1. **State the problem:** We have a line that cuts the x-axis at $A(a,0)$ and the y-axis at $B(0,b)$, where $a,b \in \mathbb{Z}$. The slope of the line is $-\frac{2}{3}$, and the area of the triangle formed by the line and the axes is 12 square units. We need to find the equations of the two lines that satisfy these conditions. 2. **Recall formulas and rules:** - The slope of a line passing through $A(a,0)$ and $B(0,b)$ is given by: $$m = \frac{0 - b}{a - 0} = -\frac{b}{a}$$ - The area of the triangle formed by the line and the coordinate axes is: $$\text{Area} = \frac{1}{2} |a \times b|$$ - Given slope $m = -\frac{2}{3}$, so: $$-\frac{b}{a} = -\frac{2}{3} \implies \frac{b}{a} = \frac{2}{3}$$ 3. **Express $b$ in terms of $a$:** $$b = \frac{2}{3}a$$ 4. **Use the area condition:** $$12 = \frac{1}{2} |a \times b| = \frac{1}{2} |a \times \frac{2}{3}a| = \frac{1}{2} \times \frac{2}{3} a^2 = \frac{1}{3} a^2$$ 5. **Solve for $a^2$:** $$12 = \frac{1}{3} a^2 \implies a^2 = 36$$ 6. **Find possible values of $a$:** $$a = \pm 6$$ 7. **Find corresponding $b$ values:** - For $a=6$: $$b = \frac{2}{3} \times 6 = 4$$ - For $a=-6$: $$b = \frac{2}{3} \times (-6) = -4$$ 8. **Write the equations of the lines:** The line passing through $(a,0)$ and $(0,b)$ can be written as: $$\frac{x}{a} + \frac{y}{b} = 1$$ - For $(6,0)$ and $(0,4)$: $$\frac{x}{6} + \frac{y}{4} = 1$$ Multiply both sides by 12: $$2x + 3y = 12$$ - For $(-6,0)$ and $(0,-4)$: $$\frac{x}{-6} + \frac{y}{-4} = 1 \implies -\frac{x}{6} - \frac{y}{4} = 1$$ Multiply both sides by -12: $$2x + 3y = -12$$ **Final answers:** $$2x + 3y = 12$$ $$2x + 3y = -12$$