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College Algebra Quiz

College Algebra Quiz: Equation Of A Line Formats

Practice Equation Of A Line Formats 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 / 5

0 of 5 answered

The equation 5x−3y=155x - 3y = 155x−3y=15 can be rewritten in slope-intercept form. If this line is then shifted so that its new y-intercept is 2 units above the original y-intercept, what is the equation of the shifted line in standard form ax+by=cax + by = cax+by=c where a>0a > 0a>0?

Select an answer to continue

What this quiz covers

This quiz focuses on Equation Of A Line Formats, 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

The equation 5x−3y=155x - 3y = 155x−3y=15 can be rewritten in slope-intercept form. If this line is then shifted so that its new y-intercept is 2 units above the original y-intercept, what is the equation of the shifted line in standard form ax+by=cax + by = cax+by=c where a>0a > 0a>0?

  1. 5x−3y=215x - 3y = 215x−3y=21
  2. 5x−3y=95x - 3y = 95x−3y=9 (correct answer)
  3. 5x+3y=215x + 3y = 215x+3y=21
  4. 5x+3y=95x + 3y = 95x+3y=9

Explanation: First convert 5x−3y=155x - 3y = 155x−3y=15 to slope-intercept form: −3y=−5x+15-3y = -5x + 15−3y=−5x+15, so y=53x−5y = \frac{5}{3}x - 5y=35​x−5. The original y-intercept is -5. The new y-intercept is −5+2=−3-5 + 2 = -3−5+2=−3. The shifted line has the same slope but y-intercept -3: y=53x−3y = \frac{5}{3}x - 3y=35​x−3. Converting back to standard form: 3y=5x−93y = 5x - 93y=5x−9, so 5x−3y=95x - 3y = 95x−3y=9. Choice A results from shifting the y-intercept down by 2 instead of up. Choice C and D have the wrong sign on the yyy coefficient.

Question 2

A line in standard form ax+by=cax + by = cax+by=c has the property that when xxx increases by 6, yyy decreases by 8. If the line passes through (2,−3)(2, -3)(2,−3), which of the following could be the equation of the line?

  1. 3x+4y=−63x + 4y = -63x+4y=−6
  2. 3x−4y=183x - 4y = 183x−4y=18
  3. 4x−3y=174x - 3y = 174x−3y=17
  4. 4x+3y=−14x + 3y = -14x+3y=−1 (correct answer)

Explanation: When you encounter a problem about how a line behaves as variables change, you're working with slope and the relationship between x and y coordinates. The key insight is that "when x increases by 6, y decreases by 8" tells you the slope is ΔyΔx=−86=−43\frac{\Delta y}{\Delta x} = \frac{-8}{6} = -\frac{4}{3}ΔxΔy​=6−8​=−34​. To find the correct equation, you need a line with slope −43-\frac{4}{3}−34​ that passes through (2,−3)(2, -3)(2,−3). For a line in standard form ax+by=cax + by = cax+by=c, the slope is −ab-\frac{a}{b}−ba​. So you need −ab=−43-\frac{a}{b} = -\frac{4}{3}−ba​=−34​, which means ab=43\frac{a}{b} = \frac{4}{3}ba​=34​. Let's check each option: Option A has ab=34\frac{a}{b} = \frac{3}{4}ba​=43​ (slope = −34-\frac{3}{4}−43​), which is incorrect. Option B gives 3−4=−34\frac{3}{-4} = -\frac{3}{4}−43​=−43​ (slope = 34\frac{3}{4}43​), also wrong. Option C has 4−3=−43\frac{4}{-3} = -\frac{4}{3}−34​=−34​ (slope = 43\frac{4}{3}34​), which has the wrong sign. Option D gives 43\frac{4}{3}34​ (slope = −43-\frac{4}{3}−34​), which matches our requirement. Now verify it passes through (2,−3)(2, -3)(2,−3): 4(2)+3(−3)=8−9=−14(2) + 3(-3) = 8 - 9 = -14(2)+3(−3)=8−9=−1 ✓ The answer is D. Study tip: When a problem describes how variables change together, immediately calculate the slope using ΔyΔx\frac{\Delta y}{\Delta x}ΔxΔy​. Remember that in standard form ax+by=cax + by = cax+by=c, the slope is −ab-\frac{a}{b}−ba​, and always verify your answer by substituting the given point.

Question 3

The equation of a line in point-slope form is y+4=−23(x−6)y + 4 = -\frac{2}{3}(x - 6)y+4=−32​(x−6). Which of the following represents the same line in slope-intercept form?

  1. y=−23x+0y = -\frac{2}{3}x + 0y=−32​x+0 (correct answer)
  2. y=−23x−8y = -\frac{2}{3}x - 8y=−32​x−8
  3. y=−23x+8y = -\frac{2}{3}x + 8y=−32​x+8
  4. y=23x−8y = \frac{2}{3}x - 8y=32​x−8

Explanation: Starting with y+4=−23(x−6)y + 4 = -\frac{2}{3}(x - 6)y+4=−32​(x−6), distribute the slope: y+4=−23x+23⋅6=−23x+4y + 4 = -\frac{2}{3}x + \frac{2}{3} \cdot 6 = -\frac{2}{3}x + 4y+4=−32​x+32​⋅6=−32​x+4. Subtracting 4 from both sides: y=−23x+4−4=−23x+0y = -\frac{2}{3}x + 4 - 4 = -\frac{2}{3}x + 0y=−32​x+4−4=−32​x+0. Choice B results from incorrectly distributing as −23⋅6=−4-\frac{2}{3} \cdot 6 = -4−32​⋅6=−4 instead of +4+4+4, then getting y=−23x−4−4=−23x−8y = -\frac{2}{3}x - 4 - 4 = -\frac{2}{3}x - 8y=−32​x−4−4=−32​x−8. Choice C comes from the error y+4=−23x+4y + 4 = -\frac{2}{3}x + 4y+4=−32​x+4 but then adding 4 instead of subtracting. Choice D has the wrong sign for the slope.

Question 4

A line has x-intercept 4 and y-intercept -6. Another line is perpendicular to this line and passes through the point (2,1)(2, 1)(2,1). What is the equation of the perpendicular line in point-slope form?

  1. y−1=−32(x−2)y - 1 = -\frac{3}{2}(x - 2)y−1=−23​(x−2)
  2. y−1=23(x−2)y - 1 = \frac{2}{3}(x - 2)y−1=32​(x−2)
  3. y−1=−23(x−2)y - 1 = -\frac{2}{3}(x - 2)y−1=−32​(x−2) (correct answer)
  4. y−1=32(x−2)y - 1 = \frac{3}{2}(x - 2)y−1=23​(x−2)

Explanation: When you encounter problems involving perpendicular lines, you need to find the slope of the original line first, then use the fact that perpendicular lines have slopes that are negative reciprocals of each other. Start by finding the slope of the line with x-intercept 4 and y-intercept -6. These intercepts give you points (4, 0) and (0, -6). Using the slope formula: m=y2−y1x2−x1=−6−00−4=−6−4=32m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-6 - 0}{0 - 4} = \frac{-6}{-4} = \frac{3}{2}m=x2​−x1​y2​−y1​​=0−4−6−0​=−4−6​=23​ Since perpendicular lines have slopes that are negative reciprocals, the perpendicular line has slope −132=−23-\frac{1}{\frac{3}{2}} = -\frac{2}{3}−23​1​=−32​. Using point-slope form with the given point (2, 1): y−1=−23(x−2)y - 1 = -\frac{2}{3}(x - 2)y−1=−32​(x−2) Let's examine why the other answers are incorrect. Choice A uses slope −32-\frac{3}{2}−23​, which is the negative of the original slope but not the negative reciprocal—this represents a common confusion between parallel and perpendicular slopes. Choice B has slope 23\frac{2}{3}32​, which is the reciprocal but not negative—this misses the "negative" part of "negative reciprocal." Choice D uses slope 32\frac{3}{2}23​, which is identical to the original line's slope, making these lines parallel rather than perpendicular. Remember this key relationship: if two lines are perpendicular, their slopes multiply to give -1. You can always check your work by verifying that 32×(−23)=−1\frac{3}{2} \times \left(-\frac{2}{3}\right) = -123​×(−32​)=−1.

Question 5

The line 3x−2y=123x - 2y = 123x−2y=12 is transformed by reflecting it across the y-axis, then shifting it down 4 units. What is the equation of the resulting line in slope-intercept form?

  1. y=32x−10y = \frac{3}{2}x - 10y=23​x−10
  2. y=−32x−10y = -\frac{3}{2}x - 10y=−23​x−10 (correct answer)
  3. y=−32x−2y = -\frac{3}{2}x - 2y=−23​x−2
  4. y=32x+2y = \frac{3}{2}x + 2y=23​x+2

Explanation: First convert 3x−2y=123x - 2y = 123x−2y=12 to slope-intercept form: −2y=−3x+12-2y = -3x + 12−2y=−3x+12, so y=32x−6y = \frac{3}{2}x - 6y=23​x−6. Reflecting across the y-axis changes xxx to −x-x−x: y=32(−x)−6=−32x−6y = \frac{3}{2}(-x) - 6 = -\frac{3}{2}x - 6y=23​(−x)−6=−23​x−6. Shifting down 4 units subtracts 4 from the y-value: y=−32x−6−4=−32x−10y = -\frac{3}{2}x - 6 - 4 = -\frac{3}{2}x - 10y=−23​x−6−4=−23​x−10. Choice A incorrectly keeps the original slope after reflection. Choice C correctly reflects but shifts down only 4 from 0 instead of from -6. Choice D incorrectly reflects (wrong sign) and shifts up instead of down.