Chapter 1: Problem 48
Find an equation of the line that is perpendicular to the given line \(l\) and passes through the given point \(P\). \(l: y+4=-\frac{3}{5}\left(x-\frac{1}{2}\right) ; P=\left(-1, \frac{1}{2}\right)\)
Short Answer
Expert verified
The equation is \( y = \frac{5}{3}x + \frac{13}{6} \).
Step by step solution
01
Identify the Slope of Given Line
The equation of the given line is in point-slope form: \( y + 4 = -\frac{3}{5}(x - \frac{1}{2}) \). From this equation, we can see that the slope \( m_1 \) of the line \( l \) is \( -\frac{3}{5} \).
02
Find the Slope of the Perpendicular Line
Since perpendicular lines have slopes that are negative reciprocals of each other, the slope \( m_2 \) of the line we want to find is the negative reciprocal of \( -\frac{3}{5} \). Calculating this gives \( m_2 = \frac{5}{3} \).
03
Write the Equation of the Perpendicular Line in Point-Slope Form
Using the point \( P(-1, \frac{1}{2}) \) and the slope \( \frac{5}{3} \), plug these into the point-slope form: \( y - y_1 = m(x - x_1) \). This results in: \( y - \frac{1}{2} = \frac{5}{3}(x + 1) \).
04
Simplify the Equation
Expand and simplify the equation. Distribute \( \frac{5}{3} \) over \( (x + 1) \) to get \( y - \frac{1}{2} = \frac{5}{3}x + \frac{5}{3} \). Then add \( \frac{1}{2} \) to both sides to isolate \( y \). The simplified equation becomes \( y = \frac{5}{3}x + \frac{13}{6} \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Point-Slope Form
The point-slope form is a very useful way to write the equation of a line. It's mostly used when you know:
Using point-slope form makes finding the equation very straightforward. Once you know a single point of where your line crosses and the direction or steepness of the line, you're all set to write the equation.
- a point on the line,
- and the slope of the line.
- \( (x_1, y_1) \) is a point on the line.
- \( m \) is the slope of the line.
Using point-slope form makes finding the equation very straightforward. Once you know a single point of where your line crosses and the direction or steepness of the line, you're all set to write the equation.
Negative Reciprocal
The concept of a negative reciprocal is important when dealing with perpendicular lines. If you know the slope of one line, you can easily find the slope of a line perpendicular to it. Here's how it works:
1. Flip the fraction.2. Change the sign.
In the problem given, the line \( l \) had a slope of \( -\frac{3}{5} \). We flipped the fraction to get \( \frac{5}{3} \) and changed its sign to find the perpendicular slope. By knowing the slope of the given line, you're equipped to find the slope of its perpendicular counterpart.
- First, understand what a reciprocal is: It's when you flip the fraction. For example, the reciprocal of \( \frac{3}{5} \) is \( \frac{5}{3} \).
- The negative reciprocal changes the sign as well. So, starting with \( -\frac{3}{5} \), the negative reciprocal is \( \frac{5}{3} \).
1. Flip the fraction.2. Change the sign.
In the problem given, the line \( l \) had a slope of \( -\frac{3}{5} \). We flipped the fraction to get \( \frac{5}{3} \) and changed its sign to find the perpendicular slope. By knowing the slope of the given line, you're equipped to find the slope of its perpendicular counterpart.
Slope of a Line
The slope of a line measures its steepness and direction. It can be seen in how much y changes for a change in x between two points on the line. The formula for finding the slope \( m \) between two points \((x_1, y_1)\) and \((x_2, y_2)\) is:\[ m = \frac{y_2 - y_1}{x_2 - x_1} \]A positive slope indicates a line rising from left to right, while a negative slope shows a line falling from left to right. A zero slope means a horizontal line, and an undefined slope refers to a vertical line (where \( x_1 = x_2 \)).
- If the line goes upwards as you move along it from left to right, it has a positive slope.
- If it goes downwards, it has a negative slope.
- If it is flat and horizontal, the slope is zero.
- If it is vertical, the slope is undefined.