The in-radius of the triangle formed by the axes and the line `4x + 3y – 12 = 0` is :

Options

A. r = 2
B. `r = (1)/(2)`
C. r = 1
D. `r = (1)/(4)`

Answer

To find the in-radius of the triangle formed by the axes and the line \(4x + 3y - 12 = 0\), we will follow these steps:

### Step 1: Find the intercepts of the line
We start with the equation of the line:
\[ 4x + 3y - 12 = 0 \]
To find the x-intercept, we set \(y = 0\):
\[ 4x - 12 = 0 \implies 4x = 12 \implies x = 3 \]
So, the x-intercept is \((3, 0)\).

Next, to find the y-intercept, we set \(x = 0\):
\[ 3y - 12 = 0 \implies 3y = 12 \implies y = 4 \]
So, the y-intercept is \((0, 4)\).

### Step 2: Identify the vertices of the triangle
The triangle is formed by the x-axis, y-axis, and the line. The vertices of the triangle are:
- \(A(0, 0)\) (the origin)
- \(B(3, 0)\) (x-intercept)
- \(C(0, 4)\) (y-intercept)

### Step 3: Calculate the area of the triangle
The area (\(\Delta\)) of the triangle can be calculated using the formula:
\[ \Delta = \frac{1}{2} \times \text{base} \times \text{height} \]
Here, the base is the x-intercept (3 units) and the height is the y-intercept (4 units):
\[ \Delta = \frac{1}{2} \times 3 \times 4 = \frac{12}{2} = 6 \text{ square units} \]

### Step 4: Calculate the lengths of the sides of the triangle
The lengths of the sides of the triangle are:
- \(AB = 3\) (base)
- \(AC = 4\) (height)
- \(BC = \sqrt{(3-0)^2 + (0-4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5\)

### Step 5: Calculate the semi-perimeter of the triangle
The semi-perimeter (\(S\)) is given by:
\[ S = \frac{AB + AC + BC}{2} = \frac{3 + 4 + 5}{2} = \frac{12}{2} = 6 \text{ units} \]

### Step 6: Calculate the in-radius of the triangle
The in-radius (\(R\)) is given by the formula:
\[ R = \frac{\Delta}{S} \]
Substituting the values we found:
\[ R = \frac{6}{6} = 1 \text{ unit} \]

### Final Answer
The in-radius of the triangle formed by the axes and the line \(4x + 3y - 12 = 0\) is \(1\) unit.

---