A line meets x - axis at A and y-axis at B such that incentre of the triangle OAB is (1,1). Equation of AB is

Options

A. `2x+y=2`
B. `3x+4y=12`
C. `2x+y=6`
D. `2x+y=4`

Answer

To find the equation of the line \( AB \) that meets the x-axis at point \( A \) and the y-axis at point \( B \), given that the incenter of triangle \( OAB \) is at the point \( (1, 1) \), we can proceed with the following steps:

### Step 1: Understand the Geometry
The points \( A \) and \( B \) can be represented as \( A(a, 0) \) on the x-axis and \( B(0, b) \) on the y-axis. The origin \( O \) is at \( (0, 0) \).

### Step 2: Incenter Formula
The coordinates of the incenter \( I \) of triangle \( OAB \) can be calculated using the formula:
\[ I = \left( \frac{a + 0 + 0}{3}, \frac{0 + b + 0}{3} \right) = \left( \frac{a}{3}, \frac{b}{3} \right) \]
Given that the incenter \( I \) is at \( (1, 1) \), we can set up the equations:
\[ \frac{a}{3} = 1 \quad \text{and} \quad \frac{b}{3} = 1 \]

### Step 3: Solve for \( a \) and \( b \)
From the equations:
1. \( \frac{a}{3} = 1 \) implies \( a = 3 \)
2. \( \frac{b}{3} = 1 \) implies \( b = 3 \)

### Step 4: Find the Points \( A \) and \( B \)
Thus, the coordinates of points \( A \) and \( B \) are:
- \( A(3, 0) \)
- \( B(0, 3) \)

### Step 5: Equation of Line \( AB \)
To find the equation of the line passing through points \( A(3, 0) \) and \( B(0, 3) \), we can use the two-point form of the equation of a line:
\[ y - y_1 = m(x - x_1) \]
where \( m \) is the slope given by:
\[ m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{3 - 0}{0 - 3} = -1 \]
Using point \( A(3, 0) \):
\[ y - 0 = -1(x - 3) \implies y = -x + 3 \]

### Step 6: Rearranging the Equation
Rearranging gives us:
\[ x + y = 3 \]

### Final Answer
Thus, the equation of line \( AB \) is:
\[ \boxed{x + y = 3} \]