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Chapter 0: Problem 8

Find all lines through \((6,-1)\) for which the product of the \(x\) - and \(y\) -intercepts is 3 .

Short Answer

Expert verified
The lines are \( y = -\frac{1}{12}x - \frac{1}{2} \) and \( y = -\frac{1}{3}x + 1 \).

Step by step solution

01

Find the general equation of the line

A line passing through the point (6, -1) can be expressed in the slope-intercept form as \( y - (-1) = m(x - 6) \) or simplified to \( y + 1 = m(x - 6) \).This can be further expanded to: \( y = mx - 6m - 1 \).
02

Determine the x- and y-intercepts

For y-intercept: Set \( x = 0 \) in the equation \( y = mx - 6m - 1 \). So,\( y = m(0) - 6m - 1 = -6m - 1 \).For x-intercept: Set \( y = 0 \), then\( 0 = mx - 6m - 1 \).Solving for x gives\( mx = 6m + 1 \) and finally \( x = \frac{6m + 1}{m} \) for \( m eq 0 \).
03

Set the product of intercepts equal to 3

The product of the x-intercept and y-intercept is:\( \left(\frac{6m + 1}{m}\right)(-6m - 1) \).Setting this equal to 3 gives:\( \frac{(6m + 1)(-6m - 1)}{m} = 3 \).
04

Simplify and solve the equation

Expanding the numerator of the product gives:\( (6m + 1)(-6m - 1) = -36m^2 - 6m - 6m - 1 = -36m^2 - 12m - 1 \).Thus, the equation becomes:\( \frac{-36m^2 - 12m - 1}{m} = 3 \).This simplifies to:\( -36m - 12 - \frac{1}{m} = 3 \), since \( m eq 0 \).
05

Further algebraic manipulation

Reorganize the equation to isolate terms without fractions:\( -36m - 12 - 3 = \frac{1}{m} \).So,\( -36m - 15 = \frac{1}{m} \).
06

Solve for m

Multiply both sides by \( m \) (ensuring \( m eq 0 \)) to eliminate the fraction:\( -36m^2 - 15m = 1 \).This equation is a quadratic: \( 36m^2 + 15m + 1 = 0 \). Use the quadratic formula: \( m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 36, b = 15, c = 1 \).Calculate:\( m = \frac{-15 \pm \sqrt{225 - 144}}{72} = \frac{-15 \pm \sqrt{81}}{72} \).So,\( m = \frac{-15 \pm 9}{72} \).The solutions are \( m = -\frac{6}{72} = -\frac{1}{12} \) and \( m = -\frac{24}{72} = -\frac{1}{3} \).
07

Write the equations of the lines

Substitute \( m = -\frac{1}{12} \):\( y = -\frac{1}{12}x + \frac{1}{2} - 1 \).Simplifying gives:\( y = -\frac{1}{12}x - \frac{1}{2} \).Now substitute \( m = -\frac{1}{3} \):\( y = -\frac{1}{3}x + 2 - 1 \).Simplifying gives:\( y = -\frac{1}{3}x + 1 \).The lines are \( y = -\frac{1}{12}x - \frac{1}{2} \) and \( y = -\frac{1}{3}x + 1 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

x-intercept
The **x-intercept** of a line is the point where the line crosses the x-axis. At this point, the y-coordinate is zero. To find the x-intercept, we set the equation of the line equal to zero and solve for the x-value.
In this exercise, the line is given in the form \[ y = mx - 6m - 1.\] To find the x-intercept, we substitute \( y = 0 \), resulting in:
  • \( 0 = mx - 6m - 1 \).
Solving for x gives:
  • \( mx = 6m + 1 \)
  • \( x = \frac{6m + 1}{m} \)
for \( m eq 0 \). This solution highlights how the slope parameter \( m \) affects the x-intercept based on the specific slope of the line through the given point.
y-intercept
The **y-intercept** of a line is the point where it crosses the y-axis. Here, the x-coordinate is zero. Finding the y-intercept involves substituting \( x = 0 \) in the equation of the line. For the exercise at hand, the line equation is:\[ y = mx - 6m - 1. \]Replacing \( x \) with zero gives:
  • \( y = m(0) - 6m - 1 = -6m - 1 \).
The calculation shows how the slope \( m \) directly influences the y-intercept. This interaction is crucial because both the x- and y-intercepts depend on \( m \) and are key to verifying the condition when their product is a given value, such as 3 in this problem.
quadratic equation
A **quadratic equation** is a polynomial equation of the form \( ax^2 + bx + c = 0 \). The exercise involves solving such an equation to find the slope \( m \). After setting the product of intercepts to 3:
  • \( \frac{(6m + 1)(-6m - 1)}{m} = 3 \).
This simplifies to:
  • \( -36m^2 - 12m - 1 = 3m \).
Further simplification gives the quadratic:
  • \( 36m^2 + 15m + 1 = 0 \).
To solve, we use the quadratic formula, \( m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 36 \), \( b = 15 \), \( c = 1 \). Substituting these values, we find:
  • \( m = \frac{-15 \pm 9}{72} \), resulting in \( m = -\frac{1}{12} \) and \( m = -\frac{1}{3} \).
This process shows how crucial it is to understand solving quadratic equations, as it directly helps find the equations of lines fulfilling given conditions.

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Most popular questions from this chapter

(a) Use the graph of \(y=\sqrt{x}\) to help sketch the graph of \(y=\sqrt{|x|}\) (b) Use the graph of \(y=\sqrt[3]{x}\) to help sketch the graph of \(y=\sqrt[3]{|x|}\)

True-False Determine whether the statement is true or false. Explain your answer. Curves in the family \(y=-5 \sin (A \pi x)\) have amplitude 5 and period \(2 /|A|\).

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Prove: (a) \(\cos ^{-1}(-x)=\pi-\cos ^{-1} x\) (b) \(\sec ^{-1}(-x)=\pi-\sec ^{-1} x\)

Sketch the graph of a function that is one-to-one on \((-\infty,+\infty)\), yet not increasing on \((-\infty,+\infty)\) and not decreasing on \((-\infty,+\infty)\)
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