Chapter 3: Problem 38
Find an equation of the line passing through each pair of points. Write the equation in the form $A x+B y=C. $$ (0,0) \text { and }\left(-\frac{1}{2}, \frac{1}{3}\right) $$
Short Answer
Expert verified
The equation is \(2x + 3y = 0\).
Step by step solution
01
Find the Slope of the Line
To find the slope \( m \) of the line passing through two points \((x_1, y_1)\) and \((x_2, y_2)\), use the formula:\[ m = \frac{y_2 - y_1}{x_2 - x_1} \]For the points \((0,0)\) and \((-\frac{1}{2}, \frac{1}{3})\), we can substitute:\[ m = \frac{\frac{1}{3} - 0}{-\frac{1}{2} - 0} = \frac{\frac{1}{3}}{-\frac{1}{2}} = -\frac{2}{3} \]
02
Use Point-Slope Form to Find Line Equation
The equation of a line in point-slope form is given by:\[ y - y_1 = m(x - x_1) \]Substituting \( m = -\frac{2}{3} \) and the point \((0,0)\), the equation becomes:\[ y - 0 = -\frac{2}{3}(x - 0) \]Thus,\[ y = -\frac{2}{3}x \]
03
Convert to Standard Form \(Ax + By = C\)
To convert \( y = -\frac{2}{3}x \) to the form \(Ax + By = C\), we can multiply through by 3 to eliminate the fraction:\[ 3y = -2x \]Rearrange to obtain the standard form:\[ 2x + 3y = 0 \]
04
Conclusion
The equation of the line passing through the given points \((0,0)\) and \((-\frac{1}{2}, \frac{1}{3})\) in the standard form is:\[ 2x + 3y = 0 \]
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Slope
To determine how steep a line is as it moves across a coordinate plane, we use a concept called "slope." Slope is essentially a measurement of how much a line rises or falls for each unit of movement horizontally. Think of it as the 'tilt' of the line. This can help us understand the direction and the steepness of a line.
The formula to calculate the slope, typically denoted by the letter \( m \), is:
For example, given points \((0,0)\) and \(\left(-\frac{1}{2}, \frac{1}{3}\right)\), using the slope formula gives us \(-\frac{2}{3}\). This negative slope indicates that the line decreases as it moves from left to right.
The formula to calculate the slope, typically denoted by the letter \( m \), is:
- \( m = \frac{y_2 - y_1}{x_2 - x_1} \)
For example, given points \((0,0)\) and \(\left(-\frac{1}{2}, \frac{1}{3}\right)\), using the slope formula gives us \(-\frac{2}{3}\). This negative slope indicates that the line decreases as it moves from left to right.
Point-Slope Form
Once we have the slope, we can use the "point-slope form" to express the equation of a line. This form is very handy when you know the slope of a line and one point through which it passes. The point-slope form is represented by the equation:
- \(y - y_1 = m(x - x_1)\)
- \(y - 0 = -\frac{2}{3}(x - 0)\)
- \(\Rightarrow y = -\frac{2}{3}x\)
Standard Form
The standard form of a linear equation is useful for a variety of algebraic manipulations. It takes the form \(Ax + By = C\), where \(A\), \(B\), and \(C\) are integers, and \(A\) should be a non-negative integer.
To convert from the point-slope form (or slope-intercept form like \(y = -\frac{2}{3}x\)), we need to rewrite the equation so that there are no fractions and all terms are on one side of the equation. We transform:
To convert from the point-slope form (or slope-intercept form like \(y = -\frac{2}{3}x\)), we need to rewrite the equation so that there are no fractions and all terms are on one side of the equation. We transform:
- Multiply the entire equation by 3 to clear fractions: \(3y = -2x\)
- Rearrange: \(2x + 3y = 0\)
Algebra
Algebra is fundamental in mathematics for solving equations and understanding variables and constants. When we find the equation of a line, we're practicing algebra.
Algebra involves operations with symbols (like \(x\) and \(y\)) to find unknowns. Steps like finding the slope, using the point-slope form, and converting to standard form are great examples.
Here’s how algebra simplifies our line equation problem:
Algebra involves operations with symbols (like \(x\) and \(y\)) to find unknowns. Steps like finding the slope, using the point-slope form, and converting to standard form are great examples.
Here’s how algebra simplifies our line equation problem:
- We use substitution to apply specific values for slope and known points.
- We rearrange terms for clarity and simplicity according to the standard form rules.