/*! 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 59 Find the distance between the po... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 6: Problem 59

Find the distance between the point and the line. Point \((1,1)\) Line \(y=x+1\)

Short Answer

Expert verified
The distance between the point (1,1) and the line y=x+1 is \(\dfrac{1}{\sqrt{2}}\).

Step by step solution

01

Identify the values of A, B, C, x1, and y1

For the line equation -x + y - 1 = 0, A = -1, B = 1, and C = -1. The point is given as (1,1), therefore x1 = 1 and y1 = 1.
02

Substitute these values into the distance formula

Using the values from Step 1, substitute into the formula to calculate the distance: \[ distance = \dfrac{|(-1)(1) + (1)(1) - 1|}{\sqrt{(-1)^2 + 1^2}} \]
03

Simplify the equation and evaluate the numerical distance

Simplify the equation and evaluate to get: \[ distance = \dfrac{1}{\sqrt{2}} \]

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.

Distance Formula
The concept of finding the distance between a point and a line involves a specific formula derived from the geometry principles. This formula is essential to calculate how far a point
  • Is located from a straight line, effectively giving the shortest path between them.
  • Provides insight into the relative position of points and lines on a plane.
The distance formula is expressed as: \[ \text{distance} = \dfrac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} \]Here,
  • \(A\), \(B\), and \(C\) are the coefficients in the standard form of the linear equation \(Ax + By + C = 0\).
  • The point is represented by \((x_1, y_1)\).
This formula efficiently calculates the perpendicular distance, which is crucial for various applications in geometry and physics.
Linear Equations
Linear equations form the basis of many calculations and are defined by equations that graph as straight lines. They have a general form of
  • \(y = mx + b\)
In this equation,
  • \(m\) is the slope of the line, representing how steep the line is, or the rate of change.
  • \(b\) is the y-intercept, which is the point where the line crosses the y-axis.
When rearranged into the standard form, an equation can transform into
  • \(Ax + By + C = 0\)
This form helps in utilizing various mathematical analyses, such as finding the distance from points.
For example, in finding the distance from a point to the line \(y=x+1\), we first rewrite it into the form \(-x + y - 1 = 0\). This allows us to identify coefficients \(A, B,\) and \(C\) necessary for the distance formula.
Coordinate Geometry
Coordinate geometry, also known as analytic geometry, involves using algebra to interpret geometric concepts.
  • This field merges the abstractness of algebra with the visuals of geometry, providing a bridge between mathematical concepts and spatial reasoning.
  • Key components include the use of coordinate planes, lines, and points, allowing us to evaluate relationships mathematically.
Understanding coordinate geometry helps visualize equations and assess various properties like distance, slope, and intersection.
This visualization is not only vital for academic purposes but also finds practical applications in fields
  • Such as engineering, computer graphics, and navigation.
  • Understanding how to position and relate geometric figures on a coordinate plane is a fundamental skill.
When tackling problems like finding the distance from a point to a line, coordinate geometry simplifies the process by offering a structured way of reasoning.

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

Convert the polar equation $$r=2(h \cos \theta+k \sin \theta)$$ to rectangular form and verify that it is the equation of a circle. Find the radius of the circle and the rectangular coordinates of the center of the circle.

A circle and a parabola can have \(0,1,2,3,\) or 4 points of intersection. Sketch the circle \(x^{2}+y^{2}=4 .\) Discuss how this circle could intersect a parabola with an equation of the form \(y=x^{2}+C .\) Then find the values of \(C\) for each of the five cases described below. Use a graphing utility to verify your results. (a) No points of intersection (b) One point of intersection (c) Two points of intersection (d) Three points of intersection (e) Four points of intersection

Determine whether the statement is true or false. Justify your answer. A line that has an inclination greater than \(\pi / 2\) radians has a negative slope.

In Exercises \(71-90,\) convert the rectangular equation to polar form. Assume \(a > 0\). $$x^{2}+y^{2}=a^{2}$$

Determine whether the statement is true or false. Justify your answer. If the vertex and focus of a parabola are on a horizontal line, then the directrix of the parabola is vertical.
See all solutions

Recommended explanations on Math Textbooks

Applied Mathematics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Decision Maths

Read Explanation

Geometry

Read Explanation

Mechanics Maths

Read Explanation

Discrete Mathematics

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.