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Difficulty: HardCoordinate Geometry: Lines, Slopes, and Distance

In the xyxy-plane, line L1L_1 has a positive slope mm and a yy-intercept of (0,4)(0, 4). The perpendicular distance from the origin (0,0)(0, 0) to line L1L_1 is 222\sqrt{2}. Line L2L_2 is perpendicular to line L1L_1 and passes through the points (2,1)(2, 1) and (6,k)(6, k). What is the value of kk?

  1. A
    −7-7
  2. B
    −5-5
  3. −3-3Answer
  4. D
    33
  5. E
    55

Answer

−3-3
The equation of line L1L_1 is mx−y+4=0mx - y + 4 = 0. Using the distance formula from the origin to a line, 4m2+1=22\frac{4}{\sqrt{m^2 + 1}} = 2\sqrt{2}, which simplifies to m2+1=2m^2 + 1 = 2, yielding m=1m = 1. The slope of line L2L_2 is the negative reciprocal of 11, which is −1-1. Using the slope equation k−16−2=−1\frac{k - 1}{6 - 2} = -1, we find k−1=−4k - 1 = -4, giving k=−3k = -3.

Step-by-Step Solution

1
Express line L1L_1 in standard form and set up the distance formula from the origin.
Line L1L_1 has slope mm and yy-intercept (0,4)(0,4), so its equation is y=mx+4y = mx + 4, or mx−y+4=0mx - y + 4 = 0. The perpendicular distance dd from (0,0)(0,0) to Ax+By+C=0Ax + By + C = 0 is d=∣C∣A2+B2=4m2+1d = \frac{|C|}{\sqrt{A^2 + B^2}} = \frac{4}{\sqrt{m^2 + 1}}.
The shortest distance from a point to a line is given by the perpendicular distance formula.
2
Solve for the positive slope mm of line L1L_1.
Setting 4m2+1=22\frac{4}{\sqrt{m^2 + 1}} = 2\sqrt{2} yields m2+1=2\sqrt{m^2 + 1} = \sqrt{2}. Squaring both sides gives m2+1=2m^2 + 1 = 2, so m2=1m^2 = 1. Since m>0m > 0, m=1m = 1.
The problem specifies that line L1L_1 has a positive slope.
3
Determine the slope of line L2L_2.
Since line L2L_2 is perpendicular to line L1L_1, its slope m2m_2 is the negative reciprocal of mm: m2=−11=−1m_2 = -\frac{1}{1} = -1.
Perpendicular lines in the coordinate plane have slopes that are negative reciprocals of each other.
4
Use the slope formula with points (2,1)(2, 1) and (6,k)(6, k) to solve for kk.
m2=k−16−2=k−14=−1  ⟹  k−1=−4  ⟹  k=−3m_2 = \frac{k - 1}{6 - 2} = \frac{k - 1}{4} = -1 \implies k - 1 = -4 \implies k = -3.
Equating the slope calculated from two points to the known perpendicular slope solves for the missing coordinate.

Key Concept

Perpendicular Distance and Perpendicular Lines in Coordinate Geometry
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Topics

Coordinate Geometry: Lines, Slopes, and Distance

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