The incentre of triangle formed by lines `x = 0, y = 0` and `3x + 4y =12` is

Options

A. (3,1)
B. (1,2)
C. (2,1)
D. (1,1)

Answer

To find the incenter of the triangle formed by the lines \( x = 0 \), \( y = 0 \), and \( 3x + 4y = 12 \), we will follow these steps:

### Step 1: Identify the vertices of the triangle
The lines \( x = 0 \) and \( y = 0 \) represent the y-axis and x-axis, respectively. The line \( 3x + 4y = 12 \) can be rewritten in slope-intercept form to find its intercepts.

1. **Find the x-intercept**:
Set \( y = 0 \):
\[ 3x + 4(0) = 12 \implies 3x = 12 \implies x = 4 \]
So, the x-intercept is \( (4, 0) \).

2. **Find the y-intercept**:
Set \( x = 0 \):
\[ 3(0) + 4y = 12 \implies 4y = 12 \implies y = 3 \]
So, the y-intercept is \( (0, 3) \).

3. **Vertices of the triangle**:
The vertices of the triangle are:
- \( A(0, 0) \) (origin)
- \( B(4, 0) \) (x-intercept)
- \( C(0, 3) \) (y-intercept)

### Step 2: Calculate the lengths of the sides of the triangle
Using the distance formula, we can find the lengths of the sides opposite to each vertex.

1. **Length of side \( a \) (opposite vertex A)**:
\[ a = BC = \sqrt{(4 - 0)^2 + (0 - 3)^2} = \sqrt{4^2 + (-3)^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \]

2. **Length of side \( b \) (opposite vertex B)**:
\[ b = AC = \sqrt{(0 - 0)^2 + (3 - 0)^2} = \sqrt{0 + 3^2} = \sqrt{9} = 3 \]

3. **Length of side \( c \) (opposite vertex C)**:
\[ c = AB = \sqrt{(4 - 0)^2 + (0 - 0)^2} = \sqrt{4^2} = 4 \]

### Step 3: Use the incenter formula
The coordinates of the incenter \( I(x, y) \) can be calculated using the formula:
\[ I_x = \frac{a x_1 + b x_2 + c x_3}{a + b + c} \]
\[ I_y = \frac{a y_1 + b y_2 + c y_3}{a + b + c} \]
Where:
- \( (x_1, y_1) = (0, 0) \)
- \( (x_2, y_2) = (4, 0) \)
- \( (x_3, y_3) = (0, 3) \)

Substituting the values:
- \( a = 5 \), \( b = 3 \), \( c = 4 \)

### Calculate \( I_x \):
\[ I_x = \frac{5(0) + 3(4) + 4(0)}{5 + 3 + 4} = \frac{0 + 12 + 0}{12} = \frac{12}{12} = 1 \]

### Calculate \( I_y \):
\[ I_y = \frac{5(0) + 3(0) + 4(3)}{5 + 3 + 4} = \frac{0 + 0 + 12}{12} = \frac{12}{12} = 1 \]

### Step 4: Conclusion
Thus, the coordinates of the incenter of the triangle are:
\[ \boxed{(1, 1)} \]