Problem 36
Question
Write the slope-intercept equation of the line that passes through the given point and that is perpendicular to the given line. $$ (-3,0), x=1-6 y $$
Step-by-Step Solution
Verified Answer
The slope-intercept equation is \( y = 6x + 18 \).
1Step 1: Identify the given line's equation
The given equation is in the form \( x = 1 - 6y \). This can be rewritten as \( 6y = -x + 1 \) and then \( y = -\frac{1}{6}x + \frac{1}{6} \). This shows the slope of the given line is \(-\frac{1}{6}\).
2Step 2: Determine the perpendicular slope
Two perpendicular lines have slopes that are negative reciprocals. So, we must find the negative reciprocal of \(-\frac{1}{6}\). This is \(6\).
3Step 3: Use the point-slope form
The point-slope form of a line is \( y - y_1 = m(x - x_1) \). We know the point \((-3,0)\) and the slope \(m = 6\). Substitute these into the point-slope formula to get: \( y - 0 = 6(x + 3) \).
4Step 4: Simplify to slope-intercept form
Expand the equation from Step 3: \( y = 6(x + 3) \). Simplify this to get \( y = 6x + 18 \). This is the equation in slope-intercept form, \( y = mx + b \) where \( m = 6 \) and \( b = 18 \).
Key Concepts
Perpendicular LinesPoint-Slope FormNegative ReciprocalLinear Equations
Perpendicular Lines
Two lines are considered perpendicular if they meet at a right angle (90 degrees).
A key characteristic of perpendicular lines when represented in a Cartesian plane is their slopes.
The slopes of two perpendicular lines are negative reciprocals of each other.
A key characteristic of perpendicular lines when represented in a Cartesian plane is their slopes.
The slopes of two perpendicular lines are negative reciprocals of each other.
- For instance, if the slope of one line is \( m \), the slope of a line perpendicular to it will be \(-\frac{1}{m}\).
Point-Slope Form
The point-slope form of a linear equation is especially useful when you're provided with a point on the line and the slope, but not the y-intercept.
This form is written as \( y - y_1 = m(x - x_1) \).
Here, \( (x_1, y_1) \) represents the coordinates of the known point and \( m \) is the slope.
Using the point-slope form can make the process of finding an equation more straightforward:
This form is written as \( y - y_1 = m(x - x_1) \).
Here, \( (x_1, y_1) \) represents the coordinates of the known point and \( m \) is the slope.
Using the point-slope form can make the process of finding an equation more straightforward:
- All you need is a point the line goes through and the slope of the line.
- This form directly incorporates both of these pieces of information into the equation.
Negative Reciprocal
Finding a negative reciprocal is a key step in determining the slope of a line that is perpendicular to another line.
To find the negative reciprocal of a fraction, first flip the fraction and then change the sign.
For example, the negative reciprocal of \(-\frac{1}{6}\) is \(6\).
To find the negative reciprocal of a fraction, first flip the fraction and then change the sign.
For example, the negative reciprocal of \(-\frac{1}{6}\) is \(6\).
- Flipping the fraction \(-\frac{1}{6}\) gives \(\frac{6}{-1}\) and changing the sign makes it simply \(6\).
Linear Equations
Linear equations form the basis of many concepts in algebra and are used to represent straight lines on a graph.
A linear equation can take various forms, but the most common are slope-intercept form \( y = mx + b \) and point-slope form \( y - y_1 = m(x - x_1) \).
Here, \( m \) represents the slope of the line, and \( b \) is the y-intercept the point where the line crosses the y-axis.
A linear equation can take various forms, but the most common are slope-intercept form \( y = mx + b \) and point-slope form \( y - y_1 = m(x - x_1) \).
Here, \( m \) represents the slope of the line, and \( b \) is the y-intercept the point where the line crosses the y-axis.
- Slope-intercept form is especially useful for quickly identifying these features from the equation.
- Understanding different forms of linear equations is crucial for interpreting and graphing linear relationships efficiently.
Other exercises in this chapter
Problem 36
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