1. **Problem statement:** Given the line equation $3x - 2y = 10$, find: a. The equation of line $L_2$ passing through $M(-5, 2)$ and parallel to $L_1$. b. The equation of line $L_3$ perpendicular to $L_2$ at $M(1, -8)$. c. The angle of inclination of $L_2$ with the horizontal. d. The magnitude of segment $MN$ (assuming $N$ is a point on $L_2$ or $L_3$, but since $N$ is not defined, we will interpret $N$ as the point on $L_3$ perpendicular from $M$). 2. **Convert $3x - 2y = 10$ to slope-intercept form $y = mx + c$:** $$3x - 2y = 10$$ Subtract $3x$ from both sides: $$-2y = -3x + 10$$ Divide both sides by $-2$: $$y = \frac{-3x + 10}{-2} = \frac{-3x}{-2} + \frac{10}{-2}$$ $$y = \frac{3}{2}x - 5$$ So, $m = \frac{3}{2}$ and $c = -5$. 3. **(a)(i) Equation of $L_2$ passing through $M(-5, 2)$ and parallel to $L_1$:** Parallel lines have the same slope, so $m_{L_2} = \frac{3}{2}$. Use point-slope form: $$y - y_1 = m(x - x_1)$$ $$y - 2 = \frac{3}{2}(x + 5)$$ Expand: $$y - 2 = \frac{3}{2}x + \frac{15}{2}$$ Add 2 to both sides: $$y = \frac{3}{2}x + \frac{15}{2} + 2$$ Convert 2 to fraction: $$2 = \frac{4}{2}$$ So: $$y = \frac{3}{2}x + \frac{15}{2} + \frac{4}{2} = \frac{3}{2}x + \frac{19}{2}$$ 4. **(a)(ii) Equation of $L_3$ perpendicular to $L_2$ at $M(1, -8)$:** Slope of $L_2$ is $m = \frac{3}{2}$. Slope of perpendicular line $L_3$ is negative reciprocal: $$m_{L_3} = -\frac{1}{m} = -\frac{1}{\frac{3}{2}} = -\frac{2}{3}$$ Use point-slope form with point $(1, -8)$: $$y - (-8) = -\frac{2}{3}(x - 1)$$ $$y + 8 = -\frac{2}{3}x + \frac{2}{3}$$ Subtract 8: $$y = -\frac{2}{3}x + \frac{2}{3} - 8$$ Convert 8 to fraction: $$8 = \frac{24}{3}$$ So: $$y = -\frac{2}{3}x + \frac{2}{3} - \frac{24}{3} = -\frac{2}{3}x - \frac{22}{3}$$ 5. **(b) Angle of inclination of $L_2$ with the horizontal:** Formula for angle of inclination $\theta$: $$\theta = \tan^{-1}(m)$$ $$\theta = \tan^{-1}\left(\frac{3}{2}\right)$$ This is approximately: $$\theta \approx 56.31^\circ$$ 6. **(c) Magnitude of $MN$:** Assuming $N$ is the foot of perpendicular from $M(-5, 2)$ to $L_3$ or $L_2$. Since $M$ is on $L_2$, and $L_3$ is perpendicular to $L_2$ at $(1, -8)$, let's find $N$ as the point on $L_3$ closest to $M$. Equation of $L_3$: $$y = -\frac{2}{3}x - \frac{22}{3}$$ Distance formula from point $(x_0, y_0)$ to line $Ax + By + C = 0$ is: $$d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}$$ Rewrite $L_3$ in standard form: $$y = -\frac{2}{3}x - \frac{22}{3}$$ Multiply both sides by 3: $$3y = -2x - 22$$ Bring all terms to one side: $$2x + 3y + 22 = 0$$ Calculate distance from $M(-5, 2)$: $$d = \frac{|2(-5) + 3(2) + 22|}{\sqrt{2^2 + 3^2}} = \frac{|-10 + 6 + 22|}{\sqrt{4 + 9}} = \frac{|18|}{\sqrt{13}} = \frac{18}{\sqrt{13}}$$ Simplify: $$d = \frac{18}{\sqrt{13}} = \frac{18\sqrt{13}}{13}$$ **Final answers:** - (a)(i) $L_2: y = \frac{3}{2}x + \frac{19}{2}$ - (a)(ii) $L_3: y = -\frac{2}{3}x - \frac{22}{3}$ - (b) Angle of inclination $\approx 56.31^\circ$ - (c) Magnitude of $MN = \frac{18\sqrt{13}}{13}$