1. Check whether the pairs of lines are parallel. (i) Lines: $2y - 6x - 1 = 0$ and $4y = 12x - 1$ Rewrite both in slope-intercept form $y = mx + c$: $2y = 6x + 1 \Rightarrow y = 3x + \frac{1}{2}$ $4y = 12x - 1 \Rightarrow y = 3x - \frac{1}{4}$ Slopes are both 3, so lines are parallel. (ii) Lines: $y - 3x = 4$ and $y - 2x + 1 = 0$ Rewrite: $y = 3x + 4$ $y = 2x - 1$ Slopes 3 and 2 are not equal, so not parallel. (iii) Lines: $2y = -x + 3$ and $y + x - 4 = 0$ Rewrite: $y = -\frac{1}{2}x + \frac{3}{2}$ $y = -x + 4$ Slopes $-\frac{1}{2}$ and $-1$ not equal, so not parallel. (iv) Lines: $y + \frac{3}{2}x = 5$ and $6y + 9x - 14 = 0$ Rewrite: $y = -\frac{3}{2}x + 5$ $6y = -9x + 14 \Rightarrow y = -\frac{3}{2}x + \frac{14}{6} = -\frac{3}{2}x + \frac{7}{3}$ Slopes equal $-\frac{3}{2}$, so lines are parallel. 2. Determine which pairs are perpendicular. (i) $y + 3x - 1 = 0$ and $3y = x + 1$ Rewrite: $y = -3x + 1$ $y = \frac{1}{3}x + \frac{1}{3}$ Slopes $-3$ and $\frac{1}{3}$ satisfy $m_1 \times m_2 = -1$, so perpendicular. (ii) $y = -x + 7$ and $y - x - 1 = 0$ Rewrite second: $y = x + 1$ Slopes $-1$ and $1$ product $-1$, so perpendicular. (iii) $y = 4x + 5$ and $4y + 6x - 1 = 0$ Rewrite second: $4y = -6x + 1 \Rightarrow y = -\frac{3}{2}x + \frac{1}{4}$ Slopes $4$ and $-\frac{3}{2}$ product $-6 \neq -1$, not perpendicular. (iv) $3y = 4x + 3$ and $4y + 3x = 20$ Rewrite: $y = \frac{4}{3}x + 1$ $4y = -3x + 20 \Rightarrow y = -\frac{3}{4}x + 5$ Slopes $\frac{4}{3}$ and $-\frac{3}{4}$ product $-1$, so perpendicular. 3. Equation of line through $(-4,3)$ parallel to $4y - 5x = 3$ Rewrite given line: $4y = 5x + 3 \Rightarrow y = \frac{5}{4}x + \frac{3}{4}$ Slope $m = \frac{5}{4}$ Use point-slope form: $y - 3 = \frac{5}{4}(x + 4)$ Simplify: $y = \frac{5}{4}x + 5 + 3 = \frac{5}{4}x + 8$ 4. Equation of line through $(3,-1)$ perpendicular to $4x - 2y + 1 = 0$ Rewrite given line: $-2y = -4x - 1 \Rightarrow y = 2x + \frac{1}{2}$ Slope $m = 2$ Perpendicular slope $m' = -\frac{1}{2}$ Use point-slope: $y + 1 = -\frac{1}{2}(x - 3)$ Simplify: $y = -\frac{1}{2}x + \frac{3}{2} - 1 = -\frac{1}{2}x + \frac{1}{2}$ 5. For what $k$ is line through $(2,k)$ and $(5,8)$ perpendicular to $4x - 5y + 11 = 0$? Slope of given line: $-5y = -4x - 11 \Rightarrow y = \frac{4}{5}x + \frac{11}{5}$ Slope $m = \frac{4}{5}$ Slope of line through points: $m' = \frac{8 - k}{5 - 2} = \frac{8 - k}{3}$ Perpendicular condition: $m \times m' = -1 \Rightarrow \frac{4}{5} \times \frac{8 - k}{3} = -1$ Solve: $\frac{4(8 - k)}{15} = -1 \Rightarrow 4(8 - k) = -15 \Rightarrow 32 - 4k = -15 \Rightarrow -4k = -47 \Rightarrow k = \frac{47}{4} = 11.75$ 6. For what $p$ is line through $(5,p)$ and $(7,11)$ parallel to $6x - 7y + 1 = 0$? Slope of given line: $-7y = -6x - 1 \Rightarrow y = \frac{6}{7}x + \frac{1}{7}$ Slope $m = \frac{6}{7}$ Slope of line through points: $m' = \frac{11 - p}{7 - 5} = \frac{11 - p}{2}$ Parallel condition: $m' = m \Rightarrow \frac{11 - p}{2} = \frac{6}{7}$ Solve: $11 - p = \frac{12}{7} \Rightarrow p = 11 - \frac{12}{7} = \frac{77 - 12}{7} = \frac{65}{7} \approx 9.29$ 7. Equation of line through intersection of $y = 3x + 1$ and $4y - 5x = 11$, parallel to $y = 6x - 15$ Find intersection: From first, $y = 3x + 1$ Substitute in second: $4(3x + 1) - 5x = 11 \Rightarrow 12x + 4 - 5x = 11 \Rightarrow 7x = 7 \Rightarrow x = 1$ $y = 3(1) + 1 = 4$ Point $(1,4)$ Slope of parallel line $m = 6$ Equation: $y - 4 = 6(x - 1) \Rightarrow y = 6x - 6 + 4 = 6x - 2$ 8. Equation of line through intersection of $y = 7x - 1$ and $y - 2x - 4 = 0$, perpendicular to $2y - 3x + 2 = 0$ Find intersection: Rewrite second: $y = 2x + 4$ Set equal: $7x - 1 = 2x + 4 \Rightarrow 5x = 5 \Rightarrow x = 1$ $y = 7(1) - 1 = 6$ Point $(1,6)$ Slope of given line: $2y = 3x - 2 \Rightarrow y = \frac{3}{2}x - 1$ Slope $m = \frac{3}{2}$ Perpendicular slope $m' = -\frac{2}{3}$ Equation: $y - 6 = -\frac{2}{3}(x - 1) \Rightarrow y = -\frac{2}{3}x + \frac{2}{3} + 6 = -\frac{2}{3}x + \frac{20}{3}$ 9. Find intersection of $y = 2x + 3$, $y + x - 6 = 0$, and $2y - 11x = -1$ From second: $y = 6 - x$ Set equal to first: $2x + 3 = 6 - x \Rightarrow 3x = 3 \Rightarrow x = 1$ $y = 2(1) + 3 = 5$ Check third: $2(5) - 11(1) = 10 - 11 = -1$ correct Intersection point $(1,5)$ 10. Equation of perpendicular bisector of $A(4,7)$ and $B(6,9)$ Midpoint: $M = \left(\frac{4+6}{2}, \frac{7+9}{2}\right) = (5,8)$ Slope of $AB$: $m = \frac{9 - 7}{6 - 4} = 1$ Slope of perpendicular bisector: $m' = -1$ Equation: $y - 8 = -1(x - 5) \Rightarrow y = -x + 5 + 8 = -x + 13$ 11. Intersection of $2y + x = 11$ and $y = x^2 - 3x + 4$ Rewrite first: $x = 11 - 2y$ Substitute $y$: $x = 11 - 2(x^2 - 3x + 4) = 11 - 2x^2 + 6x - 8 = 3 + 6x - 2x^2$ Rearranged: $2x^2 - 6x + x - 3 = 0 \Rightarrow 2x^2 - 5x - 3 = 0$ Solve quadratic: $x = \frac{5 \pm \sqrt{25 + 24}}{4} = \frac{5 \pm 7}{4}$ $x_1 = 3$, $x_2 = -\frac{1}{2}$ Find $y$: $y_1 = 3^2 - 3(3) + 4 = 9 - 9 + 4 = 4$ $y_2 = \left(-\frac{1}{2}\right)^2 - 3\left(-\frac{1}{2}\right) + 4 = \frac{1}{4} + \frac{3}{2} + 4 = \frac{1}{4} + \frac{3}{2} + 4 = \frac{1}{4} + \frac{3}{2} + 4 = \frac{1}{4} + 1.5 + 4 = 5.75$ Points: $(3,4)$ and $(-0.5, 5.75)$ Distance: $d = \sqrt{(3 + 0.5)^2 + (4 - 5.75)^2} = \sqrt{3.5^2 + (-1.75)^2} = \sqrt{12.25 + 3.0625} = \sqrt{15.3125} \approx 3.91$ 12. Intersection of $y - 5x + 3 = 0$ and $y = 2x^2 - 3x + 5$ Rewrite first: $y = 5x - 3$ Set equal: $5x - 3 = 2x^2 - 3x + 5$ Rearranged: $2x^2 - 8x + 8 = 0 \Rightarrow x^2 - 4x + 4 = 0$ $(x - 2)^2 = 0 \Rightarrow x = 2$ Find $y$: $y = 5(2) - 3 = 10 - 3 = 7$ Point of intersection: $(2,7)$ 13. Intersection of $3y + 4x = 12$ and $y = x^2 - 3x + 1$ Rewrite first: $3y = 12 - 4x \Rightarrow y = 4 - \frac{4}{3}x$ Set equal: $x^2 - 3x + 1 = 4 - \frac{4}{3}x$ Rearranged: $x^2 - 3x + 1 - 4 + \frac{4}{3}x = 0 \Rightarrow x^2 - \frac{5}{3}x - 3 = 0$ Multiply by 3: $3x^2 - 5x - 9 = 0$ Solve quadratic: $x = \frac{5 \pm \sqrt{25 + 108}}{6} = \frac{5 \pm \sqrt{133}}{6}$ Corresponding $y$ values from $y = 4 - \frac{4}{3}x$ 14. Using discriminant to check intersections: (i) $y - 2x = 13$, $y = 2x^2 - 4x + 5$ Rewrite first: $y = 2x + 13$ Set equal: $2x + 13 = 2x^2 - 4x + 5 \Rightarrow 2x^2 - 6x - 8 = 0$ Discriminant: $D = (-6)^2 - 4(2)(-8) = 36 + 64 = 100 > 0$ Two points of intersection. (ii) $y + 2x = 8$, $y = x^2 - 8x + 17$ Rewrite first: $y = 8 - 2x$ Set equal: $8 - 2x = x^2 - 8x + 17 \Rightarrow x^2 - 6x + 9 = 0$ Discriminant: $D = (-6)^2 - 4(1)(9) = 36 - 36 = 0$ One point of intersection. (iii) $3y - 2x + 6 = 0$, $y = x^2 + 6x + 10$ Rewrite first: $3y = 2x - 6 \Rightarrow y = \frac{2}{3}x - 2$ Set equal: $\frac{2}{3}x - 2 = x^2 + 6x + 10 \Rightarrow x^2 + \frac{16}{3}x + 12 = 0$ Discriminant: $D = \left(\frac{16}{3}\right)^2 - 4(1)(12) = \frac{256}{9} - 48 = \frac{256 - 432}{9} = -\frac{176}{9} < 0$ No real intersection. 15. Area of triangle $ABC$ with $A(8,1)$, $B(6,10)$, $C(4,8)$ Formula: $\text{Area} = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$ Calculate: $= \frac{1}{2} |8(10 - 8) + 6(8 - 1) + 4(1 - 10)| = \frac{1}{2} |8(2) + 6(7) + 4(-9)| = \frac{1}{2} |16 + 42 - 36| = \frac{1}{2} |22| = 11$ 16. Area of quadrilateral $ABCD$ with $A(-3,1)$, $B(3,3)$, $C(0,5)$, $D(-4,5)$ Use shoelace formula: Sum1 = $(-3)(3) + 3(5) + 0(5) + (-4)(1) = -9 + 15 + 0 - 4 = 2$ Sum2 = $1(3) + 3(0) + 5(-4) + 5(-3) = 3 + 0 - 20 - 15 = -32$ Area = $\frac{1}{2} |2 - (-32)| = \frac{1}{2} (34) = 17$ Final answers summarized: 1.(i) Parallel, (ii) Not parallel, (iii) Not parallel, (iv) Parallel 2.(i) Perpendicular, (ii) Perpendicular, (iii) Not perpendicular, (iv) Perpendicular 3. $y = \frac{5}{4}x + 8$ 4. $y = -\frac{1}{2}x + \frac{1}{2}$ 5. $k = \frac{47}{4}$ 6. $p = \frac{65}{7}$ 7. $y = 6x - 2$ 8. $y = -\frac{2}{3}x + \frac{20}{3}$ 9. Intersection at $(1,5)$ 10. $y = -x + 13$ 11. Points $(3,4)$ and $(-0.5,5.75)$, distance $\approx 3.91$ 12. Intersection at $(2,7)$ 13. Two points with $x = \frac{5 \pm \sqrt{133}}{6}$ 14.(i) Two intersections, (ii) One intersection, (iii) No intersection 15. Area = 11 16. Area = 17