1. **State the problem:** We have a line $l$ with equation $2x - 3y = 12$. - (a) Find the coordinates of point $A$ where line $l$ cuts the x-axis. - (b) Find the slope of line $l$. - (c) Find the slope of line $k$ which passes through $B(0, -4)$ and is perpendicular to $l$, and then find the equation of line $k$. 2. **Find coordinates of $A$ (x-intercept):** At the x-axis, $y=0$. Substitute $y=0$ into $2x - 3y = 12$: $$2x - 3(0) = 12$$ $$2x = 12$$ Divide both sides by 2: $$\cancel{2}x = \cancel{2}6$$ $$x = 6$$ So, $A = (6, 0)$. 3. **Find slope of line $l$:** Rewrite $2x - 3y = 12$ in slope-intercept form $y = mx + c$: $$2x - 3y = 12$$ Subtract $2x$ from both sides: $$-3y = -2x + 12$$ Divide both sides by $-3$: $$\cancel{-3}y = \cancel{-3}\frac{2x}{3} - \cancel{-3}4$$ $$y = \frac{2}{3}x - 4$$ Slope $m$ of line $l$ is the coefficient of $x$: $$m = \frac{2}{3}$$ 4. **Find slope of line $k$ (perpendicular to $l$):** The slope of a line perpendicular to another with slope $m$ is: $$m_{\perp} = -\frac{1}{m}$$ So, $$m_k = -\frac{1}{\frac{2}{3}} = -\frac{3}{2}$$ 5. **Find equation of line $k$ passing through $B(0, -4)$ with slope $-\frac{3}{2}$:** Use point-slope form: $$y - y_1 = m(x - x_1)$$ Substitute $m = -\frac{3}{2}$, $x_1 = 0$, $y_1 = -4$: $$y - (-4) = -\frac{3}{2}(x - 0)$$ $$y + 4 = -\frac{3}{2}x$$ Subtract 4 from both sides: $$y = -\frac{3}{2}x - 4$$ **Final answers:** - (a) $A = (6, 0)$ - (b) Slope of $l$ is $\frac{2}{3}$ - (c) Slope of $k$ is $-\frac{3}{2}$ and equation of $k$ is $y = -\frac{3}{2}x - 4$