1. **Problem 1: Find the equation of a straight line** Given two points $A(2,3)$ and $B(5,11)$, find the equation of the line passing through these points. 2. **Problem 2: Calculate the gradient and intercept** Find the gradient and y-intercept of the line with equation $3x - 4y + 12 = 0$. 3. **Problem 3: Graph the line and find intercepts** Graph the line $y = 2x - 3$ and find its x- and y-intercepts. 4. **Problem 4: Parallel and perpendicular lines** Find the equation of a line parallel to $y = -\frac{1}{2}x + 4$ passing through $(6,1)$. 5. **Problem 5: Distance between a point and a line** Find the distance from the point $(3,4)$ to the line $4x + 3y - 24 = 0$. --- **Solutions:** 1. Use formula for gradient $m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{11 - 3}{5 - 2} = \frac{8}{3}$. Equation: $y - 3 = \frac{8}{3}(x - 2)$ simplifies to $y = \frac{8}{3}x - \frac{7}{3}$. 2. Rewrite $3x - 4y + 12 = 0$ as $y = \frac{3}{4}x + 3$. Gradient $m = \frac{3}{4}$, y-intercept $c = 3$. 3. Intercepts: - y-intercept: set $x=0$, $y = -3$. - x-intercept: set $y=0$, $0 = 2x - 3 \Rightarrow x = \frac{3}{2}$. Graph is a straight line crossing y-axis at -3 and x-axis at 1.5. 4. Parallel lines have same gradient $m = -\frac{1}{2}$. Equation: $y - 1 = -\frac{1}{2}(x - 6)$, simplifies to $y = -\frac{1}{2}x + 4$. 5. Distance formula from point $(x_0,y_0)$ to line $Ax + By + C = 0$ is $$d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}} = \frac{|4(3) + 3(4) - 24|}{\sqrt{4^2 + 3^2}} = \frac{|12 + 12 - 24|}{5} = 0$$ The point lies on the line, so distance is 0. --- **Answers:** 1. $y = \frac{8}{3}x - \frac{7}{3}$ 2. Gradient $\frac{3}{4}$, y-intercept 3 3. x-intercept $\frac{3}{2}$, y-intercept $-3$ 4. $y = -\frac{1}{2}x + 4$ 5. Distance = 0 --- **Diagrams:** - Problem 1: Line through points (2,3) and (5,11) - Problem 3: Graph of $y=2x-3$ with intercepts - Problem 4: Parallel line through (6,1)