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 find the locus of the point P where the two lines intersect at right angles. The lines are given as:

1. \( y - 2 = m_1 (x - 5) \)
2. \( y + 4 = m_2 (x - 3) \)

### Step 1: Express the slopes in terms of \( m_1 \) and \( m_2 \)

From the first line, we can express \( m_1 \) as:
\[ m_1 = \frac{y - 2}{x - 5} \]

From the second line, we can express \( m_2 \) as:
\[ m_2 = \frac{y + 4}{x - 3} \]

### Step 2: Use the condition for perpendicular lines

For the lines to intersect at right angles, the product of their slopes must equal -1:
\[ m_1 \cdot m_2 = -1 \]
Substituting the expressions for \( m_1 \) and \( m_2 \):
\[ \left(\frac{y - 2}{x - 5}\right) \cdot \left(\frac{y + 4}{x - 3}\right) = -1 \]

### Step 3: Cross-multiply to eliminate the fractions

Cross-multiplying gives us:
\[ (y - 2)(y + 4) = - (x - 5)(x - 3) \]

### Step 4: Expand both sides

Expanding the left side:
\[ y^2 + 4y - 2y - 8 = y^2 + 2y - 8 \]
Expanding the right side:
\[ -(x^2 - 8x + 15) = -x^2 + 8x - 15 \]

### Step 5: Set the equation to zero

Now we set the equation to zero:
\[ y^2 + 2y - 8 + x^2 - 8x + 15 = 0 \]
This simplifies to:
\[ x^2 + y^2 - 8x + 2y + 7 = 0 \]

### Step 6: Identify the coefficients \( g \) and \( f \)

From the equation \( x^2 + y^2 + gx + fy + 7 = 0 \), we can identify:
- \( g = -8 \)
- \( f = 2 \)

### Step 7: Calculate the required expression

Now we need to find:
\[ \left(\frac{g}{2}\right)^2 + \left(\frac{f}{2}\right)^2 - 7 \]
Substituting the values of \( g \) and \( f \):
\[ \left(\frac{-8}{2}\right)^2 + \left(\frac{2}{2}\right)^2 - 7 = (-4)^2 + (1)^2 - 7 \]
Calculating this gives:
\[ 16 + 1 - 7 = 10 \]

### Final Answer

Thus, the value is:
\[ \boxed{10} \]