1. **Problem statement:** (a)(i) Find the slope of the line $l$ passing through points $P(-3,6)$ and $Q(9,-2)$. 2. **Formula for slope:** The slope $m$ of a line through points $(x_1,y_1)$ and $(x_2,y_2)$ is given by: $$m=\frac{y_2 - y_1}{x_2 - x_1}$$ 3. **Calculate slope of line $l$:** $$m=\frac{-2 - 6}{9 - (-3)}=\frac{-8}{12}=\frac{\cancel{-8}}{\cancel{12}}=\frac{-2}{3}$$ 4. **(a)(ii) Show equation of line $l$ is $2x + 3y - 12 = 0$:** Use point-slope form: $$y - y_1 = m(x - x_1)$$ Using point $P(-3,6)$ and slope $m=-\frac{2}{3}$: $$y - 6 = -\frac{2}{3}(x + 3)$$ Multiply both sides by 3 to clear denominator: $$3(y - 6) = -2(x + 3)$$ $$3y - 18 = -2x - 6$$ Bring all terms to one side: $$2x + 3y - 12 = 0$$ 5. **(b) Show line $k: 3x - 2y + 8 = 0$ is perpendicular to line $l$ using slopes:** Rewrite line $k$ in slope-intercept form: $$3x - 2y + 8 = 0 \Rightarrow -2y = -3x - 8 \Rightarrow y = \frac{3}{2}x + 4$$ Slope of $k$ is $m_k = \frac{3}{2}$. Slope of $l$ is $m_l = -\frac{2}{3}$. Check product of slopes: $$m_l \times m_k = -\frac{2}{3} \times \frac{3}{2} = -1$$ Since product is $-1$, lines $l$ and $k$ are perpendicular. **Final answers:** - Slope of line $l$ is $-\frac{2}{3}$. - Equation of line $l$ is $2x + 3y - 12 = 0$. - Lines $l$ and $k$ are perpendicular because their slopes multiply to $-1$.