To find the harmonic conjugate of the point \( P(7, 5) \) with respect to the points \( A(4, 2) \) and \( B(9, 7) \), we will follow these steps:
### Step 1: Find the Ratio in which P Divides AB
We will use the section formula to find the ratio in which point \( P \) divides the line segment \( AB \).
The section formula states that if a point \( P(x, y) \) divides the line segment joining points \( A(x_1, y_1) \) and \( B(x_2, y_2) \) in the ratio \( m:n \), then:
\[
x = \frac{mx_2 + nx_1}{m+n}, \quad y = \frac{my_2 + ny_1}{m+n}
\]
Let the ratio in which \( P \) divides \( AB \) be \( k:1 \). Thus, we have:
\[
x = \frac{k \cdot 9 + 4}{k + 1}, \quad y = \frac{k \cdot 7 + 2}{k + 1}
\]
Since \( P \) is given as \( (7, 5) \), we can equate the coordinates:
1. For the x-coordinate:
\[
\frac{9k + 4}{k + 1} = 7
\]
Multiplying both sides by \( k + 1 \):
\[
9k + 4 = 7(k + 1) \implies 9k + 4 = 7k + 7 \implies 2k = 3 \implies k = \frac{3}{2}
\]
### Step 2: Identify the Type of Division
Since \( k = \frac{3}{2} \) is positive, the division of \( AB \) by \( P \) is internal.
### Step 3: Find the Harmonic Conjugate
The harmonic conjugate of point \( P \) will be the point that divides the line segment \( AB \) in the same ratio but externally.
Using the external division formula:
\[
x = \frac{mx_2 - nx_1}{m - n}, \quad y = \frac{my_2 - ny_1}{m - n}
\]
Here, \( m = 3 \) and \( n = 2 \), so:
\[
x = \frac{3 \cdot 9 - 2 \cdot 4}{3 - 2}, \quad y = \frac{3 \cdot 7 - 2 \cdot 2}{3 - 2}
\]
Calculating the x-coordinate:
\[
x = \frac{27 - 8}{1} = 19
\]
Calculating the y-coordinate:
\[
y = \frac{21 - 4}{1} = 17
\]
Thus, the harmonic conjugate of \( P(7, 5) \) with respect to \( A(4, 2) \) and \( B(9, 7) \) is \( (19, 17) \).
### Final Answer
The harmonic conjugate of \( (7, 5) \) with respect to \( (4, 2) \) and \( (9, 7) \) is \( (19, 17) \).
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