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

Options

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

Answer

To find the incentre 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 vertices of the triangle can be found by determining the points where the lines intersect.

1. **Intersection of \(x = 0\) and \(y = 0\)**:
- This gives us the point \(O(0, 0)\).

2. **Intersection of \(x = 0\) and \(3x + 4y = 12\)**:
- Substitute \(x = 0\) into \(3(0) + 4y = 12\) which simplifies to \(4y = 12\).
- Solving for \(y\) gives \(y = 3\).
- Thus, the point is \(A(0, 3)\).

3. **Intersection of \(y = 0\) and \(3x + 4y = 12\)**:
- Substitute \(y = 0\) into \(3x + 4(0) = 12\) which simplifies to \(3x = 12\).
- Solving for \(x\) gives \(x = 4\).
- Thus, the point is \(B(4, 0)\).

The vertices of the triangle are \(O(0, 0)\), \(A(0, 3)\), and \(B(4, 0)\).

### Step 2: Calculate the lengths of the sides of the triangle

Using the distance formula, we can find the lengths of the sides \(OA\), \(OB\), and \(AB\):

1. **Length \(OA\)**:
\[ OA = \sqrt{(0 - 0)^2 + (3 - 0)^2} = \sqrt{0 + 9} = 3 \]

2. **Length \(OB\)**:
\[ OB = \sqrt{(4 - 0)^2 + (0 - 0)^2} = \sqrt{16 + 0} = 4 \]

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

### Step 3: Use the formula for the incentre

The coordinates of the incentre \(I(x, y)\) of a triangle with vertices \( (x_1, y_1) \), \( (x_2, y_2) \), and \( (x_3, y_3) \) and corresponding opposite sides \(a\), \(b\), and \(c\) are given by:
\[ I_x = \frac{a x_1 + b x_2 + c x_3}{a + b + c} \]
\[ I_y = \frac{a y_1 + b y_2 + c y_3}{a + b + c} \]

Here:
- \(O(0, 0)\) corresponds to \( (x_1, y_1) \)
- \(A(0, 3)\) corresponds to \( (x_2, y_2) \)
- \(B(4, 0)\) corresponds to \( (x_3, y_3) \)

Assigning the lengths:
- \(a = 5\) (length of \(AB\))
- \(b = 4\) (length of \(OB\))
- \(c = 3\) (length of \(OA\))

Substituting into the formula:
\[ I_x = \frac{5 \cdot 0 + 4 \cdot 0 + 3 \cdot 4}{5 + 4 + 3} = \frac{0 + 0 + 12}{12} = 1 \]
\[ I_y = \frac{5 \cdot 0 + 4 \cdot 3 + 3 \cdot 0}{5 + 4 + 3} = \frac{0 + 12 + 0}{12} = 1 \]

Thus, the coordinates of the incentre are \(I(1, 1)\).

### Final Answer:
The incentre of the triangle formed by \(x=0\), \(y=0\), and \(3x + 4y = 12\) is \((1, 1)\).

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