ACCUPLACER Quantitative Reasoning, Algebra & Statistics Quiz: Writing Line Equations
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Writing Line EquationsQuestion 1 of 20

A line passes through the points (−3,7)(-3, 7) and (5,7)(5, 7). What is the equation of this line?

y=x+10y = x + 10
x=7x = 7
y=7y = 7
y=5x−3y = 5x - 3
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ACCUPLACER Quantitative Reasoning, Algebra & Statistics Quiz

ACCUPLACER Quantitative Reasoning, Algebra & Statistics Quiz: Writing Line Equations

Practice Writing Line Equations in ACCUPLACER Quantitative Reasoning, Algebra & Statistics 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 Line Equations, giving you a quick way to practice the rules, question types, and explanations that matter most for ACCUPLACER Quantitative Reasoning, Algebra & Statistics.

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.

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Question 1

A line passes through the points (−3,7)(-3, 7) and (5,7)(5, 7). What is the equation of this line?

  1. y=x+10y = x + 10
  2. x=7x = 7
  3. y=7y = 7 (correct answer)
  4. y=5x−3y = 5x - 3
Explanation: Since the y-coordinates of both points are the same (both are 7), the line is a horizontal line. The equation of a horizontal line is y=ky = k, where kk is the constant y-value. Therefore, the equation of this line is y=7y = 7. Alternatively, one could calculate the slope: m=7−75−(−3)=08=0m = \frac{7 - 7}{5 - (-3)} = \frac{0}{8} = 0. Using point-slope form: y−7=0(x−5)y - 7 = 0(x - 5), which simplifies to y−7=0y - 7 = 0, or y=7y = 7.

Question 2

A line passes through points (1,1)(1, 1) and (4,−5)(4, -5). Which of the following equations represents this line in standard form, Ax+By=CAx + By = C?

  1. 2x−y=12x - y = 1
  2. x+2y=3x + 2y = 3
  3. 2x+y=−12x + y = -1
  4. 2x+y=32x + y = 3 (correct answer)
Explanation: First, find the slope of the line: m=−5−14−1=−63=−2m = \frac{-5 - 1}{4 - 1} = \frac{-6}{3} = -2. Using the point-slope form with point (1,1)(1, 1), we have y−1=−2(x−1)y - 1 = -2(x - 1). Distributing gives y−1=−2x+2y - 1 = -2x + 2. To convert to standard form Ax+By=CAx + By = C, move the x-term to the left side and the constant term to the right side: 2x+y=2+12x + y = 2 + 1, which simplifies to 2x+y=32x + y = 3.

Question 3

Which equation represents a line that is parallel to 5x−2y=85x - 2y = 8 and passes through the point (2,−1)(2, -1)?

  1. y=52x−6y = \frac{5}{2}x - 6 (correct answer)
  2. y=−25x−15y = -\frac{2}{5}x - \frac{1}{5}
  3. y=52x−4y = \frac{5}{2}x - 4
  4. y=−52x+4y = -\frac{5}{2}x + 4
Explanation: First, find the slope of the given line by converting its equation to slope-intercept form. 5x−2y=8  ⟹  −2y=−5x+8  ⟹  y=52x−45x - 2y = 8 \implies -2y = -5x + 8 \implies y = \frac{5}{2}x - 4. The slope is 52\frac{5}{2}. A parallel line must have the same slope. Now use the point-slope form with the point (2,−1)(2, -1): y−(−1)=52(x−2)y - (-1) = \frac{5}{2}(x - 2). This simplifies to y+1=52x−5y + 1 = \frac{5}{2}x - 5. Subtracting 1 from both sides gives y=52x−6y = \frac{5}{2}x - 6.

Question 4

A line is perpendicular to the line y=−5y = -5 and passes through the point (3,8)(3, 8). What is the equation of this line?

  1. y=8y = 8
  2. x=3x = 3 (correct answer)
  3. x=8x = 8
  4. y=3y = 3
Explanation: The line y=−5y = -5 is a horizontal line with a slope of 0. A line perpendicular to a horizontal line is a vertical line. A vertical line has an undefined slope and its equation is of the form x=kx = k, where kk is the x-coordinate of every point on the line. Since the line must pass through the point (3,8)(3, 8), its equation must be x=3x = 3.

Question 5

Line L passes through the origin and the point (a,b)(a, b), where a≠0a \neq 0 and b≠0b \neq 0. Line M is perpendicular to line L and also passes through the origin. Which of the following is an equation for line M?

  1. y=baxy = \frac{b}{a}x
  2. y=−baxy = -\frac{b}{a}x
  3. y=abxy = \frac{a}{b}x
  4. y=−abxy = -\frac{a}{b}x (correct answer)
Explanation: The slope of line L, which passes through (0,0)(0, 0) and (a,b)(a, b), is mL=b−0a−0=bam_L = \frac{b - 0}{a - 0} = \frac{b}{a}. Line M is perpendicular to line L, so its slope mMm_M must be the negative reciprocal of mLm_L. Therefore, mM=−abm_M = -\frac{a}{b}. Since line M also passes through the origin, its y-intercept is 0. The equation for line M is y=mMx+by = m_M x + b, which becomes y=−abx+0y = -\frac{a}{b}x + 0, or y=−abxy = -\frac{a}{b}x.

Question 6

A line passes through the points (0,−4)(0, -4) and (6,−1)(6, -1). What is the equation of the line?

  1. y=2x−4y = 2x - 4
  2. y=12x−4y = \frac{1}{2}x - 4 (correct answer)
  3. y=−56x−4y = -\frac{5}{6}x - 4
  4. y=12x+6y = \frac{1}{2}x + 6
Explanation: The point (0,−4)(0, -4) indicates that the y-intercept bb is -4. To find the slope mm, use the two points: m=−1−(−4)6−0=−1+46=36=12m = \frac{-1 - (-4)}{6 - 0} = \frac{-1 + 4}{6} = \frac{3}{6} = \frac{1}{2}. Using the slope-intercept form y=mx+by = mx + b, the equation is y=12x−4y = \frac{1}{2}x - 4.

Question 7

What is the equation of a line that is perpendicular to x+4y=8x + 4y = 8 and has the same x-intercept as this line?

  1. y=−4x+32y = -4x + 32
  2. y=4x+2y = 4x + 2
  3. y=−14x+2y = -\frac{1}{4}x + 2
  4. y=4x−32y = 4x - 32 (correct answer)
Explanation: First, find the slope of the given line. x+4y=8  ⟹  4y=−x+8  ⟹  y=−14x+2x + 4y = 8 \implies 4y = -x + 8 \implies y = -\frac{1}{4}x + 2. The slope is −14-\frac{1}{4}. The slope of a perpendicular line is the negative reciprocal, which is 44. Next, find the x-intercept of the original line by setting y=0y=0: x+4(0)=8  ⟹  x=8x + 4(0) = 8 \implies x=8. So the new line must pass through the point (8,0)(8, 0). Using the point-slope form with this point and the perpendicular slope: y−0=4(x−8)y - 0 = 4(x - 8), which simplifies to y=4x−32y = 4x - 32.

Question 8

A scuba diver is ascending to the surface at a constant rate. She starts at a depth of 120 feet. After 3 minutes, her depth is 75 feet.

Based on the passage, which equation relates her depth, dd, in feet to the time, tt, in minutes since she started ascending?

  1. d=15t+120d = 15t + 120
  2. d=−15t+75d = -15t + 75
  3. d=15t+75d = 15t + 75
  4. d=−15t+120d = -15t + 120 (correct answer)
Explanation: The starting depth gives the y-intercept (the depth at t=0t=0). So, the point is (0,120)(0, 120), and b=120b=120. After 3 minutes, the depth is 75 feet, which gives the point (3,75)(3, 75). The slope (rate of change in depth) is m=75−1203−0=−453=−15m = \frac{75 - 120}{3 - 0} = \frac{-45}{3} = -15 feet per minute. The negative sign indicates the depth is decreasing. Using the slope-intercept form d=mt+bd = mt + b, the equation is d=−15t+120d = -15t + 120.

Question 9

The graph of a linear equation has a positive slope and a negative y-intercept. Which of the following points is it impossible for the graph to pass through?

  1. (5,2)(5, 2)
  2. (−3,−10)(-3, -10)
  3. (−4,1)(-4, 1) (correct answer)
  4. (2,−1)(2, -1)
Explanation: A linear equation is of the form y=mx+by = mx + b. We are given that the slope m>0m > 0 and the y-intercept b<0b < 0. Let's test the point (−4,1)(-4, 1), which is in Quadrant II (x is negative, y is positive). Substitute the coordinates into the equation: 1=m(−4)+b1 = m(-4) + b, or 1=−4m+b1 = -4m + b. Since m>0m > 0, the term (-4m) must be negative. Since b<0b < 0, the term bb is also negative. The sum of two negative numbers ((-4m) and bb) must be a negative number. However, the equation states their sum is 1, which is positive. This is a contradiction. Therefore, the line cannot pass through (−4,1)(-4, 1). The other points are in Quadrants I, III, and IV, all of which are possible for a line with a positive slope and negative y-intercept.

Question 10

What is the equation of the line that passes through the points (4,−1)(4, -1) and (4,9)(4, 9)?

  1. y=4y = 4
  2. x=4x = 4 (correct answer)
  3. y=9y = 9
  4. y=x−5y = x - 5
Explanation: Both points have the same x-coordinate, which is 4. This indicates that the line is a vertical line. The equation of a vertical line is of the form x=kx = k, where kk is the constant x-value. In this case, k=4k=4, so the equation is x=4x = 4. Attempting to calculate the slope would result in division by zero (9−(−1)4−4=100\frac{9 - (-1)}{4 - 4} = \frac{10}{0}), which confirms the slope is undefined and the line is vertical.

Question 11

What is the equation of the line that is parallel to the x-axis and passes through the point (−2,6)(-2, 6)?

  1. y=6y = 6 (correct answer)
  2. x=−2x = -2
  3. y=−2y = -2
  4. y=0y = 0
Explanation: The x-axis is a horizontal line with the equation y=0y = 0. Any line parallel to the x-axis must also be a horizontal line, and its equation will be of the form y=ky = k, where kk is the y-coordinate for every point on the line. Since the line passes through (−2,6)(-2, 6), the y-coordinate for all points must be 6. Therefore, the equation is y=6y = 6.

Question 12

A container is being filled with water at a constant rate. After 2 minutes, the water level is 5 cm. After 6 minutes, the water level is 8 cm.

Based on the passage, which equation models the water level, hh, in centimeters as a function of time, tt, in minutes?

  1. h=43t+73h = \frac{4}{3}t + \frac{7}{3}
  2. h=34t+5h = \frac{3}{4}t + 5
  3. h=34t+72h = \frac{3}{4}t + \frac{7}{2} (correct answer)
  4. h=34t−132h = \frac{3}{4}t - \frac{13}{2}
Explanation: The problem provides two points: (2,5)(2, 5) and (6,8)(6, 8). First, calculate the slope (rate): m=8−56−2=34m = \frac{8 - 5}{6 - 2} = \frac{3}{4}. Next, use the point-slope form with the point (2,5)(2, 5): h−5=34(t−2)h - 5 = \frac{3}{4}(t - 2). Distribute the slope: h−5=34t−64h - 5 = \frac{3}{4}t - \frac{6}{4}, which simplifies to h−5=34t−32h - 5 = \frac{3}{4}t - \frac{3}{2}. Finally, add 5 to both sides: h=34t−32+5h = \frac{3}{4}t - \frac{3}{2} + 5. To combine the constants, use a common denominator: h=34t−32+102h = \frac{3}{4}t - \frac{3}{2} + \frac{10}{2}, which simplifies to h=34t+72h = \frac{3}{4}t + \frac{7}{2}.

Question 13

A company's profit was $15,000 in 2018 and grew to $35,000 in 2022. Assume the relationship between profit and time is linear.

Based on the passage, which equation models the profit, PP, in terms of the number of years, tt, after 2018?

  1. P=5000t+15000P = 5000t + 15000 (correct answer)
  2. P=15000t+5000P = 15000t + 5000
  3. P=20000t+15000P = 20000t + 15000
  4. P=5000t+35000P = 5000t + 35000
Explanation: Let t=0t=0 represent the year 2018. This gives us the point (0,15000)(0, 15000), meaning the y-intercept is 15,000. The year 2022 corresponds to t=4t=4, which gives the point (4,35000)(4, 35000). The slope (rate of change) is m=35000−150004−0=200004=5000m = \frac{35000 - 15000}{4 - 0} = \frac{20000}{4} = 5000. Using the slope-intercept form P=mt+bP = mt + b, we get P=5000t+15000P = 5000t + 15000.

Question 14

Plan A for a cell phone costs a flat fee of $20 per month plus $0.10 for each minute of calls. Plan B has no flat fee but costs $0.30 per minute.

A new Plan C is created. It has a slope (per-minute cost) equal to the average of the slopes for Plans A and B, and it passes through the point where the costs of Plan A and Plan B are equal. What is the equation for the cost, CC, of Plan C as a function of minutes, mm?

  1. C=0.20m+10C = 0.20m + 10 (correct answer)
  2. C=0.20m+20C = 0.20m + 20
  3. C=0.30mC = 0.30m
  4. C=0.10m+20C = 0.10m + 20
Explanation: First, write the cost equations for Plan A and Plan B: CA=0.10m+20C_A = 0.10m + 20 and CB=0.30mC_B = 0.30m. Find their intersection point by setting the costs equal: 0.10m+20=0.30m  ⟹  20=0.20m  ⟹  m=1000.10m + 20 = 0.30m \implies 20 = 0.20m \implies m = 100. The cost at this point is C=0.30(100)=30C = 0.30(100) = 30. So, the intersection point is (100,30)(100, 30). The slope for Plan C is the average of the slopes of A and B: mC=0.10+0.302=0.402=0.20m_C = \frac{0.10 + 0.30}{2} = \frac{0.40}{2} = 0.20. Now, use the point-slope form with the point (100,30)(100, 30) and slope 0.20: C−30=0.20(m−100)C - 30 = 0.20(m - 100). Distribute the slope: C−30=0.20m−20C - 30 = 0.20m - 20. Add 30 to both sides to get C=0.20m+10C = 0.20m + 10.

Question 15

A line is parallel to y=3x−5y = 3x - 5 and has the same y-intercept as the line 2y+4x=122y + 4x = 12. What is the equation of this line?

  1. y=3x+6y = 3x + 6 (correct answer)
  2. y=−2x−5y = -2x - 5
  3. y=3x−5y = 3x - 5
  4. y=−2x+6y = -2x + 6
Explanation: First, determine the required slope. The line is parallel to y=3x−5y = 3x - 5, so it must have the same slope, m=3m = 3. Next, determine the required y-intercept. It is the same as the y-intercept of 2y+4x=122y + 4x = 12. Convert this equation to slope-intercept form: 2y=−4x+12  ⟹  y=−2x+62y = -4x + 12 \implies y = -2x + 6. The y-intercept is b=6b = 6. Finally, combine the slope m=3m=3 and y-intercept b=6b=6 to form the new equation: y=3x+6y = 3x + 6.

Question 16

A line has an x-intercept of -4 and a y-intercept of 6. Which of the following is an equation of the line?

  1. y=−23x+6y = -\frac{2}{3}x + 6
  2. y=32x−4y = \frac{3}{2}x - 4
  3. y=−32x+6y = -\frac{3}{2}x + 6
  4. y=32x+6y = \frac{3}{2}x + 6 (correct answer)
Explanation: The x-intercept of -4 corresponds to the point (−4,0)(-4, 0). The y-intercept of 6 corresponds to the point (0,6)(0, 6). The y-intercept value directly gives b=6b=6 in the slope-intercept form y=mx+by = mx + b. To find the slope mm, use the two points: m=6−00−(−4)=64=32m = \frac{6 - 0}{0 - (-4)} = \frac{6}{4} = \frac{3}{2}. Combining the slope and y-intercept gives the equation y=32x+6y = \frac{3}{2}x + 6.

Question 17

A line segment has endpoints at A(−2,1)A(-2, 1) and B(4,5)B(4, 5). What is the equation of the perpendicular bisector of the segment AB?

  1. y=23x+73y = \frac{2}{3}x + \frac{7}{3}
  2. y=−32x+92y = -\frac{3}{2}x + \frac{9}{2} (correct answer)
  3. y=−32x+11y = -\frac{3}{2}x + 11
  4. y=32x+32y = \frac{3}{2}x + \frac{3}{2}
Explanation: A perpendicular bisector passes through the midpoint of the segment and has a slope that is the negative reciprocal of the segment's slope. First, find the midpoint: (−2+42,1+52)=(1,3)(\frac{-2+4}{2}, \frac{1+5}{2}) = (1, 3). Next, find the slope of segment AB: mAB=5−14−(−2)=46=23m_{AB} = \frac{5 - 1}{4 - (-2)} = \frac{4}{6} = \frac{2}{3}. The slope of the perpendicular bisector is the negative reciprocal, m⊥=−32m_{\perp} = -\frac{3}{2}. Finally, use the point-slope form with the midpoint (1,3)(1, 3) and perpendicular slope: y−3=−32(x−1)y - 3 = -\frac{3}{2}(x - 1). Simplifying gives y−3=−32x+32y - 3 = -\frac{3}{2}x + \frac{3}{2}, so y=−32x+32+3=−32x+92y = -\frac{3}{2}x + \frac{3}{2} + 3 = -\frac{3}{2}x + \frac{9}{2}.

Question 18

A line passes through the points (2,5)(2, 5) and (−4,1)(-4, 1). What is the equation of the line?

  1. y=23x−113y = \frac{2}{3}x - \frac{11}{3}
  2. y=23x+113y = \frac{2}{3}x + \frac{11}{3} (correct answer)
  3. y=32x+2y = \frac{3}{2}x + 2
  4. y=−23x+193y = -\frac{2}{3}x + \frac{19}{3}
Explanation: First, calculate the slope mm using the formula m=y2−y1x2−x1m = \frac{y_2 - y_1}{x_2 - x_1}. So, m=1−5−4−2=−4−6=23m = \frac{1 - 5}{-4 - 2} = \frac{-4}{-6} = \frac{2}{3}. Next, use the point-slope form, y−y1=m(x−x1)y - y_1 = m(x - x_1), with the point (2,5)(2, 5). This gives y−5=23(x−2)y - 5 = \frac{2}{3}(x - 2). Distributing the slope gives y−5=23x−43y - 5 = \frac{2}{3}x - \frac{4}{3}. Finally, add 5 (or 153\frac{15}{3}) to both sides to isolate yy: y=23x−43+153y = \frac{2}{3}x - \frac{4}{3} + \frac{15}{3}, which simplifies to y=23x+113y = \frac{2}{3}x + \frac{11}{3}.

Question 19

What is the equation of the line that passes through the point (4,−2)(4, -2) and is perpendicular to the line y=13x+5y = \frac{1}{3}x + 5?

  1. y=−3x+10y = -3x + 10 (correct answer)
  2. y=3x−14y = 3x - 14
  3. y=13x−103y = \frac{1}{3}x - \frac{10}{3}
  4. y=−13x−23y = -\frac{1}{3}x - \frac{2}{3}
Explanation: The slope of the given line y=13x+5y = \frac{1}{3}x + 5 is 13\frac{1}{3}. The slope of a line perpendicular to it is the negative reciprocal, which is −3-3. Using the point-slope form y−y1=m(x−x1)y - y_1 = m(x - x_1) with the point (4,−2)(4, -2) and slope −3-3, we get y−(−2)=−3(x−4)y - (-2) = -3(x - 4). This simplifies to y+2=−3x+12y + 2 = -3x + 12. Subtracting 2 from both sides gives the final equation y=−3x+10y = -3x + 10.

Question 20

Which of the following is an equation of the line that passes through the point (−1,3)(-1, 3) and is parallel to the line with the equation 2x+3y=62x + 3y = 6?

  1. y=32x+92y = \frac{3}{2}x + \frac{9}{2}
  2. y=−23x+2y = -\frac{2}{3}x + 2
  3. y=−23x+73y = -\frac{2}{3}x + \frac{7}{3} (correct answer)
  4. y=−23x+113y = -\frac{2}{3}x + \frac{11}{3}
Explanation: First, find the slope of the given line by converting its equation to slope-intercept form (y=mx+by = mx + b). 2x+3y=6  ⟹  3y=−2x+6  ⟹  y=−23x+22x + 3y = 6 \implies 3y = -2x + 6 \implies y = -\frac{2}{3}x + 2. The slope is −23-\frac{2}{3}. A parallel line has the same slope. Now, use the point-slope form y−y1=m(x−x1)y - y_1 = m(x - x_1) with the point (−1,3)(-1, 3) and slope −23-\frac{2}{3}: y−3=−23(x−(−1))y - 3 = -\frac{2}{3}(x - (-1)). This simplifies to y−3=−23(x+1)y - 3 = -\frac{2}{3}(x + 1), then y−3=−23x−23y - 3 = -\frac{2}{3}x - \frac{2}{3}. Adding 3 (or 93\frac{9}{3}) to both sides gives y=−23x−23+93y = -\frac{2}{3}x - \frac{2}{3} + \frac{9}{3}, which is y=−23x+73y = -\frac{2}{3}x + \frac{7}{3}.