Consider a triangle PQR with coordinates of its vertices as P(-8,5), Q(-15, -19), and R (1, -7). The bisector of the interior angle of P has the equation which can be written in the form ax+2y+c=0.
The radius of the in circle of triangle PQR is
The sum a + c is

Options

A. 129
B. 78
C. 89
D. none of these

Explanation

To solve the problem, we need to find the equation of the angle bisector of angle P in triangle PQR, and then determine the radius of the incircle of triangle PQR. Finally, we will find the sum of coefficients \( a + c \) from the equation of the angle bisector.

### Step 1: Find the lengths of the sides of triangle PQR
We will first calculate the lengths of the sides PQ, QR, and PR using the distance formula.

- **Distance PQ**:
\[ PQ = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} = \sqrt{(-15 - (-8))^2 + (-19 - 5)^2} = \sqrt{(-7)^2 + (-24)^2} = \sqrt{49 + 576} = \sqrt{625} = 25 \]

- **Distance QR**:
\[ QR = \sqrt{(1 - (-15))^2 + (-7 - (-19))^2} = \sqrt{(16)^2 + (12)^2} = \sqrt{256 + 144} = \sqrt{400} = 20 \]

- **Distance PR**:
\[ PR = \sqrt{(1 - (-8))^2 + (-7 - 5)^2} = \sqrt{(9)^2 + (-12)^2} = \sqrt{81 + 144} = \sqrt{225} = 15 \]

### Step 2: Use the angle bisector theorem
The angle bisector of angle P divides the opposite side QR in the ratio of the lengths of the other two sides, PR and PQ.
Let D be the point on QR where the angle bisector meets. According to the angle bisector theorem:
\[ \frac{QD}{DR} = \frac{PR}{PQ} = \frac{15}{25} = \frac{3}{5} \]

Let \( QD = 3x \) and \( DR = 5x \). Then:
\[ QR = QD + DR = 3x + 5x = 8x \]
Since we found \( QR = 20 \):
\[ 8x = 20 \implies x = \frac{20}{8} = 2.5 \]
Thus, \( QD = 3x = 7.5 \) and \( DR = 5x = 12.5 \).

### Step 3: Find the coordinates of point D
Using the section formula, the coordinates of point D dividing QR in the ratio 3:5:
\[ D = \left( \frac{m x_2 + n x_1}{m+n}, \frac{m y_2 + n y_1}{m+n} \right) \]
where \( m = 3 \), \( n = 5 \), \( Q(-15, -19) \), and \( R(1, -7) \):
\[ D_x = \frac{3(1) + 5(-15)}{3 + 5} = \frac{3 - 75}{8} = \frac{-72}{8} = -9 \]
\[ D_y = \frac{3(-7) + 5(-19)}{3 + 5} = \frac{-21 - 95}{8} = \frac{-116}{8} = -14.5 \]
Thus, the coordinates of D are \( D(-9, -14.5) \).

### Step 4: Find the equation of the angle bisector
The slope of line segment PD:
\[ \text{slope of PD} = \frac{-14.5 - 5}{-9 - (-8)} = \frac{-19.5}{-1} = 19.5 \]
The equation of line PD in point-slope form:
\[ y - 5 = 19.5(x + 8) \]
Rearranging gives:
\[ y - 5 = 19.5x + 156 \implies 19.5x - y + 161 = 0 \]
To convert this into the form \( ax + 2y + c = 0 \), we can multiply through by a suitable factor to make the coefficients integers.

### Step 5: Find the radius of the incircle
The radius \( r \) of the incircle can be calculated using the formula:
\[ r = \frac{A}{s} \]
where \( A \) is the area of the triangle and \( s \) is the semi-perimeter.
- Semi-perimeter \( s = \frac{PQ + QR + PR}{2} = \frac{25 + 20 + 15}{2} = 30 \).
- Area \( A \) can be calculated using the determinant method:
\[ A = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \]
Substituting the coordinates:
\[ A = \frac{1}{2} \left| -8(-19 + 7) + (-15)(-7 - 5) + 1(5 + 19) \right| = \frac{1}{2} \left| -8(-12) + (-15)(-12) + 24 \right| = \frac{1}{2} \left| 96 + 180 + 24 \right| = \frac{1}{2} \cdot 300 = 150 \]
Thus, the radius \( r = \frac{150}{30} = 5 \).

### Step 6: Find the sum \( a + c \)
From the equation of the angle bisector \( 19.5x - y + 161 = 0 \), we can express it in the form \( ax + 2y + c = 0 \) by multiplying by 2:
\[ 39x - 2y + 322 = 0 \]
Here, \( a = 39 \) and \( c = 322 \). Therefore, \( a + c = 39 + 322 = 361 \).

### Final Answer:
The sum \( a + c \) is \( 361 \).