The incentre of the triangle whose vertices are (-36, 7), (20, 7) and (0, -8) is

Options

A. (0, -1)
B. (-1, 0)
C. (1, 1)
D. `((1)/(2), 1)`

Answer

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

### Step 1: Identify the vertices of the triangle
The vertices of the triangle are:
- A = (-36, 7)
- B = (20, 7)
- C = (0, -8)

### Step 2: Calculate the lengths of the sides of the triangle
We will use the distance formula to find the lengths of the sides opposite to each vertex.

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

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

### Step 3: Use the incenter formula
The incenter \(I\) of a triangle with vertices \((x_1, y_1)\), \((x_2, y_2)\), and \((x_3, y_3)\) is given by:
\[ I_x = \frac{a x_1 + b x_2 + c x_3}{a + b + c} \]
\[ I_y = \frac{a y_1 + b y_2 + c y_3}{a + b + c} \]

Substituting the values:
- \(x_1 = -36\), \(y_1 = 7\)
- \(x_2 = 20\), \(y_2 = 7\)
- \(x_3 = 0\), \(y_3 = -8\)
- \(a = 25\), \(b = 39\), \(c = 56\)

### Step 4: Calculate \(I_x\)
\[ I_x = \frac{25 \cdot (-36) + 39 \cdot 20 + 56 \cdot 0}{25 + 39 + 56} \]
\[ = \frac{-900 + 780 + 0}{120} = \frac{-120}{120} = -1 \]

### Step 5: 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 6: Conclusion
Thus, the incenter of the triangle is:
\[ I = (-1, 0) \]