Find the coordinates of the in-centre of the triangle with vertices `A(-1,12), B(-1,0)` and C (4,0).

Answer

To find the coordinates of the in-center of the triangle with vertices A(-1, 12), B(-1, 0), and C(4, 0), we can follow these steps:

### Step 1: Identify the vertices of the triangle
The vertices of the triangle are:
- A(x1, y1) = A(-1, 12)
- B(x2, y2) = B(-1, 0)
- C(x3, y3) = C(4, 0)

### Step 2: Calculate the lengths of the sides of the triangle
To find the in-center, we first need to calculate the lengths of the sides opposite to each vertex.

1. **Length of side BC (a)**:
\[ a = \sqrt{(x2 - x3)^2 + (y2 - y3)^2} = \sqrt{((-1) - 4)^2 + (0 - 0)^2} = \sqrt{(-5)^2} = 5 \]

2. **Length of side AC (b)**:
\[ b = \sqrt{(x1 - x3)^2 + (y1 - y3)^2} = \sqrt{((-1) - 4)^2 + (12 - 0)^2} = \sqrt{(-5)^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13 \]

3. **Length of side AB (c)**:
\[ c = \sqrt{(x1 - x2)^2 + (y1 - y2)^2} = \sqrt{((-1) - (-1))^2 + (12 - 0)^2} = \sqrt{0 + 12^2} = 12 \]

### Step 3: Use the formula for the in-center coordinates
The coordinates of the in-center (I) can be calculated using the formula:
\[ I_x = \frac{a \cdot x1 + b \cdot x2 + c \cdot x3}{a + b + c} \]
\[ I_y = \frac{a \cdot y1 + b \cdot y2 + c \cdot y3}{a + b + c} \]

### Step 4: Substitute the values into the formula
1. **Calculate \(I_x\)**:
\[ I_x = \frac{5 \cdot (-1) + 13 \cdot (-1) + 12 \cdot 4}{5 + 13 + 12} \]
\[ = \frac{-5 - 13 + 48}{30} = \frac{30}{30} = 1 \]

2. **Calculate \(I_y\)**:
\[ I_y = \frac{5 \cdot 12 + 13 \cdot 0 + 12 \cdot 0}{5 + 13 + 12} \]
\[ = \frac{60 + 0 + 0}{30} = \frac{60}{30} = 2 \]

### Step 5: Combine the results
Thus, the coordinates of the in-center \(I\) are:
\[ I(1, 2) \]

### Final Answer:
The coordinates of the in-center of the triangle are \( (1, 2) \).

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