line segment \( AB \) in the ratio \( \lambda : 1 \), then the coordinates of \( P \) can be expressed as:

\[ P\left( \frac{\lambda x_2 + x_1}{\lambda + 1}, \frac{\lambda y_2 + y_1}{\lambda + 1} \right) \]

Substituting the coordinates of \( A \) and \( B \):
- \( x_1 = 1, y_1 = 1 \)
- \( x_2 = 6, y_2 = 16 \)

So we have:
\[ P\left( \frac{\lambda \cdot 6 + 1}{\lambda + 1}, \frac{\lambda \cdot 16 + 1}{\lambda + 1} \right) = (3, 7) \]

### Step 3: Set up equations for x and y coordinates
From the x-coordinate:
\[ \frac{6\lambda + 1}{\lambda + 1} = 3 \]

From the y-coordinate:
\[ \frac{16\lambda + 1}{\lambda + 1} = 7 \]

### Step 4: Solve the equations
**For the x-coordinate:**
\[ 6\lambda + 1 = 3(\lambda + 1) \]
\[ 6\lambda + 1 = 3\lambda + 3 \]
\[ 6\lambda - 3\lambda = 3 - 1 \]
\[ 3\lambda = 2 \implies \lambda = \frac{2}{3} \]

**For the y-coordinate:**
\[ 16\lambda + 1 = 7(\lambda + 1) \]
\[ 16\lambda + 1 = 7\lambda + 7 \]
\[ 16\lambda - 7\lambda = 7 - 1 \]
\[ 9\lambda = 6 \implies \lambda = \frac{2}{3} \]

Both equations confirm that \( \lambda = \frac{2}{3} \).

### Step 5: Determine the ratio for the harmonic conjugate Q
Since \( P \) divides \( AB \) in the ratio \( 2:3 \) internally, the harmonic conjugate \( Q \) will divide the same line segment in the ratio \( -2:3 \) (internally) or \( 2:3 \) (externally).

### Step 6: Use the section formula to find the coordinates of Q
Using the section formula again for point \( Q \):
\[ Q\left( \frac{-2 \cdot 6 + 3 \cdot 1}{-2 + 3}, \frac{-2 \cdot 16 + 3 \cdot 1}{-2 + 3} \right) \]

Calculating the x-coordinate:
\[ Q_x = \frac{-12 + 3}{1} = -9 \]

Calculating the y-coordinate:
\[ Q_y = \frac{-32 + 3}{1} = -29 \]

Thus, the coordinates of point \( Q \) are \( (-9, -29) \).

### Final Answer
The coordinates of the harmonic conjugate \( Q \) are \( (-9, -29) \).