Chapter 8.4, Problem 43E

Solution

Find the value of $x$ .

  $x=-10$ .

Given information:

Given points are $\left(x,7\right)\text{ and }\left(-4,1\right)$ .

Distance ‘ d’between the given two points is $6\sqrt{2}$ .

Formula used:

• The distance between two points, $\left(x_1,y{1}_{}\right),\text{ and }\left(x_2,y{2}_{}\right)$ , is given by:

  $\sqrt{{\left(x_2-x_1\right)}^2+{\left(y{2}_{}-y{1}_{}\right)}^2}$

Calculation:

To calculate the value of $x$ , use the formula for the distance between two points $\left(x_1,y{1}_{}\right),\text{ and }\left(x_2,y{2}_{}\right)$ that is

  $d=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y{2}_{}-y{1}_{}\right)}^2}$

Here, distance is $6\sqrt{2}$ , thus the value of $x$ is calculated as:

  $\begin{array}{c}6\sqrt{2}=\sqrt{{\left(-4-x\right)}^2+{\left(1-7\right)}^2}\\ =\sqrt{{\left(-4-x\right)}^2+{\left(-6\right)}^2}\\ 6\sqrt{2}=\sqrt{{\left(-4-x\right)}^2+36}\\ {\left(6\sqrt{2}\right)}^2={\left(\sqrt{{\left(-4-x\right)}^2+36}\right)}^2\\ 36\times 2={\left(-4-x\right)}^2+36\\ 72={\left(-4-x\right)}^2+36\\ {\left(-4-x\right)}^2=72-36\\ {\left(-4-x\right)}^2=36\\ -4-x=\sqrt{36}\\ -4-x=6\\ -x=6+4\\ -x=10\\ x=-10\end{array}$

Thus, the value of $x$ is $-10$ .