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College Algebra Quiz

College Algebra Quiz: Parallel And Perpendicular Lines

Practice Parallel And Perpendicular Lines in College Algebra with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

Question 1 / 3

0 of 3 answered

A line passes through points (−3,7)(-3, 7)(−3,7) and (2,−8)(2, -8)(2,−8). Which of the following represents the equation of a line that is perpendicular to this line and passes through the point (4,−1)(4, -1)(4,−1)?

Select an answer to continue

What this quiz covers

This quiz focuses on Parallel And Perpendicular Lines, giving you a quick way to practice the rules, question types, and explanations that matter most for College Algebra.

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 (−3,7)(-3, 7)(−3,7) and (2,−8)(2, -8)(2,−8). Which of the following represents the equation of a line that is perpendicular to this line and passes through the point (4,−1)(4, -1)(4,−1)?

  1. y=13x−73y = \frac{1}{3}x - \frac{7}{3}y=31​x−37​ (correct answer)
  2. y=−3x+11y = -3x + 11y=−3x+11
  3. y=13x+73y = \frac{1}{3}x + \frac{7}{3}y=31​x+37​
  4. y=−13x+13y = -\frac{1}{3}x + \frac{1}{3}y=−31​x+31​

Explanation: First, find the slope of the given line: m=−8−72−(−3)=−155=−3m = \frac{-8-7}{2-(-3)} = \frac{-15}{5} = -3m=2−(−3)−8−7​=5−15​=−3. The slope of a perpendicular line is the negative reciprocal: m⊥=13m_{\perp} = \frac{1}{3}m⊥​=31​. Using point-slope form with (4,−1)(4, -1)(4,−1): y−(−1)=13(x−4)y - (-1) = \frac{1}{3}(x - 4)y−(−1)=31​(x−4), which simplifies to y=13x−43−1=13x−73y = \frac{1}{3}x - \frac{4}{3} - 1 = \frac{1}{3}x - \frac{7}{3}y=31​x−34​−1=31​x−37​. Choice B uses the original slope instead of the perpendicular slope. Choice C has the correct slope but wrong y-intercept (forgot to subtract 1). Choice D uses the negative reciprocal incorrectly.

Question 2

Line L1L_1L1​ has equation 3x−4y=123x - 4y = 123x−4y=12 and line L2L_2L2​ has equation ax+by=6ax + by = 6ax+by=6. If L1L_1L1​ and L2L_2L2​ are parallel, and aaa and bbb are integers with gcd⁡(a,b)=1\gcd(a,b) = 1gcd(a,b)=1, what is the value of a+ba + ba+b?

  1. −1-1−1 (correct answer)
  2. 111
  3. 777
  4. −7-7−7

Explanation: For parallel lines, the slopes must be equal. From 3x−4y=123x - 4y = 123x−4y=12, we get y=34x−3y = \frac{3}{4}x - 3y=43​x−3, so the slope is 34\frac{3}{4}43​. From ax+by=6ax + by = 6ax+by=6, we get y=−abx+6by = -\frac{a}{b}x + \frac{6}{b}y=−ba​x+b6​, so the slope is −ab-\frac{a}{b}−ba​. Setting slopes equal: −ab=34-\frac{a}{b} = \frac{3}{4}−ba​=43​, which gives ab=−34\frac{a}{b} = -\frac{3}{4}ba​=−43​. Since gcd⁡(a,b)=1\gcd(a,b) = 1gcd(a,b)=1, we have a=−3a = -3a=−3 and b=4b = 4b=4, so a+b=−3+4=−1a + b = -3 + 4 = -1a+b=−3+4=−1. Choice B incorrectly uses a=3,b=−4a = 3, b = -4a=3,b=−4. Choice C uses a=3,b=4a = 3, b = 4a=3,b=4. Choice D uses a=−3,b=−4a = -3, b = -4a=−3,b=−4.

Question 3

The equation 4x+3y−12=04x + 3y - 12 = 04x+3y−12=0 represents line mmm. Which of the following could be the equation of a line that is parallel to line mmm and passes through the origin?

  1. 3x−4y=03x - 4y = 03x−4y=0
  2. 4x+3y=04x + 3y = 04x+3y=0 (correct answer)
  3. 3x+4y=03x + 4y = 03x+4y=0
  4. 4x−3y=04x - 3y = 04x−3y=0

Explanation: Line mmm has equation 4x+3y−12=04x + 3y - 12 = 04x+3y−12=0, which can be written as y=−43x+4y = -\frac{4}{3}x + 4y=−34​x+4, so its slope is −43-\frac{4}{3}−34​. A parallel line must have the same slope. A line through the origin has the form y=mxy = mxy=mx where mmm is the slope. So we need y=−43xy = -\frac{4}{3}xy=−34​x, which can be written as 4x+3y=04x + 3y = 04x+3y=0. Choice A gives slope 34\frac{3}{4}43​, which is the negative reciprocal (perpendicular). Choice C gives slope −34-\frac{3}{4}−43​. Choice D gives slope 43\frac{4}{3}34​, which is the negative of the correct slope.