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Difficulty: Very hardParallel and Perpendicular Lines

In the standard (x,y)(x, y) coordinate plane, line L1L_1 is defined by the equation 3x−ky=83x - ky = 8, where kk is a non-zero constant. Line L2L_2 is perpendicular to L1L_1 and passes through the points (k,2)(k, 2) and (−1,k+5)(-1, k+5). What is the sum of all possible real values of kk?

  1. A
    −9-9
  2. B
    −6-6
  3. C
    −2-2
  4. D
    00
  5. 22Answer

Answer

The sum of all possible real values of kk is 22.
The slope of line L1L_1 is found by converting its equation to slope-intercept form, yielding m1=3km_1 = \frac{3}{k}. Since line L2L_2 is perpendicular, its slope must be the negative reciprocal, m2=−k3m_2 = -\frac{k}{3}. Using the slope formula with the given points on line L2L_2, we also have m2=(k+5)−2−1−k=−k+3k+1m_2 = \frac{(k+5) - 2}{-1 - k} = -\frac{k+3}{k+1}. Setting these two expressions for m2m_2 equal to each other and solving the resulting proportion gives the quadratic equation k2−2k−9=0k^2 - 2k - 9 = 0. The discriminant of this equation is positive (40>040 > 0), confirming the existence of two real solutions. By Vieta's formulas, the sum of the roots is −ba=−−21=2-\frac{b}{a} = -\frac{-2}{1} = 2.

Step-by-Step Solution

1
Determine the slope of line L1L_1 in terms of kk.
The slope of line L1L_1 is m1=3km_1 = \frac{3}{k}.
Rewriting the equation 3x−ky=83x - ky = 8 in slope-intercept form (y=mx+by = mx + b) gives ky=3x−8ky = 3x - 8, which simplifies to y=3kx−8ky = \frac{3}{k}x - \frac{8}{k}. The coefficient of xx represents the slope of the line.
2
Find the perpendicular slope m2m_2 of line L2L_2 in terms of kk.
The slope of line L2L_2 is m2=−k3m_2 = -\frac{k}{3}.
Since line L2L_2 is perpendicular to line L1L_1, its slope must be the negative reciprocal of m1m_1 (i.e., m2=−1m1m_2 = -\frac{1}{m_1}).
3
Express the slope of line L2L_2 using the two given points (k,2)(k, 2) and (−1,k+5)(-1, k+5).
The slope of line L2L_2 is m2=−k+3k+1m_2 = -\frac{k+3}{k+1}.
Using the slope formula m=y2−y1x2−x1m = \frac{y_2 - y_1}{x_2 - x_1} with the points (k,2)(k, 2) and (−1,k+5)(-1, k+5), we get m2=(k+5)−2−1−k=k+3−(k+1)=−k+3k+1m_2 = \frac{(k+5) - 2}{-1 - k} = \frac{k+3}{-(k+1)} = -\frac{k+3}{k+1}.
4
Set the two expressions for the slope of line L2L_2 equal to each other and solve for kk.
The quadratic equation is k2−2k−9=0k^2 - 2k - 9 = 0.
Equating the slopes gives −k3=−k+3k+1-\frac{k}{3} = -\frac{k+3}{k+1}. Multiplying both sides by −1-1 gives k3=k+3k+1\frac{k}{3} = \frac{k+3}{k+1}. Cross-multiplying yields k(k+1)=3(k+3)  ⟹  k2+k=3k+9  ⟹  k2−2k−9=0k(k+1) = 3(k+3) \implies k^2 + k = 3k + 9 \implies k^2 - 2k - 9 = 0.
5
Calculate the sum of all possible real values of kk.
The sum of all possible real values of kk is 22.
The discriminant of the quadratic equation k2−2k−9=0k^2 - 2k - 9 = 0 is (−2)2−4(1)(−9)=40>0(-2)^2 - 4(1)(-9) = 40 > 0, confirming that two distinct real solutions for kk exist. According to Vieta's formulas, the sum of the roots of a quadratic equation ax2+bx+c=0ax^2 + bx + c = 0 is −ba-\frac{b}{a}. Here, a=1a=1 and b=−2b=-2, so the sum is −−21=2-\frac{-2}{1} = 2.

Key Concept

Perpendicular lines have slopes that are negative reciprocals of each other (m1⋅m2=−1m_1 \cdot m_2 = -1).
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Topics

Parallel and Perpendicular Lines

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