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Difficulty: MediumParallel and Perpendicular Lines

In the standard (x,y)(x, y) coordinate plane, line L1L_1 passes through the points (k,5)(k, 5) and (4,−1)(4, -1). Line L2L_2 is perpendicular to line L1L_1 and is represented by the equation 3x−2y=63x - 2y = 6. What is the value of kk?

  1. A
    -2
  2. B
    0
  3. -5Answer
  4. D
    8
  5. E
    13

Answer

-5
The correct answer is −5-5. First, find the slope of line L2L_2 by rewriting the equation 3x−2y=63x - 2y = 6 in slope-intercept form: y=32x−3y = \frac{3}{2}x - 3, which gives a slope of 32\frac{3}{2}. Because line L1L_1 is perpendicular to line L2L_2, its slope must be the negative reciprocal of 32\frac{3}{2}, which is −23-\frac{2}{3}. Using the slope formula with the points (k,5)(k, 5) and (4,−1)(4, -1) gives −1−54−k=−23\frac{-1 - 5}{4 - k} = -\frac{2}{3}. Solving this equation gives −64−k=−23\frac{-6}{4 - k} = -\frac{2}{3}, which simplifies to 18=8−2k18 = 8 - 2k, leading to k=−5k = -5.

Step-by-Step Solution

1
Convert the equation of line L2L_2 to slope-intercept form to find its slope.
3x−2y=6  ⟹  −2y=−3x+6  ⟹  y=32x−33x - 2y = 6 \implies -2y = -3x + 6 \implies y = \frac{3}{2}x - 3. Thus, the slope of L2L_2 is 32\frac{3}{2}.
Converting to slope-intercept form (y=mx+by = mx + b) isolates the slope as the coefficient of xx.
2
Find the slope of line L1L_1 using the relationship between perpendicular lines.
The slope of L1L_1 is the negative reciprocal of 32\frac{3}{2}, which is −23-\frac{2}{3}.
Perpendicular lines have slopes that are negative reciprocals of each other.
3
Use the slope formula with the points (k,5)(k, 5) and (4,−1)(4, -1) to solve for kk.
−1−54−k=−23  ⟹  −64−k=−23  ⟹  18=2(4−k)  ⟹  18=8−2k  ⟹  2k=−10  ⟹  k=−5\frac{-1 - 5}{4 - k} = -\frac{2}{3} \implies \frac{-6}{4 - k} = -\frac{2}{3} \implies 18 = 2(4 - k) \implies 18 = 8 - 2k \implies 2k = -10 \implies k = -5.
Setting the calculated slope of L1L_1 equal to the slope formula expression allows us to solve for the missing coordinate parameter.

Key Concept

Finding a coordinate parameter by using the negative reciprocal relationship between the slopes of perpendicular lines.
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Topics

Parallel and Perpendicular Lines

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