1. **State the problem:** We have line $l$ with equation $2x - 3y = 12$. Point $A$ is where $l$ cuts the x-axis. Point $B$ is given as $(0, -4)$ on the y-axis. We need to find: (a) Coordinates of $A$. (b) Slope of line $l$. (c) Slope and equation of line $k$ perpendicular to $l$ passing through $B$. (d) Coordinates of $C$, where $k$ crosses the x-axis. (e) Area of triangle $ABC$. 2. **Find coordinates of $A$ (x-intercept of $l$):** At the x-axis, $y=0$. Substitute $y=0$ into $2x - 3y = 12$: $$2x - 3(0) = 12 \implies 2x = 12$$ Divide both sides by 2: $$\cancel{2}x = \frac{12}{\cancel{2}} \implies x = 6$$ So, $A = (6, 0)$. 3. **Find slope of line $l$:** Rewrite $2x - 3y = 12$ in slope-intercept form $y = mx + c$: $$2x - 3y = 12 \implies -3y = -2x + 12$$ Divide both sides by $-3$: $$\cancel{-3}y = \frac{-2x + 12}{\cancel{-3}} \implies y = \frac{2}{3}x - 4$$ Slope $m_l = \frac{2}{3}$. 4. **Find slope and equation of line $k$ perpendicular to $l$ passing through $B(0, -4)$:** Slope of $k$ is negative reciprocal of $m_l$: $$m_k = -\frac{1}{m_l} = -\frac{1}{\frac{2}{3}} = -\frac{3}{2}$$ Equation of $k$ using point-slope form: $$y - y_1 = m_k(x - x_1)$$ Substitute $B(0, -4)$: $$y - (-4) = -\frac{3}{2}(x - 0) \implies y + 4 = -\frac{3}{2}x$$ Simplify: $$y = -\frac{3}{2}x - 4$$ 5. **Find coordinates of $C$ (x-intercept of $k$):** At x-axis, $y=0$. Substitute $y=0$ into $y = -\frac{3}{2}x - 4$: $$0 = -\frac{3}{2}x - 4$$ Add 4 to both sides: $$4 = -\frac{3}{2}x$$ Divide both sides by $-\frac{3}{2}$: $$x = \frac{4}{-\frac{3}{2}} = 4 \times -\frac{2}{3} = -\frac{8}{3}$$ So, $C = \left(-\frac{8}{3}, 0\right)$. 6. **Find area of triangle $ABC$:** Points are $A(6,0)$, $B(0,-4)$, $C\left(-\frac{8}{3},0\right)$. Base $AC$ lies on x-axis, length: $$AC = 6 - \left(-\frac{8}{3}\right) = 6 + \frac{8}{3} = \frac{18}{3} + \frac{8}{3} = \frac{26}{3}$$ Height is vertical distance from $B$ to x-axis: $$|y_B| = |-4| = 4$$ Area formula: $$\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times \frac{26}{3} \times 4 = \frac{1}{2} \times \frac{104}{3} = \frac{104}{6} = \frac{52}{3}$$ **Final answers:** (a) $A = (6, 0)$ (b) Slope of $l$ is $\frac{2}{3}$ (c) Slope of $k$ is $-\frac{3}{2}$ and equation is $y = -\frac{3}{2}x - 4$ (d) $C = \left(-\frac{8}{3}, 0\right)$ (e) Area of triangle $ABC$ is $\frac{52}{3}$ square units.