/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 36 Write the slope-intercept equati... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 1: Problem 36

Write the slope-intercept equation of the line that passes through the given point and that is perpendicular to the given line. $$ (-3,0), x=1-6 y $$

Short Answer

Expert verified
The slope-intercept equation is \( y = 6x + 18 \).

Step by step solution

01

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 \(-\frac{1}{6}\).
02

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 \(6\).
03

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: \( y - 0 = 6(x + 3) \).
04

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 \( b = 18 \).

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 91影视!

Key Concepts

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

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.
  • For instance, if the slope of one line is \( m \), the slope of a line perpendicular to it will be \(-\frac{1}{m}\).
By identifying that two lines are perpendicular, you can determine one slope if you know the other.
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:
  • 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.
After obtaining the equation in point-slope form, you can easily convert it to slope-intercept form, which is more straightforward to work with for graphing purposes.
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 \(6\).
  • Flipping the fraction \(-\frac{1}{6}\) gives \(\frac{6}{-1}\) and changing the sign makes it simply \(6\).
This concept helps in quickly solving problems involving perpendicular lines on a coordinate grid.
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.
  • 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.
Knowing how to manipulate these forms can aid in solving a wide variety of algebraic problems.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Graph the curves \(\mathcal{C}\) and \(\mathcal{C}^{\prime}\) in the same viewing window. \(\mathcal{C}=\left\\{(x, y): y=(x+1) /\left(x^{4}+1\right)\right\\} ; \mathcal{C}^{\prime}\) is obtained by reflecting \(\mathcal{C}\) about the origin.

Describe and sketch the curve that has the given parametric equations. \(x=t^{2}, y=t^{2}\)

A function \(f: S \rightarrow T\) is specified. Determine if \(f\) is invertible. If it is, state the formula for \(f^{-1}(t) .\) Otherwise, state whether \(f\) fails to be one-to-one, onto, or both. \(S=(1,4), T=(4,5), f(s)=\sqrt{s}+3\)

In Exercises \(48-52,\) describe the curve that is the graph of the given parametric equations. \(x=7, y=t^{2}+1,-1 \leq t \leq 2\)

Plot the given region. \(\left\\{(x, y): 1 \leq x^{2}-2 x-\sqrt{3} y^{2}+2 \sqrt{3} y\right.\) and \(\left.y \leq \sqrt{5}+\sqrt{3} x-x^{2}\right\\}\) Plot the parabola with Cartesian equation \(y=2+x+x^{2}\)
See all solutions

Recommended explanations on Math Textbooks

Applied Mathematics

Read Explanation

Geometry

Read Explanation

Discrete Mathematics

Read Explanation

Logic and Functions

Read Explanation

Mechanics Maths

Read Explanation

Decision Maths

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.