Find incentre (I) of triangle whose vertices are A (– 36, 7), B (20, 7), C (0, – 8).

Options

A. (2,0)
B. (5,0)
C. (0,8)
D. (-1,0)

Answer

To find the incenter (I) of the triangle with vertices A (–36, 7), B (20, 7), and C (0, –8), we will follow these steps:

### Step 1: Calculate the lengths of the sides of the triangle
We need to find the lengths of the sides opposite to each vertex. The lengths of the sides can be calculated using the distance formula:

1. **Length of side BC (a)**:
\[ a = \sqrt{(x_B - x_C)^2 + (y_B - y_C)^2} = \sqrt{(20 - 0)^2 + (7 - (-8))^2} = \sqrt{20^2 + 15^2} = \sqrt{400 + 225} = \sqrt{625} = 25 \]

2. **Length of side CA (b)**:
\[ b = \sqrt{(x_C - x_A)^2 + (y_C - y_A)^2} = \sqrt{(0 - (-36))^2 + (-8 - 7)^2} = \sqrt{36^2 + (-15)^2} = \sqrt{1296 + 225} = \sqrt{1521} = 39 \]

3. **Length of side AB (c)**:
\[ c = \sqrt{(x_A - x_B)^2 + (y_A - y_B)^2} = \sqrt{(-36 - 20)^2 + (7 - 7)^2} = \sqrt{(-56)^2 + 0^2} = \sqrt{3136} = 56 \]

### Step 2: Use the incenter formula
The coordinates of the incenter (I) can be found using the formula:
\[ I_x = \frac{a x_A + b x_B + c x_C}{a + b + c} \]
\[ I_y = \frac{a y_A + b y_B + c y_C}{a + b + c} \]

### Step 3: Substitute the values into the formula
1. **Calculate \(I_x\)**:
\[ I_x = \frac{25 \cdot (-36) + 39 \cdot 20 + 56 \cdot 0}{25 + 39 + 56} \]
\[ = \frac{-900 + 780 + 0}{120} = \frac{-120}{120} = -1 \]

2. **Calculate \(I_y\)**:
\[ I_y = \frac{25 \cdot 7 + 39 \cdot 7 + 56 \cdot (-8)}{25 + 39 + 56} \]
\[ = \frac{175 + 273 - 448}{120} = \frac{0}{120} = 0 \]

### Step 4: Conclusion
The coordinates of the incenter (I) of the triangle are:
\[ I(-1, 0) \]