Chapter 10.5, Problem 28HP

To determine

To Evaluate: The value of x if the distance between $\left(1,2\right)$ and $\left(x,7\right)$ is 13 units.

Answer to Problem 28HP

The value of x is 13.

Explanation of Solution

Given information:

The distance between $\left(1,2\right)$ and $\left(x,7\right)$ is 13 units.

Calculation:

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

(Write the distance formula)

  $\Rightarrow \sqrt{{\left(x-1\right)}^2+{\left(7-2\right)}^2}=13$

Substituting the values of, $\left(x_1,y_1\right)=\left(1,2\right)$ , $\left(x_2,y_2\right)=\left(x,7\right)$ and $d=13$

  $\Rightarrow \sqrt{{\left(x-1\right)}^2+5^2}=13$

  $Evaluate\left(7-2\right)$ i.e.5

  $\Rightarrow \sqrt{{\left(x-1\right)}^2+25}=13$

(Evaluate $5^2$ )

  $\Rightarrow {\left[\sqrt{{\left(x-1\right)}^2+25}\right]}^2={13}^2$

(Squaring both sides)

  $\Rightarrow {\left(x-1\right)}^2+25=169$

(Evaluate ${13}^2$ )

  $\Rightarrow {\left(x-1\right)}^2+25-25=169-25$

(Subtract 25 from both sides)

  $\Rightarrow {\left(x-1\right)}^2=144$

(Simplify)

  $\Rightarrow \sqrt{{\left(x-1\right)}^2}=\sqrt{144}$

(Perform square root on both sides)

  $\Rightarrow x-1=12$

(Simplify)

  $\Rightarrow x-1+1=12+1$

(Adding 1 on both sides)

  $\Rightarrow x=13$

(Simplify)

Thus the value of x is 13..