Chapter I, Problem 1DE

Find the distance between each pair of points. Where appropriate, give an approximation to three decimal places.

$\left(2,6\right)\text{ and }\left(-4,-2\right)$

To determine

To calculate: The distance between points $\left(2,6\right)$ and $\left(-4,-\text{2}\right)$.

Answer to Problem 1DE

Solution:

The distance between points $\left(2,6\right)$ and $\left(-4,-\text{2}\right)$ is $\underset{\_}{10\text{ units}}$.

Explanation of Solution

Given Information:

The provided points are, $\left(2,6\right)$ and $\left(-4,-\text{2}\right)$.

Formula Used:

The distance between any two points $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$ is given by

$d=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$

Calculation:

Consider the provided points $\left(2,6\right)$ and $\left(-4,-\text{2}\right)$

Plot two points on a graph paper as follows:

Use distance formula to compute the distance between two given points,

$d=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$

Substitute $\left(x_1,y_1\right)=\left(2,6\right)$ and $\left(x_2,y_2\right)=\left(-4,-2\right)$,

$\begin{array}{c}d=\sqrt{{\left(-2-6\right)}^2+{\left\{-4-2\right\}}^2}\\ =\sqrt{{\left(-8\right)}^2+{\left(-6\right)}^2}\\ =\sqrt{64+36}\\ =\sqrt{100}\end{array}$

Take square root,

$\begin{array}{c}d=\sqrt{100}\\ =10\text{ units}\end{array}$

Therefore, the distance between the given points is $10\text{ units}$.