1. **Problem 33:** A line from the point $(2,3)$ is perpendicular to the line $y=\frac{1}{3}x+1$. Find the coordinates of the intersection point $P$. 2. **Formula and rules:** - The slope of the given line is $m=\frac{1}{3}$. - The slope of a line perpendicular to it is the negative reciprocal: $m_\perp = -3$. - Equation of the perpendicular line through $(2,3)$ is $y - 3 = -3(x - 2)$. 3. **Find the equation of the perpendicular line:** $$y - 3 = -3x + 6 \implies y = -3x + 9$$ 4. **Find intersection point $P$ by solving system:** Given lines: $$y = \frac{1}{3}x + 1$$ $$y = -3x + 9$$ Set equal: $$\frac{1}{3}x + 1 = -3x + 9$$ Multiply both sides by 3: $$x + 3 = -9x + 27$$ $$x + 9x = 27 - 3$$ $$10x = 24$$ $$x = \frac{24}{10} = 2.4$$ 5. **Find $y$ coordinate:** $$y = \frac{1}{3}(2.4) + 1 = 0.8 + 1 = 1.8$$ 6. **Coordinates of $P$ are:** $$(2.4, 1.8)$$ --- 7. **Problem 34(a):** Gradient of line $l$ with equation $y=5x+12$ is the coefficient of $x$. 8. **Answer:** $$\text{Gradient} = 5$$ --- 9. **Problem 34(b):** Find where line $l$ crosses the x-axis. 10. At x-axis, $y=0$, so solve: $$0 = 5x + 12$$ $$5x = -12$$ $$x = -\frac{12}{5} = -2.4$$ 11. Coordinates of x-intercept: $$(-2.4, 0)$$ --- 12. **Problem 34(c):** Find gradient $k$ of line perpendicular to $l$. 13. Since gradient of $l$ is $5$, perpendicular gradient is negative reciprocal: $$k = -\frac{1}{5}$$ --- **Final answers:** - Problem 33: $P = (2.4, 1.8)$ - Problem 34(a): Gradient $= 5$ - Problem 34(b): x-intercept $= (-2.4, 0)$ - Problem 34(c): $k = -\frac{1}{5}$