Math 1 Quiz: Writing Linear Equations
13 questions · exam conditions
0:00
Writing Linear EquationsQuestion 1 of 13

A line passes through points (−2,7)(-2, 7) and (1,−5)(1, -5). What is the equation of the line that is parallel to this line and passes through the origin?

y=−4x+3y = -4x + 3
y=−4xy = -4x
y=4xy = 4x
y=−14xy = -\frac{1}{4}x
← Back to quizzes

Math 1 Quiz

Math 1 Quiz: Writing Linear Equations

Practice Writing Linear Equations in Math 1 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Writing Linear Equations, giving you a quick way to practice the rules, question types, and explanations that matter most for Math 1.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

A line passes through points (−2,7)(-2, 7) and (1,−5)(1, -5). What is the equation of the line that is parallel to this line and passes through the origin?

  1. y=−4x+3y = -4x + 3
  2. y=−4xy = -4x (correct answer)
  3. y=4xy = 4x
  4. y=−14xy = -\frac{1}{4}x
Explanation: When you encounter parallel line problems, remember that parallel lines have identical slopes but different y-intercepts. Your first step is always to find the slope of the given line. To find the slope of the line through (−2,7)(-2, 7) and (1,−5)(1, -5), use the slope formula: m=y2−y1x2−x1=−5−71−(−2)=−123=−4m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-5 - 7}{1 - (-2)} = \frac{-12}{3} = -4. Since parallel lines have the same slope, your target line also has slope −4-4. The line you're looking for passes through the origin (0,0)(0, 0) with slope −4-4. Using point-slope form: y−0=−4(x−0)y - 0 = -4(x - 0), which simplifies to y=−4xy = -4x. This confirms that choice B is correct. Let's examine why the other answers are wrong. Choice A gives y=−4x+3y = -4x + 3, which has the correct slope but the wrong y-intercept—this line doesn't pass through the origin since when x=0x = 0, y=3y = 3. Choice C offers y=4xy = 4x, which passes through the origin but has slope 44 instead of −4-4, making it perpendicular to the original line, not parallel. Choice D presents y=−14xy = -\frac{1}{4}x, which passes through the origin but has the wrong slope—this would actually be perpendicular to the original line since (−4)×(−14)=1(-4) \times (-\frac{1}{4}) = 1. Remember: for parallel line problems, match the slope exactly, then use the given point to find the y-intercept. Lines through the origin always have y-intercept zero.

Question 2

Line mm is perpendicular to the line 3x+4y=83x + 4y = 8 and passes through the point where this line intersects the y-axis. What is the equation of line mm?

  1. y=43x−2y = \frac{4}{3}x - 2
  2. y=−34x+2y = -\frac{3}{4}x + 2
  3. y=43x+2y = \frac{4}{3}x + 2 (correct answer)
  4. y=−43x+2y = -\frac{4}{3}x + 2
Explanation: When you see a question about perpendicular lines, you need two key pieces of information: the slope relationship between perpendicular lines, and a point the new line passes through. First, find the slope of the given line 3x+4y=83x + 4y = 8. Convert to slope-intercept form by solving for yy: 4y=−3x+84y = -3x + 8 y=−34x+2y = -\frac{3}{4}x + 2 The slope is −34-\frac{3}{4}. Since perpendicular lines have slopes that are negative reciprocals of each other, line mm has slope 43\frac{4}{3}. Next, find where the original line intersects the y-axis. From the equation y=−34x+2y = -\frac{3}{4}x + 2, when x=0x = 0, y=2y = 2. So the y-intercept is (0,2)(0, 2). Since line mm passes through (0,2)(0, 2) with slope 43\frac{4}{3}, its equation is y=43x+2y = \frac{4}{3}x + 2. This matches choice C. Choice A has the correct slope but wrong y-intercept (−2-2 instead of 22). Choice B uses the original line's slope instead of the perpendicular slope. Choice D has the wrong slope (−43-\frac{4}{3} instead of 43\frac{4}{3}) — this would be perpendicular to a line with slope 34\frac{3}{4}, not −34-\frac{3}{4}. Study tip: Always remember that perpendicular slopes multiply to −1-1. When finding a perpendicular slope, flip the fraction and change the sign. Also, carefully track your positive and negative signs throughout the problem.

Question 3

A car rental company charges a base fee plus a rate per mile driven. After driving 50 miles, the total cost is $45. After driving 120 miles, the total cost is $80.

Which equation represents the total cost CC (in dollars) for driving mm miles?

  1. C=1.4m+15C = 1.4m + 15
  2. C=0.5m+45C = 0.5m + 45
  3. C=35m+0.5C = 35m + 0.5
  4. C=0.5m+20C = 0.5m + 20 (correct answer)
Explanation: This problem tests your ability to create a linear equation from two data points, which is a fundamental skill in algebra. When you see a scenario involving a base fee plus a rate per mile (or per unit), you're dealing with a linear relationship where the total cost equals the fixed cost plus the variable cost. To find the equation, you need to determine both the rate per mile (slope) and the base fee (y-intercept). Using the two given points: (50, 45) and (120, 80), calculate the slope: slope=80−45120−50=3570=0.5\text{slope} = \frac{80 - 45}{120 - 50} = \frac{35}{70} = 0.5 dollars per mile. Now use either point to find the base fee. With (50, 45): 45=0.5(50)+b45 = 0.5(50) + b, so 45=25+b45 = 25 + b, which gives b=20b = 20. Therefore, C=0.5m+20C = 0.5m + 20. Choice A (C=1.4m+15C = 1.4m + 15) uses an incorrect slope calculation—likely from dividing incorrectly or mixing up the coordinate values. Choice B (C=0.5m+45C = 0.5m + 45) correctly identifies the slope but mistakenly uses one of the total costs (45) as the base fee instead of calculating the y-intercept. Choice C (C=35m+0.5C = 35m + 0.5) appears to use the difference in costs (35) as the slope and the correct slope (0.5) as the y-intercept—a complete reversal of the correct values. When solving linear word problems, always identify your two points clearly, calculate slope carefully using the slope formula, then substitute back to find the y-intercept. Double-check by testing both original points in your final equation.

Question 4

A line has equation y=mx+by = mx + b where mm and bb are constants. If the line passes through (−2,7)(-2, 7) and has the same slope as the line connecting (0,3)(0, 3) and (4,11)(4, 11), what is the value of m+bm + b?

  1. 1111
  2. 99
  3. 1313 (correct answer)
  4. 1515
Explanation: When you encounter a linear equation problem involving a point and slope information, you need to find both the slope mm and y-intercept bb to determine the line's equation. First, find the slope using the two given points (0,3)(0, 3) and (4,11)(4, 11). The slope formula is m=y2−y1x2−x1m = \frac{y_2 - y_1}{x_2 - x_1}, so m=11−34−0=84=2m = \frac{11 - 3}{4 - 0} = \frac{8}{4} = 2. Now you know the line has slope m=2m = 2 and passes through (−2,7)(-2, 7). Substitute these values into y=mx+by = mx + b: 7=2(−2)+b7 = 2(-2) + b, which gives 7=−4+b7 = -4 + b, so b=11b = 11. Therefore, m+b=2+11=13m + b = 2 + 11 = 13. Looking at the wrong answers: Choice A (1111) is just the value of bb alone—you might get this if you forgot to add mm. Choice B (99) could result from a sign error when solving for bb, perhaps calculating 7=−4+b7 = -4 + b as b=7−4=3b = 7 - 4 = 3, then getting m+b=2+3=5m + b = 2 + 3 = 5... actually, this likely comes from miscalculating the slope as 84=1\frac{8}{4} = 1 instead of 22, giving m+b=1+8=9m + b = 1 + 8 = 9. Choice D (1515) might result from an error in the slope calculation, perhaps getting m=4m = 4 instead of 22. Remember: always double-check your slope calculation using the slope formula, and verify your final equation by substituting the given point back into y=mx+by = mx + b.

Question 5

A scientist is studying bacterial growth. She observes that the bacteria population grows linearly over time. At 2 hours, there are 150 bacteria. At 5 hours, there are 240 bacteria.

If the linear pattern continues, which equation represents the bacteria population PP after tt hours?

  1. P=45t+60P = 45t + 60
  2. P=30t+150P = 30t + 150
  3. P=90t+30P = 90t + 30
  4. P=30t+90P = 30t + 90 (correct answer)
Explanation: When you encounter a problem about linear growth with two data points, you're looking to find the equation of a line in the form P=mt+bP = mt + b, where mm is the rate of change (slope) and bb is the y-intercept. First, calculate the slope using the two given points: (2, 150) and (5, 240). The slope is m=240−1505−2=903=30m = \frac{240 - 150}{5 - 2} = \frac{90}{3} = 30 bacteria per hour. This means the population increases by 30 bacteria each hour. Now use the point-slope form with either data point. Using (2, 150): P−150=30(t−2)P - 150 = 30(t - 2). Expanding: P−150=30t−60P - 150 = 30t - 60, so P=30t+90P = 30t + 90. You can verify this works with the second point: P=30(5)+90=240P = 30(5) + 90 = 240 ✓ Looking at the wrong answers: Choice A gives a slope of 45, which doesn't match our calculated rate of 30 bacteria per hour. Choice B has the correct slope of 30, but using the y-intercept of 150 would mean 150 bacteria at time zero—however, plugging in t=2t = 2 gives P=210P = 210, not the required 150. Choice C has an inflated slope of 90 and would predict far too many bacteria at any given time. The key strategy here is always to calculate the slope first from your two points, then use point-slope form to find the equation. Don't be tempted by answer choices that simply use one of the given data points as the y-intercept—the y-intercept represents the population at t=0t = 0, not at your given time points.

Question 6

A rental car company charges a flat fee plus a rate per mile driven. The total cost for driving 100 miles is $85, and the total cost for driving 250 miles is $145. If someone drives exactly 200 miles, what would be their total cost?

  1. The total cost would be $120 for this distance
  2. The total cost would be $125 for this distance (correct answer)
  3. The total cost would be $130 for this distance
  4. The total cost would be $135 for this distance
Explanation: Let CC = total cost and mm = miles driven. We have points (100,85)(100, 85) and (250,145)(250, 145). The rate per mile is 145−85250−100=60150=0.4\frac{145-85}{250-100} = \frac{60}{150} = 0.4 dollars per mile. Using point-slope form with (100,85)(100, 85): C−85=0.4(m−100)=0.4m−40C - 85 = 0.4(m - 100) = 0.4m - 40, so C=0.4m+45C = 0.4m + 45. When m=200m = 200: C=0.4(200)+45=80+45=125C = 0.4(200) + 45 = 80 + 45 = 125. Choice B is correct.

Question 7

The relationship between temperature TT (in degrees Fahrenheit) and the number of cricket chirps cc per minute is approximately linear. When the temperature is 68°F, there are 120 chirps per minute. When the temperature is 80°F, there are 180 chirps per minute. According to this model, at what temperature would you expect to hear exactly 200 chirps per minute?

  1. The temperature would be approximately 84°F at this chirp rate (correct answer)
  2. The temperature would be approximately 86°F at this chirp rate
  3. The temperature would be approximately 88°F at this chirp rate
  4. The temperature would be approximately 90°F at this chirp rate
Explanation: We have points (68,120)(68, 120) and (80,180)(80, 180) where the first coordinate is temperature and second is chirps. The slope is 180−12080−68=6012=5\frac{180-120}{80-68} = \frac{60}{12} = 5 chirps per degree. Using point-slope form with (68,120)(68, 120): c−120=5(T−68)c - 120 = 5(T - 68), so c=5T−220c = 5T - 220. When c=200c = 200: 200=5T−220200 = 5T - 220, so 5T=4205T = 420 and T=84T = 84°F. Choice A is correct.

Question 8

Maria is tracking the temperature of her greenhouse. At 6 AM, the temperature was 58°F. By 10 AM, it had risen to 74°F. The temperature continues to change at the same rate throughout the morning.

Which equation represents the temperature TT (in °F) as a function of time hh hours after 6 AM?

  1. T=4h+58T = 4h + 58 (correct answer)
  2. T=16h+58T = 16h + 58
  3. T=4h+74T = 4h + 74
  4. T=2h+58T = 2h + 58
Explanation: The temperature rises from 58°F to 74°F in 4 hours (6 AM to 10 AM), so the rate is 74−584=4\frac{74-58}{4} = 4 degrees per hour. At h=0h = 0 (6 AM), T=58T = 58, so the equation is T=4h+58T = 4h + 58. Choice B uses the total temperature change as the rate. Choice C uses 74 as the y-intercept instead of the starting temperature. Choice D uses half the correct rate.

Question 9

A line passes through points (2,−3)(2, -3) and (6,9)(6, 9). If this line is shifted vertically so that it passes through the point (4,2)(4, 2), what is the equation of the new line?

  1. y=3x−10y = 3x - 10 (correct answer)
  2. y=3x−4y = 3x - 4
  3. y=3x+2y = 3x + 2
  4. y=3x−8y = 3x - 8
Explanation: First, find the slope of the original line: m=9−(−3)6−2=124=3m = \frac{9-(-3)}{6-2} = \frac{12}{4} = 3. The new line has the same slope but passes through (4,2)(4, 2). Using point-slope form: y−2=3(x−4)y - 2 = 3(x - 4), which simplifies to y=3x−10y = 3x - 10. Choice B results from forgetting to distribute the 3 in point-slope form. Choice C uses the y-coordinate as the y-intercept. Choice D comes from using the wrong point coordinates.

Question 10

The points (−3,2)(-3, 2), (1,6)(1, 6), and (5,k)(5, k) are collinear. What is the equation of the line containing these points?

  1. y=2x+5y = 2x + 5
  2. y=x+2y = x + 2
  3. y=x+5y = x + 5 (correct answer)
  4. y=2x+8y = 2x + 8
Explanation: When you see three points described as collinear, you need to find the line they all lie on. Since you already have two complete points, you can find the slope and equation without even solving for k. Start by finding the slope using the two complete points (−3,2)(-3, 2) and (1,6)(1, 6). Using the slope formula: m=y2−y1x2−x1=6−21−(−3)=44=1m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{6 - 2}{1 - (-3)} = \frac{4}{4} = 1 Now use point-slope form with either point. Using (1,6)(1, 6): y−6=1(x−1)y - 6 = 1(x - 1), which simplifies to y=x+5y = x + 5. You can verify this works with the other point: when x=−3x = -3, y=−3+5=2y = -3 + 5 = 2 ✓ Looking at the wrong answers: Choice A (y=2x+5y = 2x + 5) uses an incorrect slope of 2 instead of 1 - this might come from miscalculating 44\frac{4}{4} as 42\frac{4}{2}. Choice B (y=x+2y = x + 2) has the right slope but wrong y-intercept, likely from an error in the point-slope algebra. Choice D (y=2x+8y = 2x + 8) combines both errors - wrong slope and wrong y-intercept. Study tip: For collinear point problems, you often don't need to find the missing coordinate first. Calculate the slope from the two complete points, then use point-slope form. Always double-check by substituting both known points back into your final equation.

Question 11

The line passing through points A(2,k)A(2, k) and B(6,3k−1)B(6, 3k-1) has slope 12\frac{1}{2}. What is the equation of this line?

  1. y=12x−12y = \frac{1}{2}x - \frac{1}{2}
  2. y=12x+12y = \frac{1}{2}x + \frac{1}{2} (correct answer)
  3. y=12x−1y = \frac{1}{2}x - 1
  4. y=12x+1y = \frac{1}{2}x + 1
Explanation: First, find kk using the slope formula: 3k−1−k6−2=12\frac{3k-1-k}{6-2} = \frac{1}{2}, so 2k−14=12\frac{2k-1}{4} = \frac{1}{2}. Cross-multiplying: 2(2k−1)=42(2k-1) = 4, so 4k−2=44k-2 = 4, giving k=32k = \frac{3}{2}. Point AA is (2,32)(2, \frac{3}{2}). Using point-slope form: y−32=12(x−2)y - \frac{3}{2} = \frac{1}{2}(x - 2), which gives y=12x+12y = \frac{1}{2}x + \frac{1}{2}. Choice A results from a sign error in the calculation. Choice C uses k=1k = 1 instead of the correct value. Choice D uses k=2k = 2.

Question 12

Line LL has slope 23\frac{2}{3} and passes through the point where the lines y=5x−7y = 5x - 7 and x=2x = 2 intersect. What is the equation of line LL?

  1. y=23x+53y = \frac{2}{3}x + \frac{5}{3} (correct answer)
  2. y=23x+3y = \frac{2}{3}x + 3
  3. y=23x−53y = \frac{2}{3}x - \frac{5}{3}
  4. y=23x−3y = \frac{2}{3}x - 3
Explanation: First, find the intersection point. When x=2x = 2, y=5(2)−7=3y = 5(2) - 7 = 3, so the point is (2,3)(2, 3). Using point-slope form with slope 23\frac{2}{3}: y−3=23(x−2)=23x−43y - 3 = \frac{2}{3}(x - 2) = \frac{2}{3}x - \frac{4}{3}, so y=23x−43+3=23x+53y = \frac{2}{3}x - \frac{4}{3} + 3 = \frac{2}{3}x + \frac{5}{3}. Choice B results from using 3 directly as the y-intercept. Choice C comes from a sign error when solving for the y-intercept. Choice D uses -3 as the y-intercept.

Question 13

A line passes through the point (4,−1)(4, -1) and has the same y-intercept as the line 2x−3y=122x - 3y = 12. What is the equation of this line?

  1. y=−34x+4y = -\frac{3}{4}x + 4
  2. y=23x−4y = \frac{2}{3}x - 4
  3. y=34x+4y = \frac{3}{4}x + 4
  4. y=34x−4y = \frac{3}{4}x - 4 (correct answer)
Explanation: When you encounter a problem about finding a line equation with specific conditions, break it down into parts: identify what you know and what you need to find. First, find the y-intercept of the given line 2x−3y=122x - 3y = 12. Convert to slope-intercept form by solving for y: −3y=−2x+12-3y = -2x + 12, so y=23x−4y = \frac{2}{3}x - 4. The y-intercept is -4. Now you need a line that passes through (4,−1)(4, -1) and has y-intercept -4. Using the slope-intercept form y=mx+by = mx + b, you know b=−4b = -4, so your equation looks like y=mx−4y = mx - 4. To find the slope mm, substitute the point (4,−1)(4, -1): −1=m(4)−4-1 = m(4) - 4. Solving: −1+4=4m-1 + 4 = 4m, so 3=4m3 = 4m, which gives m=34m = \frac{3}{4}. Therefore, the equation is y=34x−4y = \frac{3}{4}x - 4, which is answer D. Looking at the wrong answers: A has the wrong slope (−34-\frac{3}{4} instead of 34\frac{3}{4}) and wrong y-intercept (4 instead of -4). B has the slope from the original line (23\frac{2}{3}) rather than calculating the new slope needed. C has the correct slope but uses the wrong y-intercept (4 instead of -4), possibly from misreading the original equation. Remember: when a line must pass through a specific point and have a given y-intercept, always use the point to find the slope after identifying the y-intercept. Don't assume the slopes will be the same between different lines.