Let the lines `(y-2)=m_(1)(x-5) and (y+4)=m_(2)(x-3)` intersect at right angles at P (where `m_(1)` and `m_(2)` are parameters). If locus of P is `x^(2)+y^(2)+gx+fy+7=0`, then `((g)/(2))^(2)+((f)/(2))^(2)-7` is equal to

Options

A. 1
B. 2
C. 8
D. 10

Answer

To solve the problem, we need to follow these steps:

### Step 1: Write the equations of the lines
The equations of the lines are given in point-slope form:
1. \( y - 2 = m_1(x - 5) \)
2. \( y + 4 = m_2(x - 3) \)

### Step 2: Find the slopes
From the equations, we can express the slopes:
- For the first line:
\[ m_1 = \frac{y - 2}{x - 5} \]
- For the second line:
\[ m_2 = \frac{y + 4}{x - 3} \]

### Step 3: Use the condition for perpendicular lines
Since the lines intersect at right angles, we have:
\[ m_1 \cdot m_2 = -1 \]
Substituting the expressions for \(m_1\) and \(m_2\):
\[ \frac{y - 2}{x - 5} \cdot \frac{y + 4}{x - 3} = -1 \]

### Step 4: Cross-multiply and simplify
Cross-multiplying gives:
\[ (y - 2)(y + 4) = - (x - 5)(x - 3) \]
Expanding both sides:
\[ y^2 + 4y - 2y - 8 = - (x^2 - 8x + 15) \]
This simplifies to:
\[ y^2 + 2y - 8 = -x^2 + 8x - 15 \]
Rearranging gives:
\[ x^2 + y^2 - 8x + 2y + 7 = 0 \]

### Step 5: Identify coefficients
The equation of the locus of point P is given as:
\[ x^2 + y^2 + gx + fy + 7 = 0 \]
From our derived equation, we can identify:
- \(g = -8\)
- \(f = 2\)

### Step 6: Calculate the required expression
We need to find:
\[ \left(\frac{g}{2}\right)^2 + \left(\frac{f}{2}\right)^2 - 7 \]
Substituting the values of \(g\) and \(f\):
\[ \left(\frac{-8}{2}\right)^2 + \left(\frac{2}{2}\right)^2 - 7 \]
Calculating each term:
\[ (-4)^2 + (1)^2 - 7 = 16 + 1 - 7 = 10 \]

### Final Answer
Thus, the value of \(\left(\frac{g}{2}\right)^2 + \left(\frac{f}{2}\right)^2 - 7\) is \(10\).