/*! 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 82 Find the distance between the pa... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 82

Find the distance between the pair of points. $$(4,2) \text { and }(-2,-4)$$

Short Answer

Expert verified
\The distance between \(4, 2\) and \((-2, -4)\) is \ 6\sqrt{2}}

Step by step solution

01

Determine the Distance Formula

The distance between two points \(x_1, y_1\) and \(x_2, y_2\) on a coordinate plane is given by the formula \[d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\]
02

Substitute the Coordinates

Substitute \((x_1, y_1) = (4, 2)\) and \((x_2, y_2) = (-2, -4)\) into the distance formula. \[d = \sqrt{((-2) - 4)^2 + ((-4) - 2)^2}\]
03

Simplify the Expressions

First, simplify the expressions inside the parentheses: \((-2) - 4 = -6\) and \((-4) - 2 = -6\). The distance formula now becomes: \[d = \sqrt{(-6)^2 + (-6)^2}\]
04

Square and Sum

Next, square each term and find their sum: \[(-6)^2 = 36\] \[36 + 36 = 72\]
05

Take the Square Root

Finally, take the square root of the sum to find the distance: \[d = \sqrt{72} \] Simplify further to get the exact value: \[d = 6\sqrt{2}\]

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.

Coordinate plane
The coordinate plane is a fundamental concept in geometry. It's a two-dimensional plane formed by the intersection of a horizontal axis (called the x-axis) and a vertical axis (called the y-axis). These axes divide the plane into four quadrants. Each point on the plane is defined by an ordered pair \((x, y)\), where \(x\) represents the horizontal position and \(y\) represents the vertical position.

Understanding the coordinate plane is essential for solving problems involving distance between points. Points like \((4, 2)\) and \((-2, -4)\) are plotted based on their respective \(x\) and \(y\) values.
Distance calculation
Calculating the distance between two points on a coordinate plane uses the Distance Formula. This formula is derived from the Pythagorean theorem, which relates the length of the sides of a right triangle to the length of its hypotenuse.

The Distance Formula is given by \[d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\right. \] Here, \(d\) stands for distance, \( (x_1, y_1) \), and \( (x_2, y_2) \) are the coordinates of the two points.

Using this formula, you first find the differences between the corresponding \(x\) and \(y\) values, then square those differences, add them together, and take the square root of the sum.
Square root
A square root is a value that, when multiplied by itself, gives the original number. In the context of the Distance Formula, taking the square root of the sum of the squared distances yields the actual distance between two points.

For example, if you need to find the distance between points \((4, 2)\) and \((-2, -4)\), you calculate the sum of the squares of \(-6\) and \(-6\) as follows: \[d = \sqrt{(-6)^2 + (-6)^2} = \sqrt{36 + 36} = \sqrt{72} = 6\sqrt{2} \]
By simplifying the square root, you achieve the precise distance between these points, \(6\sqrt{2}\).

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

Use synthetic division and the remainder theorem to find the zeros. $$f(x)=-x^{3}+3 x^{2}+6 x-8$$

Factor the polynomial function \(f(x) .\) Then solve the equation \(f(x)=0\) $$f(x)=x^{3}-x^{2}-14 x+24$$

Solve. $$2 x^{3}+6 \leq 5 x^{2}+x$$

Determine the degree and the leading term of the polynomial function. $$f(x)=\left(x^{5}-1\right)^{2}\left(x^{2}+2\right)^{3}$$

Fill in the blank with the correct term. Some of the given choices will not be used. Others will be used more than once. $$\begin{array}{ll}x \text { -intercept } & \text { midpoint formula } \\ y \text { -intercept } & \text { horizontal lines } \\ \text { odd function } & \text { vertical lines } \\ \text { even function } & \text { point-slope equation } \\ \text { domain } & \text { slope-intercept equation } \\ \text { range } & \text { difference quotient } \\ \text { slope } & f(x)=f(-x) \\\ \text { distance formula } & f(-x)=-f(x)\end{array}$$ The _______ of the line with slope \(m\) and \(y\) -intercept \((0, b)\) is \(y=m x+b\)
See all solutions

Recommended explanations on Math Textbooks

Probability and Statistics

Read Explanation

Logic and Functions

Read Explanation

Statistics

Read Explanation

Discrete Mathematics

Read Explanation

Decision Maths

Read Explanation

Geometry

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.