1. **State the problem:** Find the coordinates of points A and B where the line $2x + 3y - 6 = 0$ intersects the x-axis and y-axis respectively, and then find the area of triangle $OAB$ where $O$ is the origin. 2. **Find coordinates of A (x-intercept):** On the x-axis, $y=0$. Substitute $y=0$ into the line equation: $$2x + 3(0) - 6 = 0 \implies 2x - 6 = 0$$ Solve for $x$: $$2x = 6$$ $$x = \frac{6}{2}$$ Show cancellation: $$x = \frac{\cancel{6}}{\cancel{2}} = 3$$ So, $A = (3,0)$. 3. **Find coordinates of B (y-intercept):** On the y-axis, $x=0$. Substitute $x=0$ into the line equation: $$2(0) + 3y - 6 = 0 \implies 3y - 6 = 0$$ Solve for $y$: $$3y = 6$$ $$y = \frac{6}{3}$$ Show cancellation: $$y = \frac{\cancel{6}}{\cancel{3}} = 2$$ So, $B = (0,2)$. 4. **Find area of triangle $OAB$:** The triangle is formed by points $O(0,0)$, $A(3,0)$, and $B(0,2)$. The area of a triangle with vertices on axes is: $$\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$$ Here, base = length $OA = 3$, height = length $OB = 2$. Calculate area: $$\text{Area} = \frac{1}{2} \times 3 \times 2 = 3$$ **Final answer:** The coordinates are $A(3,0)$ and $B(0,2)$, and the area of triangle $OAB$ is $3$ square units.