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Parallel and Perpendicular Lines - College Algebra

  1. Line L1L_1 has equation 3x−4y=123x - 4y = 12 and line L2L_2 has equation ax+by=6ax + by = 6. If L1L_1 and L2L_2 are parallel, and aa and bb are integers with gcd⁡(a,b)=1\gcd(a,b) = 1, what is the value of a+ba + b?
    Answer: −1-1 For parallel lines, the slopes must be equal. From 3x−4y=123x - 4y = 12, we get y=34x−3y = \frac{3}{4}x - 3, so the slope is 34\frac{3}{4}. From ax+by=6ax + by = 6, we get y=−abx+6by = -\frac{a}{b}x + \frac{6}{b}, so the slope is −ab-\frac{a}{b}. Setting slopes equal: −ab=34-\frac{a}{b} = \frac{3}{4}, which gives ab=−34\frac{a}{b} = -\frac{3}{4}. Since gcd⁡(a,b)=1\gcd(a,b) = 1, we have a=−3a = -3 and b=4b = 4, so a+b=−3+4=−1a + b = -3 + 4 = -1. Choice B incorrectly uses a=3,b=−4a = 3, b = -4. Choice C uses a=3,b=4a = 3, b = 4. Choice D uses a=−3,b=−4a = -3, b = -4.
  2. The equation 4x+3y−12=04x + 3y - 12 = 0 represents line mm. Which of the following could be the equation of a line that is parallel to line mm and passes through the origin?
    Answer: 4x+3y=04x + 3y = 0 Line mm has equation 4x+3y−12=04x + 3y - 12 = 0, which can be written as y=−43x+4y = -\frac{4}{3}x + 4, so its slope is −43-\frac{4}{3}. A parallel line must have the same slope. A line through the origin has the form y=mxy = mx where mm is the slope. So we need y=−43xy = -\frac{4}{3}x, which can be written as 4x+3y=04x + 3y = 0. Choice A gives slope 34\frac{3}{4}, which is the negative reciprocal (perpendicular). Choice C gives slope −34-\frac{3}{4}. Choice D gives slope 43\frac{4}{3}, which is the negative of the correct slope.
  3. A line passes through points (−3,7)(-3, 7) and (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)?
    Answer: y=13x−73y = \frac{1}{3}x - \frac{7}{3} First, find the slope of the given line: m=−8−72−(−3)=−155=−3m = \frac{-8-7}{2-(-3)} = \frac{-15}{5} = -3. The slope of a perpendicular line is the negative reciprocal: m⊥=13m_{\perp} = \frac{1}{3}. Using point-slope form with (4,−1)(4, -1): y−(−1)=13(x−4)y - (-1) = \frac{1}{3}(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}. 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.

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