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Difficulty: Very hardLinear Equations and Graphing

In the standard (x,y)(x,y) coordinate plane, the set of all points equidistant from the line 3x−4y=83x - 4y = 8 and the line 5x+12y=135x + 12y = 13 consists of two perpendicular lines. What is the slope of the line in this set that has a positive slope?

  1. A
    -8
  2. B
    219\frac{2}{19}
  3. C
    47\frac{4}{7}
  4. 18\frac{1}{8}Answer
  5. E
    8

Answer

The slope of the line with a positive slope is 1/8.
By using the point-to-line distance formula, the equidistant relationship ∣3x−4y−8∣5=∣5x+12y−13∣13\frac{|3x - 4y - 8|}{5} = \frac{|5x + 12y - 13|}{13} yields two linear equations: 14x−112y−39=014x - 112y - 39 = 0 and 64x+8y−169=064x + 8y - 169 = 0. The slopes of these lines are 1/8 and -8, respectively. The line with the positive slope has a slope of 1/8.

Step-by-Step Solution

1
Set up the distance formula from a point (x,y)(x,y) to both given lines.
The distance d1d_1 to the first line is ∣3x−4y−8∣32+(−4)2=∣3x−4y−8∣5\frac{|3x - 4y - 8|}{\sqrt{3^2 + (-4)^2}} = \frac{|3x - 4y - 8|}{5}. The distance d2d_2 to the second line is ∣5x+12y−13∣52+122=∣5x+12y−13∣13\frac{|5x + 12y - 13|}{\sqrt{5^2 + 12^2}} = \frac{|5x + 12y - 13|}{13}.
Points equidistant from both lines must satisfy d1=d2d_1 = d_2.
2
Equate the two distance expressions to represent the equidistant relationship.
∣3x−4y−8∣5=∣5x+12y−13∣13\frac{|3x - 4y - 8|}{5} = \frac{|5x + 12y - 13|}{13}
This represents the geometric condition of being equidistant from both lines.
3
Solve for Case 1 where the expressions inside the absolute values have the same sign.
13(3x−4y−8)=5(5x+12y−13)⇒39x−52y−104=25x+60y−65⇒14x−112y−39=013(3x - 4y - 8) = 5(5x + 12y - 13) \Rightarrow 39x - 52y - 104 = 25x + 60y - 65 \Rightarrow 14x - 112y - 39 = 0.
One of the two bisecting lines is found when the signs match.
4
Solve for Case 2 where the expressions inside the absolute values have opposite signs.
13(3x−4y−8)=−5(5x+12y−13)⇒39x−52y−104=−25x−60y+65⇒64x+8y−169=013(3x - 4y - 8) = -5(5x + 12y - 13) \Rightarrow 39x - 52y - 104 = -25x - 60y + 65 \Rightarrow 64x + 8y - 169 = 0.
The other bisecting line is found when the signs are opposite.
5
Find the slope of each resulting linear equation to identify the positive one.
For 14x−112y−39=014x - 112y - 39 = 0, the slope is −14−112=18-\frac{14}{-112} = \frac{1}{8}. For 64x+8y−169=064x + 8y - 169 = 0, the slope is −648=−8-\frac{64}{8} = -8.
The slope of a line in the standard form Ax+By+C=0Ax + By + C = 0 is given by −AB-\frac{A}{B}.

Key Concept

The set of points equidistant from two intersecting lines forms two perpendicular lines representing the angle bisectors of the original lines, which can be determined by equating their point-to-line distance formulas.

Alternative Method

Instead of using the distance formula, one could find the angle of inclination of each line using trigonometry: θ1=arctan⁡(3/4)\theta_1 = \arctan(3/4) and θ2=arctan⁡(−5/12)\theta_2 = \arctan(-5/12). The angle bisectors have inclinations at the average of these two angles plus or minus 90 degrees. Converting back to slopes using m=tan⁡(θ)m = \tan(\theta) yields the same results.
Estimated Time:3m 0s
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Topics

Linear Equations and Graphing

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