1. **State the problem:** We are given the line equation $3x - 2y = 10$ (call this line $L1$). We need to: a) Find the equation of line $L2$ passing through $M(-5, 2)$ and parallel to $L1$, and the equation of line $L3$ perpendicular to $L2$ passing through $M(1, -8)$. b) Find the angle of inclination of $L2$ with the horizontal. c) Find the magnitude of $MN$ (assuming $N$ is a point related to the problem, but since $N$ is not defined, we interpret $N$ as the foot of perpendicular from $M$ to $L1$ or $L2$; here we assume $N$ is the projection of $M(-5,2)$ onto $L2$). --- 2. **Convert $L1$ to slope-intercept form $y = mx + c$:** Given: $$3x - 2y = 10$$ Rearranged: $$-2y = -3x + 10$$ Divide both sides by $-2$: $$y = \frac{-3x + 10}{-2} = \frac{\cancel{-3}x}{\cancel{-2}} - \frac{10}{2} = \frac{3}{2}x - 5$$ So, slope $m_1 = \frac{3}{2}$ and intercept $c_1 = -5$. --- 3. **Find equation of $L2$ parallel to $L1$ passing through $M(-5, 2)$:** Parallel lines have the same slope, so $m_2 = m_1 = \frac{3}{2}$. Use point-slope form: $$y - y_1 = m(x - x_1)$$ Substitute $m = \frac{3}{2}$, $x_1 = -5$, $y_1 = 2$: $$y - 2 = \frac{3}{2}(x + 5)$$ Simplify: $$y = \frac{3}{2}x + \frac{3}{2} \times 5 + 2 = \frac{3}{2}x + \frac{15}{2} + 2$$ 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. **Find equation of $L3$ perpendicular to $L2$ passing through $M(1, -8)$:** Slope of $L2$ is $m_2 = \frac{3}{2}$. Slope of perpendicular line $L3$ is negative reciprocal: $$m_3 = -\frac{1}{m_2} = -\frac{1}{\frac{3}{2}} = -\frac{2}{3}$$ Use point-slope form with point $(1, -8)$: $$y - (-8) = -\frac{2}{3}(x - 1)$$ Simplify: $$y + 8 = -\frac{2}{3}x + \frac{2}{3}$$ 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. **Find angle of inclination of $L2$ with horizontal:** Angle of inclination $\theta$ is given by: $$\theta = \tan^{-1}(m_2) = \tan^{-1}\left(\frac{3}{2}\right)$$ This is the angle between the line and the positive x-axis. --- 6. **Find magnitude of $MN$:** Assuming $N$ is the foot of perpendicular from $M(-5, 2)$ to $L2$. Equation of $L2$: $$y = \frac{3}{2}x + \frac{19}{2}$$ Slope of $L2$ is $m_2 = \frac{3}{2}$. Slope of perpendicular from $M$ to $L2$ is $m_p = -\frac{2}{3}$. Equation of perpendicular line through $M(-5, 2)$: $$y - 2 = -\frac{2}{3}(x + 5)$$ Simplify: $$y = -\frac{2}{3}x - \frac{10}{3} + 2 = -\frac{2}{3}x - \frac{10}{3} + \frac{6}{3} = -\frac{2}{3}x - \frac{4}{3}$$ Find intersection $N$ of $L2$ and this perpendicular: Set equal: $$\frac{3}{2}x + \frac{19}{2} = -\frac{2}{3}x - \frac{4}{3}$$ Multiply both sides by 6 (LCM of denominators 2 and 3): $$6 \times \left(\frac{3}{2}x + \frac{19}{2}\right) = 6 \times \left(-\frac{2}{3}x - \frac{4}{3}\right)$$ $$3 \times 3x + 3 \times 19 = 2 \times (-2x) + 2 \times (-4)$$ $$9x + 57 = -4x - 8$$ Bring terms to one side: $$9x + 4x = -8 - 57$$ $$13x = -65$$ $$x = -5$$ Substitute $x = -5$ into $L2$: $$y = \frac{3}{2}(-5) + \frac{19}{2} = -\frac{15}{2} + \frac{19}{2} = \frac{4}{2} = 2$$ So, $N = (-5, 2)$ which is the same as $M$. Therefore, the magnitude $MN = 0$. If $N$ is different, please clarify. --- **Final answers:** - $L2: y = \frac{3}{2}x + \frac{19}{2}$ - $L3: y = -\frac{2}{3}x - \frac{22}{3}$ - Angle of inclination of $L2$: $$\theta = \tan^{-1}\left(\frac{3}{2}\right)$$ - Magnitude $MN = 0$ (since $M$ lies on $L2$).