Question

Difficulty: HardSlope of a Line

In the standard (x,y)(x, y) coordinate plane, a line LL passes through the point (2,−3)(2, -3) and does not pass through the origin. The distance from the origin to the line's xx-intercept is twice the distance from the origin to the line's yy-intercept. Which of the following could be the slope of line LL?

  1. A
    −2-2
  2. −12-\frac{1}{2}Answer
  3. C
    23\frac{2}{3}
  4. D
    11
  5. E
    22

Answer

The slope of the line could be −12-\frac{1}{2}.
The correct answer is −12-\frac{1}{2}. By setting the xx-intercept of the line to (a,0)(a, 0) and the yy-intercept to (0,b)(0, b), the distance condition gives ∣a∣=2∣b∣|a| = 2|b|. Substituting the point (2,−3)(2, -3) into the intercept equation of a line, xa+yb=1\frac{x}{a} + \frac{y}{b} = 1, yields 2a−3b=1\frac{2}{a} - \frac{3}{b} = 1. Solving the two cases (a=2ba = 2b and a=−2ba = -2b) gives the intercept pairs (−4,−2)(-4, -2) and (8,−4)(8, -4). Calculating the slope m=−bam = -\frac{b}{a} for both pairs results in −12-\frac{1}{2} and 12\frac{1}{2}. Since −12-\frac{1}{2} is one of these possible values, it is the correct choice.

Step-by-Step Solution

1
Represent the line LL using its intercepts. Let the xx-intercept of LL be (a,0)(a, 0) and the yy-intercept be (0,b)(0, b). Since the line does not pass through the origin, we have a≠0a \neq 0 and b≠0b \neq 0.
The equation of the line can be written in intercept form as: xa+yb=1\frac{x}{a} + \frac{y}{b} = 1
Using intercept form allows us to directly relate the coordinates of the intercepts to the given point and the distance condition.
2
Translate the distance condition and the given point into equations. The distance from the origin to the xx-intercept is ∣a∣|a|, and the distance to the yy-intercept is ∣b∣|b|. The problem states that the distance to the xx-intercept is twice the distance to the yy-intercept, so ∣a∣=2∣b∣|a| = 2|b|. Additionally, the line passes through (2,−3)(2, -3), so we substitute x=2x = 2 and y=−3y = -3 into the intercept equation.
∣a∣=2∣b∣|a| = 2|b| and 2a−3b=1\frac{2}{a} - \frac{3}{b} = 1
This sets up a system of equations to solve for the unknown intercepts aa and bb.
3
Solve the system of equations by analyzing the two cases for the absolute value: a=2ba = 2b and a=−2ba = -2b.
Case 1: If a=2ba = 2b, then 22b−3b=1  ⟹  1b−3b=1  ⟹  −2b=1  ⟹  b=−2\frac{2}{2b} - \frac{3}{b} = 1 \implies \frac{1}{b} - \frac{3}{b} = 1 \implies -\frac{2}{b} = 1 \implies b = -2, which gives a=−4a = -4. Case 2: If a=−2ba = -2b, then 2−2b−3b=1  ⟹  −1b−3b=1  ⟹  −4b=1  ⟹  b=−4\frac{2}{-2b} - \frac{3}{b} = 1 \implies -\frac{1}{b} - \frac{3}{b} = 1 \implies -\frac{4}{b} = 1 \implies b = -4, which gives a=8a = 8.
Resolving the absolute value yields the exact coordinates of the intercepts for both valid scenarios.
4
Calculate the slope mm for both cases using the formula m=−bam = -\frac{b}{a}.
For Case 1, m=−−2−4=−12m = -\frac{-2}{-4} = -\frac{1}{2}. For Case 2, m=−−48=12m = -\frac{-4}{8} = \frac{1}{2}.
The slope of a line with intercepts (a,0)(a, 0) and (0,b)(0, b) is given by −ba-\frac{b}{a}.

Key Concept

Slope of a Line
Estimated Time:2m 0s
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