Appendix A — Answer Key, Chapter 6: Writing Equations of Lines

SOL A.F.1 (d, e) · Covers textbook Chapter 6 and the companion workbook. Item numbers match the textbook; workbook items are the same problems, so this key serves both. Item numbers run continuously from 1 to 123 across the chapter. Reasoning answers show an acceptable response, not the only wording.

Conventions used in every answer below: point-slope form is $y - y_1 = m(x - x_1)$ and slope-intercept form is $y = mx + b$; unless an item says otherwise, the final equation is reported in slope-intercept form. A vertical line is $x = a$, has no slope, and is not a function; a horizontal line is $y = c$ and has slope $0$. Parallel lines have equal slopes; perpendicular lines have slopes whose product is $-1$. Every equation below has been checked by substituting the given point or points, and those checks are shown.

The figures used repeatedly in the chapter, for reference:

• Figure 1 is $y = 3x - 7$, built from the slope $3$ and the point $(2,-1)$, with $y$-intercept $(0,-7)$
• Figure 2 is $y = -2x - 2$, built from $(-3,4)$ and $(1,-4)$, with $y$-intercept $(0,-2)$
• Figure 3 is $y = \tfrac23 x - 2$, with the lattice points $(0,-2)$, $(3,0)$, and $(6,2)$
• Figure 4 is $y = -\tfrac32 x + 14$ on a window from $0$ to $10$, with the lattice points $(4,8)$ and $(6,5)$ and no visible $y$-intercept
• Figure 5 is the horizontal line $y = -3$ beside the vertical line $x = 4$
• Figure 6 is the plumber, $C(h) = 50h + 70$, through the billed jobs $(2,170)$ and $(5,320)$
• Figure 7 is $y = 2x - 3$ with the parallel line $y = 2x + 2$ through $(1,4)$
• Figure 8 is $y = 2x - 3$ with the perpendicular line $y = -\tfrac12 x + 3$ through $(4,1)$
• Figure 9 is $y = \tfrac32 x$ beside $y = -\tfrac23 x$, with their slope triangles
• Figure 10 is $y = \tfrac12 x + 1$ through $(-4,-1)$ and $(2,2)$

Lesson 6.1 — From a Slope and a Point

Guided practice

1. The slope is $m = 3$ and the point is $(x_1, y_1) = (2, -1)$. In $y - y_1 = m(x - x_1)$, the slope $3$ fills the $m$ slot, the $2$ fills the $x_1$ slot, and the $-1$ fills the $y_1$ slot.
2. $y - (-1) = 3(x - 2)$, which tidies to $y + 1 = 3(x - 2)$. The form subtracts $y_1$, and $y_1$ is itself $-1$; subtracting a negative is adding, so $y - (-1)$ becomes $y + 1$.
3. Distribute: $y + 1 = 3x - 6$. Subtract $1$: $y = 3x - 7$. The $y$-intercept is $(0, -7)$, which is marked in blue on the figure and was never mentioned in the problem.
4. Slope is rise over run, so a run of $1$ with a rise of $3$ gives $m = \tfrac31 = 3$. The triangle starts at the given point $(2,-1)$ and lands at $(3,2)$, which is back on the line — so the line really does climb $3$ units for every $1$ unit right.
5. Point-slope: $y - 5 = -4(x - 1)$. Distribute: $y - 5 = -4x + 4$. Add $5$: $y = -4x + 9$. Check: $-4(1) + 9 = 5$ ✓
6. Point-slope: $y - 2 = \tfrac12(x + 6)$, because $x - (-6) = x + 6$. Distribute: $y - 2 = \tfrac12 x + 3$. Add $2$: $y = \tfrac12 x + 5$. Check: $\tfrac12(-6) + 5 = -3 + 5 = 2$ ✓

Independent practice

7. a) $y - 1 = 2(x - 3)$, so $y = 2x - 5$. Check: $2(3) - 5 = 1$ ✓ b) $y - 4 = -3(x + 2)$, so $y - 4 = -3x - 6$ and $y = -3x - 2$. Check: $-3(-2) - 2 = 4$ ✓ c) $y + 1 = \tfrac23(x - 6)$, so $y + 1 = \tfrac23 x - 4$ and $y = \tfrac23 x - 5$. Check: $\tfrac23(6) - 5 = -1$ ✓ d) $y - 3 = -\tfrac14(x + 8)$, so $y - 3 = -\tfrac14 x - 2$ and $y = -\tfrac14 x + 1$. Check: $-\tfrac14(-8) + 1 = 3$ ✓
8. $y = -2$. A slope of $0$ means the output never changes, so every point on the line has $y$-coordinate $-2$: it is a horizontal line, and it is a function.
9. $y = -5x + 7$. The given point $(0,7)$ has $x$-coordinate $0$, so it is the $y$-intercept and $b = 7$ can be written down directly. Point-slope would give $y - 7 = -5(x - 0)$, which simplifies to the same thing with an extra step.
10. Slope-intercept form was built to display the $y$-intercept, so the only point it can accept is the one on the $y$-axis: $b$ has no meaning except as the output at $x = 0$. Point-slope form was built to display a slope and a point, and its two slots $x_1$ and $y_1$ accept the coordinates of any point at all. Since a slope plus any point on a line determines that line completely, point-slope form never has to wait for the special point.
11. $y + 6 = 4(x + 1)$, so $y + 6 = 4x + 4$ and $y = 4x - 2$. Check: $4(-1) - 2 = -6$ ✓
12. The error is the sign of $y_1$. The point is $(2,-1)$, so $y_1 = -1$ and $y - y_1$ is $y - (-1) = y + 1$, not $y - 1$. Correct equation: $y + 1 = 3(x-2)$, so $y = 3x - 7$. The check catches the student's version at once: substituting $x = 2$ into $y = 3x - 5$ gives $3(2) - 5 = 1$, but the point says the output there must be $-1$. The student's line is parallel to the right one and two units too high.
13. The rate is the slope, $m = -12$ gallons per minute, negative because the water is leaving. The measurement is the point $(4, 200)$. Then $W - 200 = -12(t - 4)$, so $W - 200 = -12t + 48$ and $W(t) = -12t + 248$. Check: $-12(4) + 248 = 200$ ✓ The slope $-12$ means the pool loses $12$ gallons every minute. The $y$-intercept $(0,248)$ means the pool held $248$ gallons at the moment draining began — a number the story never stated.
14. The rate is $m = 35$ dollars per month and the point is $(6, 260)$. Then $C - 260 = 35(m - 6)$, so $C - 260 = 35m - 210$ and $C(m) = 35m + 50$. Check: $35(6) + 50 = 260$ ✓ The $y$-intercept $(0,50)$ is the amount already paid at zero months of membership, so it cannot be a monthly charge — nothing monthly has been billed yet. It is a one-time joining fee of $\$50$.
15. Graphing $y = 2x - 5$ and tracing to $x = 3$ gives $y = 1$, so the point $(3,1)$ lies on the line, and the graph agrees with the algebra. If the point did not lie on the graph, the equation and the point would be describing different lines, and one of the two steps is wrong: check the substitution first, since it is exact arithmetic, then re-enter the equation in case it was typed wrong. A disagreement is not something to average out or ignore — it means the work is not finished.
16. $y + 4 = -\tfrac35(x - 10)$, so $y + 4 = -\tfrac35 x + 6$ and $y = -\tfrac35 x + 2$. Check: $-\tfrac35(10) + 2 = -6 + 2 = -4$ ✓

Exit ticket 6.1

17. $y - 5 = 6(x - 2)$, so $y = 6x - 7$. Check: $6(2) - 7 = 5$ ✓
18. $y - 1 = -\tfrac12(x + 4)$, so $y - 1 = -\tfrac12 x - 2$ and $y = -\tfrac12 x - 1$. Check: $-\tfrac12(-4) - 1 = 1$ ✓
19. Point-slope: $y + 1 = 3(x - 2)$, which displays the given point $(2,-1)$. Slope-intercept: $y = 3x - 7$, which displays the $y$-intercept $(0,-7)$. Both describe the same line; they advertise different points on it.
20. Substituting the given point tests the finished equation against the one fact the problem guaranteed, and it catches the sign errors that this lesson's two traps produce — a wrong sign on $x_1$ or $y_1$ changes the constant term but leaves the equation looking perfectly reasonable. It costs one line of arithmetic and it is the only step that can tell you your answer is wrong.

Lesson 6.2 — From Two Points

Guided practice

21. The marked points are $(-3, 4)$ and $(1, -4)$. $m = \dfrac{-4 - 4}{1 - (-3)} = \dfrac{-8}{4} = -2$. The dashed path shows the same computation as a picture: a run of $4$ and a rise of $-8$.
22. Using $(1,-4)$: $y + 4 = -2(x - 1)$, so $y + 4 = -2x + 2$ and $y = -2x - 2$. (Using $(-3,4)$ instead: $y - 4 = -2(x+3)$, so $y - 4 = -2x - 6$ and $y = -2x - 2$ — the same equation.)
23. Checking with $(-3,4)$, the point not used above: $-2(-3) - 2 = 6 - 2 = 4$, which matches the point's $y$-coordinate ✓ This is the strongest check available, because the point being tested played no part in building the equation.
24. $m = \dfrac{13 - 5}{4 - 0} = \dfrac84 = 2$, and the point $(0,5)$ has $x$-coordinate $0$, so it is the $y$-intercept and $b = 5$. The equation is $y = 2x + 5$. No substitution was needed because one of the two given points was already the special point slope-intercept form asks for.
25. $m = \dfrac{-3 - 7}{3 - (-2)} = \dfrac{-10}{5} = -2$. Using $(-2,7)$: $y - 7 = -2(x + 2)$, so $y = -2x + 3$. Check with $(3,-3)$: $-2(3) + 3 = -3$ ✓
26. The figure shows a substitution check — $\tfrac12(-4) + 1 = -1$ and $\tfrac12(2) + 1 = 2$ — and a graphical check, that both points land on the drawn line. If they disagreed, trust the substitution: it is exact arithmetic, while a graph can be misread, mis-scaled, or drawn from an equation that was typed in wrong.

Independent practice

27. a) $m = \dfrac{8-2}{3-1} = 3$; $y - 2 = 3(x-1)$, so $y = 3x - 1$. Check with $(3,8)$: $3(3) - 1 = 8$ ✓ b) $m = \dfrac{4-1}{2-(-4)} = \dfrac36 = \tfrac12$; $y - 4 = \tfrac12(x - 2)$, so $y = \tfrac12 x + 3$. Check with $(-4,1)$: $\tfrac12(-4) + 3 = 1$ ✓ c) $m = \dfrac{4-(-5)}{2-(-1)} = \dfrac93 = 3$; $y - 4 = 3(x - 2)$, so $y = 3x - 2$. Check with $(-1,-5)$: $3(-1) - 2 = -5$ ✓ d) $m = \dfrac{10-(-2)}{-1-5} = \dfrac{12}{-6} = -2$; $y + 2 = -2(x - 5)$, so $y = -2x + 8$. Check with $(-1,10)$: $-2(-1) + 8 = 10$ ✓
28. $m = \dfrac{-2 - 6}{3 - (-3)} = \dfrac{-8}{6} = -\dfrac43$. Using $(3,-2)$: $y + 2 = -\tfrac43(x - 3)$, so $y + 2 = -\tfrac43 x + 4$ and $y = -\tfrac43 x + 2$. Checks: at $x = 3$, $-4 + 2 = -2$ ✓; at $x = -3$, $4 + 2 = 6$ ✓
29. $m = \dfrac{7 - 7}{5 - 2} = \dfrac03 = 0$. A slope of $0$ makes the line horizontal, and every point on it has $y$-coordinate $7$, so the equation is $y = 7$. It is a function.
30. $m = \dfrac{6 - 1}{-4 - (-4)} = \dfrac{5}{0}$, which is undefined — the run is zero, and a rise cannot be divided by zero. The line is vertical, its equation is $x = -4$, and it is not a function, since the single input $-4$ is paired with every output at once.
31. $m = \dfrac{320 - 170}{5 - 2} = \dfrac{150}{3} = 50$. Using $(2,170)$: $C - 170 = 50(h - 2)$, so $C(h) = 50h + 70$. Check with $(5,320)$: $50(5) + 70 = 320$ ✓ The slope $50$ is the hourly rate, $\$50$ per hour of work. The $y$-intercept $(0,70)$ is the charge for a job of zero hours: a $\$70$ service fee for showing up, owed before any work is done.
32. The points are $(1, 10)$ and $(4, 4)$. $m = \dfrac{4 - 10}{4 - 1} = \dfrac{-6}{3} = -2$. Using $(1,10)$: $H - 10 = -2(t - 1)$, so $H(t) = -2t + 12$. Check with $(4,4)$: $-2(4) + 12 = 4$ ✓ The slope $-2$ means the candle loses $2$ inches of height every hour. The $y$-intercept $(0,12)$ means the candle was $12$ inches tall when it was lit.
33. The points are $(3, 5)$ and $(7, 13)$. $m = \dfrac{13-5}{7-3} = \dfrac84 = 2$. Using $(3,5)$: $h - 5 = 2(a - 3)$, so $h(a) = 2a - 1$. Check with $(7,13)$: $2(7) - 1 = 13$ ✓ The slope $2$ means the tree grows $2$ feet per year. The $y$-intercept $(0,-1)$ would say the tree was $-1$ foot tall when it was planted, which is not a fact about any tree — the model is only trustworthy across the ages where growth actually was steady, and extending it back to age $0$ leaves that range.
34. Graphing $y = -2x + 8$ shows the line passing through both $(5,-2)$ and $(-1,10)$, confirming the algebra. If the graph passed through only one of the two points, the slope is the suspect: a line through one given point with the wrong slope will miss the other. If it missed both, the constant term is the suspect, since a wrong $b$ shifts the whole line off without changing its steepness.
35. From $(-2,7)$: $y - 7 = -2(x + 2)$, so $y - 7 = -2x - 4$ and $y = -2x + 3$. From $(3,-3)$: $y + 3 = -2(x - 3)$, so $y + 3 = -2x + 6$ and $y = -2x + 3$. They are identical. This had to happen because both points lie on the same line, and a slope together with any one point of a line determines that line completely — there is only one line with slope $-2$ through $(-2,7)$, and it is the same one that has slope $-2$ through $(3,-3)$.
36. The student inverted the slope formula, dividing the change in $x$ by the change in $y$ instead of the other way around. Slope is $\dfrac{\text{change in } y}{\text{change in } x}$, so $m = \dfrac{8-2}{3-1} = \dfrac62 = 3$, and the equation is $y = 3x - 1$. The student's $\tfrac13$ would describe a much flatter line: it fails the check, since $\tfrac13(3) + b$ cannot give $8$ and $\tfrac13(1) + b$ give $2$ with the same $b$.

Exit ticket 6.2

37. $m = \dfrac{1 - (-3)}{2 - 0} = \dfrac42 = 2$, and $(0,-3)$ is the $y$-intercept, so $y = 2x - 3$. Check with $(2,1)$: $2(2) - 3 = 1$ ✓
38. $m = \dfrac{-4 - 2}{1 - (-5)} = \dfrac{-6}{6} = -1$. Using $(1,-4)$: $y + 4 = -(x - 1)$, so $y = -x - 3$. Check with $(-5,2)$: $5 - 3 = 2$ ✓
39. The $x$-coordinates are equal, so the run is $0$ and the slope is undefined — the line has no slope. Its equation is $x = 4$, and it is not a function.
40. The points are $(3, 11)$ and $(7, 19)$. $m = \dfrac{19 - 11}{7 - 3} = \dfrac84 = 2$. Using $(3,11)$: $C - 11 = 2(d - 3)$, so $C(d) = 2d + 5$. Check with $(7,19)$: $2(7) + 5 = 19$ ✓ The slope $2$ is the rate, $\$2$ per mile. The $y$-intercept $(0,5)$ is a fixed $\$5$ charged for a ride of zero miles — the fee for the pickup itself.

Lesson 6.3 — From a Graph

Guided practice

41. The line crosses the $y$-axis at $(0, -2)$, which gives $b = -2$ directly, with no computation.
42. The triangle runs from $(0,-2)$ right $3$ to $(3,-2)$ and up $2$ to $(3,0)$, so $m = \dfrac{\text{rise}}{\text{run}} = \dfrac23$. It runs between the lattice points $(0,-2)$ and $(3,0)$.
43. $y = \tfrac23 x - 2$.
44. The window does not include the place where the line crosses the $y$-axis — the line leaves the top of the picture first — so there is no crossing point to read, and the first step, "find $b$," has nothing to look at. The $y$-intercept turns out to be $(0,14)$, far above the drawn grid.
45. The marked lattice points are $(4, 8)$ and $(6, 5)$. $m = \dfrac{5 - 8}{6 - 4} = \dfrac{-3}{2} = -\dfrac32$.
46. $y - 8 = -\tfrac32(x - 4)$, so $y - 8 = -\tfrac32 x + 6$ and $y = -\tfrac32 x + 14$. The $y$-intercept predicted by that answer is $(0, 14)$, which is off the top of the drawn window — exactly why it could not be read.

Independent practice

47. Step 1: find $b$ by locating where the line crosses the $y$-axis. Step 2: find $m$ by counting rise over run between two lattice points. Step 3: write $y = mx + b$ and check against a third point on the graph. Step 1 fails when the $y$-axis crossing is off the drawn window; then you take any lattice point on the line together with the slope and use point-slope form instead.
48. $b = 4$ and $m = \dfrac{6-4}{3-0} = \dfrac23$, so $y = \tfrac23 x + 4$. Check: $\tfrac23(3) + 4 = 6$ ✓
49. $b = -1$ and $m = \dfrac{-7-(-1)}{2-0} = \dfrac{-6}{2} = -3$, so $y = -3x - 1$. Check: $-3(2) - 1 = -7$ ✓
50. $m = \dfrac{2 - 5}{2 - (-4)} = \dfrac{-3}{6} = -\dfrac12$. Using $(2,2)$: $y - 2 = -\tfrac12(x - 2)$, so $y = -\tfrac12 x + 3$. Check with $(-4,5)$: $-\tfrac12(-4) + 3 = 5$ ✓
51. $\tfrac23(6) - 2 = 4 - 2 = 2$, which is the $y$-coordinate of $(6,2)$ ✓ The point was not used to build the equation, so it is a genuine test of the whole answer.
52. $-\tfrac32(6) + 14 = -9 + 14 = 5$, which is the $y$-coordinate of $(6,5)$ ✓
53. $m = \dfrac{-4}{1} = -4$ and $b = 2$, so $y = -4x + 2$.
54. A lattice point sits exactly on a grid corner, so both of its coordinates are known exactly and the rise and the run are whole numbers you can count. A point read off the middle of a square is an estimate in both coordinates, and a slope built from two estimates is an estimate — one that is usually a little wrong in a way that produces an equation missing the drawn line entirely a few units away.
55. The student inverted rise over run, reporting $\tfrac{\text{run}}{\text{rise}} = \tfrac32$ instead of $\tfrac{\text{rise}}{\text{run}} = \tfrac23$. Correct equation: $y = \tfrac23 x - 2$. The marked point $(6,2)$ shows the student's line is wrong: $\tfrac32(6) - 2 = 7$, but the graph passes through $(6,2)$, not $(6,7)$.
56. Graph $y = \tfrac23 x - 2$ and check two features against the figure: the $y$-axis crossing, which should be at $(0,-2)$ in both, and a second lattice point, such as $(3,0)$ — both pictures should pass through it. Matching one feature is not enough, since a line with the right intercept and the wrong slope still crosses correctly at one point.

Exit ticket 6.3

57. $b = 5$ and $m = \dfrac{3-5}{4-0} = \dfrac{-2}{4} = -\dfrac12$, so $y = -\tfrac12 x + 5$. Check: $-\tfrac12(4) + 5 = 3$ ✓
58. $m = \dfrac{-1 - (-5)}{2 - (-6)} = \dfrac{4}{8} = \dfrac12$. Using $(2,-1)$: $y + 1 = \tfrac12(x - 2)$, so $y = \tfrac12 x - 2$. Check with $(-6,-5)$: $\tfrac12(-6) - 2 = -5$ ✓
59. $y = -\tfrac32 x + 14$, with $y$-intercept $(0, 14)$.
60. You can read $b$ straight off the graph whenever the drawn window includes the point where the line crosses the $y$-axis. When it does not, read any two lattice points instead, compute the slope from them, and use point-slope form with either point — the $y$-intercept then falls out of the algebra rather than being read.

Lesson 6.4 — Horizontal and Vertical Lines

Guided practice

61. The blue points $(-4,-3)$ and $(3,-3)$ share their $y$-coordinate, $-3$. Every point on the line has $y$-coordinate $-3$, so the equation is $y = -3$, a horizontal line with slope $0$.
62. The red points $(4,-1)$ and $(4,5)$ share their $x$-coordinate, $4$. Every point on the line has $x$-coordinate $4$, so the equation is $x = 4$, a vertical line with no slope.
63. The horizontal line $y = -3$ is a function: every input has exactly one output, which happens to be $-3$ every time. The vertical line $x = 4$ is not a function, because the single input $4$ is paired with every output at once — the vertical line test fails against the line itself.
64. $y = -6$. A horizontal line fixes the $y$-coordinate of the given point.
65. $x = 7$. A vertical line fixes the $x$-coordinate of the given point.
66. Slope-intercept form has a slot for $m$, and Chapter 5 established that a vertical line has no slope: the run between any two of its points is $0$, and a rise cannot be divided by $0$. With no number to put in the $m$ slot, no equation of the form $y = mx + b$ can describe it. In standard form it is $1x + 0y = 7$.

Independent practice

67. a) $y = 9$ b) $x = -5$ c) The $y$-coordinates match, so $y = 4$ d) The $x$-coordinates match, so $x = 3$
68. a) slope $0$ b) no slope c) slope $0$ d) no slope
69. The functions are a and c, the two horizontal lines. Test: a line is a function when no input is paired with more than one output, which is what the vertical line test checks — and a vertical line fails it against itself.
70. The horizontal line through $(0,-7)$ is $y = -7$, and it crosses the $y$-axis at that point. The vertical line through $(5,0)$ is $x = 5$, and it crosses the $x$-axis at that point.
71. $T(h) = 68$, a horizontal line. The slope is $0$ degrees per hour: the temperature is not changing as the hours pass. The graph is a flat line at height $68$ across the whole day.
72. The wall is $x = 12$. It is not a function of $x$ because the single input $12$ is paired with every $y$-value along the wall at once — asking "what is the output when $x = 12$?" has infinitely many answers. That is not a defect in the wall; a boundary simply is not the kind of object $f(x)$ describes.
73. Two mistakes at once. First, the student used the wrong letter: a horizontal line is a statement about height, so it is $y = \ldots$, not $x = \ldots$. Second, the student took the wrong coordinate: even for a vertical line the answer would have used $4$, the $x$-coordinate, not $-1$. The correct answer is $y = -1$. (The student's $x = -1$ is a vertical line through a point the problem never mentioned.)
74. $y = -3$ graphs immediately, as a flat line three units below the $x$-axis. $x = 4$ is refused by most graphing calculators, because the entry line accepts only equations of the form $y = \ldots$ — that is, only functions. The refusal is the tool restating the mathematics: a vertical line is not a function, so a function grapher has no way to accept it. If your tool has a separate vertical-line or relation command, use it; otherwise plot two points with the same $x$-coordinate and draw the line yourself.

Exit ticket 6.4

75. $y = 5$
76. $x = -2$
77. The $y$-coordinates are both $-4$, so the line is $y = -4$, with slope $0$.
78. $x = a$ is a vertical line: every point on it has $x$-coordinate $a$, it has no slope, and it is not a function. $y = c$ is a horizontal line: every point on it has $y$-coordinate $c$, its slope is $0$, and it is a function. The quickest way to keep them straight is that a horizontal line is a statement about height, and height is $y$.

Lesson 6.5 — Parallel and Perpendicular Lines

Guided practice

79. The given black line $y = 2x - 3$ has slope $2$, and the blue line $y = 2x + 2$ also has slope $2$. They must be equal: two lines that never meet have to climb at the same rate, since if one were steeper it would eventually catch and cross the other.
80. Slope $2$ through $(1,4)$: $y - 4 = 2(x - 1)$, so $y - 4 = 2x - 2$ and $y = 2x + 2$. Check: $2(1) + 2 = 4$ ✓
81. The blue line is $y = -\tfrac12 x + 3$, with slope $-\tfrac12$. The product with the given slope is $2 \times \left(-\tfrac12\right) = -1$ ✓, which is the perpendicular condition.
82. Slope $-\tfrac12$ through $(4,1)$: $y - 1 = -\tfrac12(x - 4)$, so $y - 1 = -\tfrac12 x + 2$ and $y = -\tfrac12 x + 3$. Check: $-\tfrac12(4) + 3 = 1$ ✓
83. The slopes are $\tfrac32$ and $-\tfrac23$, and $\tfrac32 \times \left(-\tfrac23\right) = -\tfrac{6}{6} = -1$ ✓
84. The first triangle is run $2$, rise $3$. Turning it a quarter turn makes the leg that ran across stand up and the leg that stood up lie across, so the second triangle is run $3$, rise $2$ — the two numbers have traded places, which is the "flip the fraction" part. The quarter turn also carries one of the legs in the opposite direction, so one of the two signs reverses and the rise becomes $-2$ — that is the "change the sign" part. The result, $-\tfrac23$, is the negative reciprocal of $\tfrac32$.

Independent practice

85. Parallel means the same slope, $m = 3$: $y - 4 = 3(x - 2)$, so $y = 3x - 2$. Check: $3(2) - 2 = 4$ ✓
86. Same slope, $m = -\tfrac34$: $y + 2 = -\tfrac34(x - 8)$, so $y + 2 = -\tfrac34 x + 6$ and $y = -\tfrac34 x + 4$. Check: $-\tfrac34(8) + 4 = -6 + 4 = -2$ ✓
87. The negative reciprocal of $4$ is $-\tfrac14$: $y - 3 = -\tfrac14(x - 8)$, so $y - 3 = -\tfrac14 x + 2$ and $y = -\tfrac14 x + 5$. Check: $-\tfrac14(8) + 5 = 3$ ✓ Product check: $4 \times \left(-\tfrac14\right) = -1$ ✓
88. The negative reciprocal of $-\tfrac23$ is $\tfrac32$: $y - 0 = \tfrac32(x + 4)$, so $y = \tfrac32 x + 6$. Check: $\tfrac32(-4) + 6 = 0$ ✓ Product check: $-\tfrac23 \times \tfrac32 = -1$ ✓
89. Convert first: $2x + y = 6$ gives $y = -2x + 6$, so the slope is $-2$. Parallel means the same slope: $y - 3 = -2(x + 1)$, so $y - 3 = -2x - 2$ and $y = -2x + 1$. Check: $-2(-1) + 1 = 3$ ✓
90. Convert first: $x - 3y = 9$ gives $-3y = -x + 9$, so $y = \tfrac13 x - 3$ and the slope is $\tfrac13$. The negative reciprocal is $-3$: $y + 5 = -3(x - 2)$, so $y + 5 = -3x + 6$ and $y = -3x + 1$. Check: $-3(2) + 1 = -5$ ✓
91. Parallel to the horizontal line $y = 2$ means another horizontal line, and through $(5,-8)$ that is $y = -8$. Perpendicular to a horizontal line means a vertical line, and through $(5,-8)$ that is $x = 5$.
92. Parallel to the vertical line $x = -3$ means another vertical line, and through $(7,1)$ that is $x = 7$. Perpendicular to a vertical line means a horizontal line, and through $(7,1)$ that is $y = 1$.
93. No. Their slopes are $5$ and $-5$, and $5 \times (-5) = -25$, not $-1$. The two lines are reflections of each other across a horizontal line — equally steep, opposite directions — and that is a different relationship from perpendicular. A line perpendicular to $y = 5x - 2$ would have slope $-\tfrac15$.
94. The given slope is $\tfrac12$. Parallel street: same slope through $(6,1)$: $y - 1 = \tfrac12(x - 6)$, so $y - 1 = \tfrac12 x - 3$ and $y = \tfrac12 x - 2$. Check: $\tfrac12(6) - 2 = 1$ ✓ Access road: the negative reciprocal of $\tfrac12$ is $-2$, so $y - 1 = -2(x - 6)$, giving $y = -2x + 13$. Check: $-2(6) + 13 = 1$ ✓ Product check: $\tfrac12 \times (-2) = -1$ ✓
95. On a square window, $y = -\tfrac23 x + 5$ and $y = \tfrac32 x + 6$ cross at a visible right angle, and the second line passes through $(-4,0)$ as required. On a window that is wider than it is tall, the picture is stretched horizontally, which flattens every slope on screen by the same factor but does not preserve angles — so a genuine right angle can appear noticeably wide or narrow, and a correct answer can look wrong. The product of the slopes, $-\tfrac23 \times \tfrac32 = -1$, does not depend on the window at all, so that is the check to trust.
96. The student changed the sign but did not flip the fraction; "negative reciprocal" requires both. The reciprocal of $\tfrac25$ is $\tfrac52$, and with the sign changed the perpendicular slope is $-\tfrac52$. Verify: $\tfrac25 \times \left(-\tfrac52\right) = -1$ ✓ The student's slope fails the same test: $\tfrac25 \times \left(-\tfrac25\right) = -\tfrac{4}{25}$, which is not $-1$.

Exit ticket 6.5

97. Same slope, $m = -4$: $y + 3 = -4(x - 1)$, so $y + 3 = -4x + 4$ and $y = -4x + 1$. Check: $-4(1) + 1 = -3$ ✓
98. The negative reciprocal of $\tfrac12$ is $-2$: $y - 4 = -2(x - 3)$, so $y = -2x + 10$. Check: $-2(3) + 10 = 4$ ✓
99. $x = 8$ is vertical, so a line parallel to it is also vertical, and through $(-2,6)$ that line is $x = -2$.
100. Parallel: two lines are parallel exactly when their slopes are equal, so copy the given slope and use the given point to pick out which parallel line is wanted. Perpendicular: two lines are perpendicular exactly when the product of their slopes is $-1$, so flip the given slope over and change its sign. The pair the product rule cannot be used on is a vertical line with a horizontal line: they meet at a right angle, but $x = a$ has no slope, so there is nothing to multiply.

Chapter 6 Review

Part A — From a slope and a point

101. $y - 3 = -2(x - 4)$, so $y - 3 = -2x + 8$ and $y = -2x + 11$. Check: $-2(4) + 11 = 3$ ✓
102. $y - 2 = \tfrac35(x + 5)$, so $y - 2 = \tfrac35 x + 3$ and $y = \tfrac35 x + 5$. Check: $\tfrac35(-5) + 5 = -3 + 5 = 2$ ✓
103. The rate is $m = 8$ meters per second and the measurement is the point $(4, 50)$. Then $h - 50 = 8(t - 4)$, so $h - 50 = 8t - 32$ and $h(t) = 8t + 18$. Check: $8(4) + 18 = 50$ ✓ The slope $8$ means the drone gains $8$ meters of altitude every second. The $y$-intercept $(0,18)$ means it was already $18$ meters above the ground when the climb began — it launched from a rooftop or a hill rather than from the ground.
104. The error is the sign of $x_1$. The point is $(-2,5)$, so $x_1 = -2$ and $x - x_1$ is $x - (-2) = x + 2$, not $x - 2$. Correct: $y - 5 = -3(x + 2)$, so $y - 5 = -3x - 6$ and $y = -3x - 1$. Check: $-3(-2) - 1 = 6 - 1 = 5$ ✓ (The student's version simplifies to $y = -3x + 11$, which gives $17$ at $x = -2$, not $5$.)

Part B — From two points

105. $m = \dfrac{5 - 1}{2 - (-6)} = \dfrac48 = \dfrac12$. Using $(2,5)$: $y - 5 = \tfrac12(x - 2)$, so $y = \tfrac12 x + 4$. Check with $(-6,1)$: $\tfrac12(-6) + 4 = 1$ ✓
106. $m = \dfrac{8 - (-4)}{-3 - 3} = \dfrac{12}{-6} = -2$. Using $(3,-4)$: $y + 4 = -2(x - 3)$, so $y = -2x + 2$. Check with $(-3,8)$: $-2(-3) + 2 = 8$ ✓
107. From $(2,170)$ and $(5,320)$: $m = \dfrac{150}{3} = 50$, and $C - 170 = 50(h-2)$ gives $C(h) = 50h + 70$. Check: $50(5) + 70 = 320$ ✓ The slope $50$ is $\$50$ per hour of work. The $y$-intercept $(0,70)$ is a $\$70$ service fee charged for a job of zero hours — the cost of the visit itself.
108. The points are $(100, 45)$ and $(300, 95)$. $m = \dfrac{95 - 45}{300 - 100} = \dfrac{50}{200} = 0.25$. Using $(100,45)$: $C - 45 = 0.25(n - 100)$, so $C - 45 = 0.25n - 25$ and $C(n) = 0.25n + 20$. Check with $(300,95)$: $0.25(300) + 20 = 95$ ✓ The slope $0.25$ is $\$0.25$ per flyer. The $y$-intercept $(0,20)$ is a $\$20$ setup fee, charged before a single flyer is printed.