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Step-by-Step Solution
Step 1: Identify the Points of Intersection
The line given is
$3x + 4y - 24 = 0$.
To find its intersection with the x-axis (point A), set $y=0$:
$$
3x - 24 = 0 \quad \Longrightarrow \quad x = 8.
$$
Hence, $A(8, 0)$.
To find its intersection with the y-axis (point B), set $x=0$:
$$
4y - 24 = 0 \quad \Longrightarrow \quad y = 6.
$$
Hence, $B(0, 6)$.
Step 2: Write Down the Coordinates of the Triangle
We now have the triangle $OAB$ with vertices:
$$
O(0, 0),\quad A(8, 0),\quad B(0, 6).
$$
Step 3: Calculate the Side Lengths
1. $OA = \text{distance between } O(0,0) \text{ and } A(8,0)$
$$
OA = \sqrt{(8 - 0)^2 + (0 - 0)^2} = \sqrt{64} = 8.
$$
2. $OB = \text{distance between } O(0,0) \text{ and } B(0,6)$
$$
OB = \sqrt{(0 - 0)^2 + (6 - 0)^2} = \sqrt{36} = 6.
$$
3. $AB = \text{distance between } A(8,0) \text{ and } B(0,6)$
$$
AB = \sqrt{(0 - 8)^2 + (6 - 0)^2}
= \sqrt{(-8)^2 + 6^2}
= \sqrt{64 + 36}
= \sqrt{100}
= 10.
$$
Step 4: Use the Formula for the Incenter
Recall that the incenter of a triangle ($x_I, y_I$) whose vertices are $(x_A, y_A)$, $(x_B, y_B)$, and $(x_C, y_C)$ and side lengths (opposite these vertices) $a, b, c$ is given by:
$$
x_I = \frac{a\,x_A + b\,x_B + c\,x_C}{a + b + c}, \quad
y_I = \frac{a\,y_A + b\,y_B + c\,y_C}{a + b + c}.
$$
Here:
$a = BC$ is the side opposite vertex $A$
$b = AC$ is the side opposite vertex $B$
$c = AB$ is the side opposite vertex $C$
In our case, we label the triangle $OAB$ such that:
Side opposite $O$ is $AB$, of length 10.
Side opposite $A$ is $OB$, of length 6.
Side opposite $B$ is $OA$, of length 8.
So the incenter coordinates become:
$$
x_I = \frac{10 \cdot 0 + 6 \cdot 8 + 8 \cdot 0}{10 + 6 + 8}
= \frac{48}{24}
= 2,
$$
$$
y_I = \frac{10 \cdot 0 + 6 \cdot 0 + 8 \cdot 6}{10 + 6 + 8}
= \frac{48}{24}
= 2.
$$
Therefore, the incenter is at $(2,2)$.
Step 5: Final Answer
The incenter of triangle $OAB$ is $(2, 2)$.
