1. **Problem statement:** (a)(i) Find the angle from line $l_1: y = -5x + 7$ to line $l_2: y = 3x - 8$. (a)(ii) Find the equation of a line parallel to $l_2$ passing through $(2, -3)$. (b)(i) Find the perpendicular distance from point $(5, -7)$ to line $2x - 3y - 13 = 0$. (b)(ii) Find the equation of a line perpendicular to $y = 5x + 9$ and cutting the $y$-axis at 3. (c) Find the equation of the perpendicular bisector of the line joining points $(-3, 10)$ and $(-5, 14)$ and find the coordinates of its intersection $Q$ with the $x$-axis. --- 2. **Solution:** **(a)(i) Angle between lines:** - Slopes: $m_1 = -5$, $m_2 = 3$. - Formula for angle $\theta$ between two lines: $$\tan \theta = \left| \frac{m_2 - m_1}{1 + m_1 m_2} \right| = \left| \frac{3 - (-5)}{1 + (-5)(3)} \right| = \left| \frac{8}{1 - 15} \right| = \left| \frac{8}{-14} \right| = \frac{4}{7}.$$ - Therefore, $$\theta = \arctan \frac{4}{7}.$$ - Numerically, $\theta \approx 29.74^\circ$. **(a)(ii) Equation of line parallel to $l_2$ through $(2, -3)$:** - Slope of $l_2$ is $m = 3$. - Equation using point-slope form: $$y - (-3) = 3(x - 2) \implies y + 3 = 3x - 6 \implies y = 3x - 9.$$ **(b)(i) Perpendicular distance from $(5, -7)$ to line $2x - 3y - 13 = 0$:** - Distance formula: $$d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}} = \frac{|2(5) - 3(-7) - 13|}{\sqrt{2^2 + (-3)^2}} = \frac{|10 + 21 - 13|}{\sqrt{4 + 9}} = \frac{18}{\sqrt{13}} = \frac{18\sqrt{13}}{13}.$$ **(b)(ii) Equation of line perpendicular to $y = 5x + 9$ and cuts $y$-axis at 3:** - Slope of given line: $m = 5$. - Slope of perpendicular line: $m_{\perp} = -\frac{1}{5}$. - Since it cuts $y$-axis at 3, $y$-intercept $b = 3$. - Equation: $$y = -\frac{1}{5}x + 3.$$ **(c) Perpendicular bisector of segment joining $(-3, 10)$ and $(-5, 14)$:** - Midpoint $M$: $$\left( \frac{-3 + (-5)}{2}, \frac{10 + 14}{2} \right) = (-4, 12).$$ - Slope of segment: $$m = \frac{14 - 10}{-5 - (-3)} = \frac{4}{-2} = -2.$$ - Slope of perpendicular bisector: $$m_{\perp} = \frac{1}{2}.$$ - Equation of perpendicular bisector through $M$: $$y - 12 = \frac{1}{2}(x + 4) \implies y = \frac{1}{2}x + 2 + 12 = \frac{1}{2}x + 14.$$ - To find $Q$ where it meets $x$-axis, set $y=0$: $$0 = \frac{1}{2}x + 14 \implies \frac{1}{2}x = -14 \implies x = -28.$$ - Coordinates of $Q$: $$(-28, 0).$$ --- 3. **Final answers:** - (a)(i) Angle $\theta = \arctan \frac{4}{7} \approx 29.74^\circ$. - (a)(ii) $y = 3x - 9$. - (b)(i) Distance $= \frac{18\sqrt{13}}{13}$. - (b)(ii) $y = -\frac{1}{5}x + 3$. - (c) Perpendicular bisector: $y = \frac{1}{2}x + 14$, point $Q = (-28, 0)$.