1. Foundations of Linear Equations 1.a) Given the line equation $$3x - 4y + 12 = 0$$, rewrite it in slope-intercept form $$y = mx + b$$. Start by isolating $$y$$: $$3x - 4y + 12 = 0 \implies -4y = -3x - 12 \implies y = \frac{3}{4}x + 3$$. 1.b) The slope $$m$$ is the coefficient of $$x$$, so $$m = \frac{3}{4}$$. The y-intercept $$b$$ is the constant term, so $$b = 3$$. 1.c) Convert to intercept form $$\frac{x}{a} + \frac{y}{b} = 1$$. Find x-intercept by setting $$y=0$$: $$3x - 4(0) + 12 = 0 \implies 3x = -12 \implies x = -4$$. Find y-intercept by setting $$x=0$$: $$3(0) - 4y + 12 = 0 \implies -4y = -12 \implies y = 3$$. Intercept form: $$\frac{x}{-4} + \frac{y}{3} = 1$$. 2. Two-Point & Point-Slope Forms Given points $$A(-2,5)$$ and $$B(4,-1)$$. 2.a) Two-point form equation: $$\frac{y - y_1}{x - x_1} = \frac{y_2 - y_1}{x_2 - x_1}$$. Calculate slope: $$m = \frac{-1 - 5}{4 - (-2)} = \frac{-6}{6} = -1$$. Equation: $$\frac{y - 5}{x + 2} = -1 \implies y - 5 = -1(x + 2) \implies y = -x + 3$$. 2.b) Point-slope form using point A: $$y - 5 = -1(x + 2)$$. 3. Parallel Lines Given line $$2x + 3y = 6$$, find line parallel passing through $$(4,-2)$$. Slope of given line: $$3y = -2x + 6 \implies y = -\frac{2}{3}x + 2$$, so $$m = -\frac{2}{3}$$. Equation of parallel line: $$y - (-2) = -\frac{2}{3}(x - 4) \implies y + 2 = -\frac{2}{3}x + \frac{8}{3} \implies y = -\frac{2}{3}x + \frac{8}{3} - 2 = -\frac{2}{3}x + \frac{2}{3}$$. 4. Perpendicular Lines Given $$5x - y = 8$$. 4.a) Find perpendicular line at $$(1,-3)$$. Slope of given line: $$-y = -5x + 8 \implies y = 5x - 8$$, so $$m = 5$$. Slope of perpendicular line: $$m_{\perp} = -\frac{1}{5}$$. Equation: $$y - (-3) = -\frac{1}{5}(x - 1) \implies y + 3 = -\frac{1}{5}x + \frac{1}{5} \implies y = -\frac{1}{5}x + \frac{1}{5} - 3 = -\frac{1}{5}x - \frac{14}{5}$$. 4.b) Verify perpendicularity: $$m \times m_{\perp} = 5 \times -\frac{1}{5} = -1$$, confirming perpendicular lines. 5. Perpendicular Bisector Given $$P(3,7)$$ and $$Q(-1,1)$$. 5.a) Midpoint: $$M = \left( \frac{3 + (-1)}{2}, \frac{7 + 1}{2} \right) = (1,4)$$. 5.b) Slope of $$PQ$$: $$m_{PQ} = \frac{1 - 7}{-1 - 3} = \frac{-6}{-4} = \frac{3}{2}$$. Slope of perpendicular bisector: $$m_{\perp} = -\frac{2}{3}$$. Equation: $$y - 4 = -\frac{2}{3}(x - 1) \implies y = -\frac{2}{3}x + \frac{2}{3} + 4 = -\frac{2}{3}x + \frac{14}{3}$$. 6. Equation of Medians Triangle $$ABC$$ with $$A(1,2), B(5,6), C(3,0)$$. 6.a) Median from $$A$$ to $$BC$$. Midpoint of $$BC$$: $$M_{BC} = \left( \frac{5 + 3}{2}, \frac{6 + 0}{2} \right) = (4,3)$$. Slope of median: $$m = \frac{3 - 2}{4 - 1} = \frac{1}{3}$$. Equation: $$y - 2 = \frac{1}{3}(x - 1) \implies y = \frac{1}{3}x + \frac{5}{3}$$. 6.b) All medians intersect at centroid: Calculate medians from $$B$$ and $$C$$ and solve system to find intersection. Centroid coordinates: $$G = \left( \frac{1 + 5 + 3}{3}, \frac{2 + 6 + 0}{3} \right) = (3, \frac{8}{3})$$. 7. Distance from Point to Line Point $$(4,-1)$$ to line $$4x - 3y + 5 = 0$$. Distance formula: $$d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}$$. Calculate numerator: $$|4(4) - 3(-1) + 5| = |16 + 3 + 5| = 24$$. Denominator: $$\sqrt{4^2 + (-3)^2} = \sqrt{16 + 9} = 5$$. Distance: $$d = \frac{24}{5} = 4.8$$. 8. Distance Between Parallel Lines Lines: $$6x + 8y = 10$$ and $$3x + 4y = 8$$. Rewrite second line to match first's coefficients: Multiply second by 2: $$6x + 8y = 16$$. Distance formula between parallel lines: $$d = \frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}}$$. Here, $$C_1 = 10$$, $$C_2 = 16$$, $$A=6$$, $$B=8$$. Calculate: $$d = \frac{|10 - 16|}{\sqrt{36 + 64}} = \frac{6}{10} = 0.6$$. 9. Urban Planning Scenario 9.a) Line $$R$$ with slope $$-\frac{1}{2}$$ through $$(2,3)$$. Equation: $$y - 3 = -\frac{1}{2}(x - 2) \implies y = -\frac{1}{2}x + 1 + 3 = -\frac{1}{2}x + 4$$. 9.b) Distance from hospital $$(5,-2)$$ to line $$y = -\frac{1}{2}x + 4$$. Rewrite line in standard form: $$y = -\frac{1}{2}x + 4 \implies \frac{1}{2}x + y - 4 = 0 \implies x + 2y - 8 = 0$$. Distance: $$d = \frac{|1(5) + 2(-2) - 8|}{\sqrt{1^2 + 2^2}} = \frac{|5 - 4 - 8|}{\sqrt{5}} = \frac{7}{\sqrt{5}} = \frac{7\sqrt{5}}{5} \approx 3.13$$. 9.c) Perpendicular bike path through $$(2,3)$$. Slope of $$R$$ is $$-\frac{1}{2}$$, so perpendicular slope is $$2$$. Equation: $$y - 3 = 2(x - 2) \implies y = 2x - 4 + 3 = 2x - 1$$.