Find the harmonic conjugates of the point R(5, 1) with respect to the points P(2, 10) and Q(6, -2)

Answer

To find the harmonic conjugates of the point R(5, 1) with respect to the points P(2, 10) and Q(6, -2), we will follow these steps:

### Step 1: Understand the Concept of Harmonic Conjugates
Harmonic conjugates are points that divide a segment in a specific ratio. In this case, we are looking for the point R' that is harmonic conjugate to R with respect to points P and Q.

### Step 2: Use the Section Formula
The section formula states that if a point R divides the line segment joining points P(x1, y1) and Q(x2, y2) in the ratio m:n, then the coordinates of R can be given as:
\[ R\left(\frac{mx_2 + nx_1}{m+n}, \frac{my_2 + ny_1}{m+n}\right) \]
Here, we will let R divide PQ in the ratio \( \lambda : 1 \).

### Step 3: Set Up the Equations
Given:
- P(2, 10)
- Q(6, -2)
- R(5, 1)

Using the section formula, we can express the coordinates of R as:
\[ x_R = \frac{6\lambda + 2}{\lambda + 1} \quad \text{and} \quad y_R = \frac{-2\lambda + 10}{\lambda + 1} \]

### Step 4: Set Up the Equations for x and y
Now we set up the equations based on the coordinates of R:
1. For x-coordinate:
\[ \frac{6\lambda + 2}{\lambda + 1} = 5 \]
2. For y-coordinate:
\[ \frac{-2\lambda + 10}{\lambda + 1} = 1 \]

### Step 5: Solve for λ
**For the x-coordinate:**
\[ 6\lambda + 2 = 5(\lambda + 1) \]
\[ 6\lambda + 2 = 5\lambda + 5 \]
\[ 6\lambda - 5\lambda = 5 - 2 \]
\[ \lambda = 3 \]

**For the y-coordinate:**
\[ -2\lambda + 10 = 1(\lambda + 1) \]
\[ -2\lambda + 10 = \lambda + 1 \]
\[ -2\lambda - \lambda = 1 - 10 \]
\[ -3\lambda = -9 \]
\[ \lambda = 3 \]

### Step 6: Find the Harmonic Conjugate
The harmonic conjugate of R can be found using the negative of the ratio. Thus, the ratio becomes \(-\lambda : 1\) or \(-3 : 1\).

### Step 7: Use the Section Formula Again
Now we can find the coordinates of the harmonic conjugate R' using the section formula:
\[ R'\left(\frac{6(-3) + 2}{-3 + 1}, \frac{-2(-3) + 10}{-3 + 1}\right) \]
Calculating the x-coordinate:
\[ x_{R'} = \frac{-18 + 2}{-2} = \frac{-16}{-2} = 8 \]
Calculating the y-coordinate:
\[ y_{R'} = \frac{6 + 10}{-2} = \frac{16}{-2} = -8 \]

### Final Answer
Thus, the harmonic conjugate R' is:
\[ R'(8, -8) \]