Chapter 14.5, Problem 26WE

To determine

To show that $P\text{'}Q\text{'}=\left|k\right|\sqrt{{\left(x_1-x_2\right)}^2+{\left(y_1-y_2\right)}^2}=\left|k\right|\cdot PQ$ by using the distance formula.

Explanation of Solution

Given information:

The dilation $D_{O,k}$ maps $\overline{PQ}\text{ to }\overline{P\text{'}Q\text{'}}$ .

  

Formula used:

The following formula is used:

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

Proof:

Calculate the distance $PQ$ :

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

Calculate the distance $P\text{'}Q\text{'}$ :

  $\begin{array}{l}P\text{'}Q\text{'}=\sqrt{{\left(k{x}_1-k{x}_2\right)}^2+{\left(k{y}_1-k{y}_2\right)}^2}\\ P\text{'}Q\text{'}=\sqrt{k^2{\left(x_1-x_2\right)}^2+k^2{\left(y_1-y_2\right)}^2}\\ P\text{'}Q\text{'}=\pm k\sqrt{{\left(x_1-x_2\right)}^2+{\left(y_1-y_2\right)}^2}\\ P\text{'}Q\text{'}=\left|k\right|\cdot PQ\left[PQ=\sqrt{{\left(x_1-x_2\right)}^2+{\left(y_1-y_2\right)}^2}\right]\end{array}$

Hence, it is proved that $P\text{'}Q\text{'}=\left|k\right|\cdot PQ$ .