The incentre of the triangle formed by the line `3x + 4y-12 = 0` with the coordinate axis is

Options

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

Answer

To find the incenter of the triangle formed by the line \(3x + 4y - 12 = 0\) with the coordinate axes, we can follow these steps:

### Step 1: Find the intercepts of the line with the axes
The line can be rewritten as:
\[ 3x + 4y = 12 \]

**Hint:** To find the x-intercept, set \(y = 0\) and solve for \(x\). To find the y-intercept, set \(x = 0\) and solve for \(y\).

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

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

### Step 2: Identify the vertices of the triangle
The triangle formed by the line and the coordinate axes has vertices at:
- \( O(0, 0) \) (origin)
- \( A(4, 0) \) (x-intercept)
- \( B(0, 3) \) (y-intercept)

### Step 3: Calculate the lengths of the sides of the triangle
Let:
- \( OA = 4 \) (length along the x-axis)
- \( OB = 3 \) (length along the y-axis)
- \( AB \) can be calculated using the distance formula.

**Hint:** Use the Pythagorean theorem to find the length of the side \( AB \).

3. **Finding the length of \( AB \):**
\[ AB = \sqrt{(4 - 0)^2 + (0 - 3)^2} = \sqrt{4^2 + (-3)^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \]

### Step 4: Use the incenter formula
The incenter \( I \) of triangle \( OAB \) can be found 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:
- \( a = OB = 3 \)
- \( b = OA = 4 \)
- \( c = AB = 5 \)
- \( (x_1, y_1) = (0, 0) \)
- \( (x_2, y_2) = (4, 0) \)
- \( (x_3, y_3) = (0, 3) \)

**Hint:** Substitute the values of \( a, b, c \) and the coordinates of the vertices into the formulas for \( I_x \) and \( I_y \).

4. **Calculating \( I_x \):**
\[ I_x = \frac{3 \cdot 0 + 4 \cdot 4 + 5 \cdot 0}{3 + 4 + 5} = \frac{0 + 16 + 0}{12} = \frac{16}{12} = \frac{4}{3} \]

5. **Calculating \( I_y \):**
\[ I_y = \frac{3 \cdot 0 + 4 \cdot 0 + 5 \cdot 3}{3 + 4 + 5} = \frac{0 + 0 + 15}{12} = \frac{15}{12} = \frac{5}{4} \]

### Step 5: Conclusion
Thus, the incenter of the triangle formed by the line \(3x + 4y - 12 = 0\) with the coordinate axes is:
\[ I\left(\frac{4}{3}, \frac{5}{4}\right) \]