Let the coordinates of P be (x, y) and of Q be `(alpha, beta)` where `alpha` is the geometric and `beta` is the arithmetic mean of the coordinates of P. If the mid point of PQ is (42, 31) the coordinates of P are

Options

A. (61, 21)
B. (49, 25)
C. (31, 31)
D. none of these

Answer

To find the coordinates of point P, we will follow these steps:

### Step 1: Define the coordinates of points P and Q
Let the coordinates of point P be \( P(x, y) \) and the coordinates of point Q be \( Q(\alpha, \beta) \), where:
- \( \alpha \) is the geometric mean of the coordinates of P.
- \( \beta \) is the arithmetic mean of the coordinates of P.

### Step 2: Express the means
The geometric mean \( \alpha \) and the arithmetic mean \( \beta \) can be expressed as:
- \( \alpha = \sqrt{xy} \)
- \( \beta = \frac{x + y}{2} \)

### Step 3: Use the midpoint formula
The midpoint of line segment PQ is given by:
\[ \left( \frac{x + \alpha}{2}, \frac{y + \beta}{2} \right) = (42, 31) \]
From this, we can set up the following equations:
1. \( \frac{x + \alpha}{2} = 42 \)
2. \( \frac{y + \beta}{2} = 31 \)

### Step 4: Solve for \( x + \alpha \) and \( y + \beta \)
From the equations above, we can multiply both sides by 2:
1. \( x + \alpha = 84 \)
2. \( y + \beta = 62 \)

### Step 5: Substitute \( \alpha \) and \( \beta \)
Now we substitute \( \alpha \) and \( \beta \) into these equations:
1. \( x + \sqrt{xy} = 84 \) (Substituting \( \alpha \))
2. \( y + \frac{x + y}{2} = 62 \) (Substituting \( \beta \))

### Step 6: Simplify the second equation
From the second equation:
\[ y + \frac{x + y}{2} = 62 \]
Multiply through by 2:
\[ 2y + x + y = 124 \]
This simplifies to:
\[ x + 3y = 124 \tag{1} \]

### Step 7: Solve for \( \sqrt{xy} \)
From the first equation:
\[ \sqrt{xy} = 84 - x \tag{2} \]
Now square both sides:
\[ xy = (84 - x)^2 \]
Expanding the right side:
\[ xy = 7056 - 168x + x^2 \tag{3} \]

### Step 8: Substitute \( y \) from equation (1) into equation (3)
From equation (1), we can express \( y \) in terms of \( x \):
\[ y = \frac{124 - x}{3} \]
Substituting this into equation (3):
\[ x \left( \frac{124 - x}{3} \right) = 7056 - 168x + x^2 \]
Multiplying through by 3 to eliminate the fraction:
\[ x(124 - x) = 21168 - 504x + 3x^2 \]
This simplifies to:
\[ 124x - x^2 = 21168 - 504x + 3x^2 \]
Rearranging gives:
\[ 0 = 4x^2 - 628x + 21168 \]

### Step 9: Solve the quadratic equation
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
- Here, \( a = 4, b = -628, c = 21168 \).
- Calculate \( b^2 - 4ac \):
\[ (-628)^2 - 4 \cdot 4 \cdot 21168 = 394384 - 338688 = 55796 \]
Now calculate \( x \):
\[ x = \frac{628 \pm \sqrt{55796}}{8} \]
Calculating \( \sqrt{55796} \approx 236.8 \):
\[ x = \frac{628 \pm 236.8}{8} \]
This gives two possible values for \( x \):
1. \( x_1 = \frac{864.8}{8} = 108.1 \) (approximately)
2. \( x_2 = \frac{391.2}{8} = 48.9 \) (approximately)

### Step 10: Find corresponding \( y \) values
Using \( x + 3y = 124 \):
1. For \( x_1 \approx 108.1 \):
\[ 3y = 124 - 108.1 \Rightarrow y \approx \frac{15.9}{3} \approx 5.3 \]
2. For \( x_2 \approx 48.9 \):
\[ 3y = 124 - 48.9 \Rightarrow y \approx \frac{75.1}{3} \approx 25.0 \]

### Final Coordinates of P
Thus, the coordinates of point P can be approximately:
1. \( (108.1, 5.3) \)
2. \( (48.9, 25.0) \)

### Conclusion
The coordinates of point P are approximately \( (49, 25) \) when rounded.