Chapter 14.5, Problem 28WE

To determine

To prove: The Theorem $14-5$ by the coordinate definition of a dilation $D_{O,k}:\left(x,y\right)\to \left(kx,ky\right)$ .

Explanation of Solution

Given information: The dilation is $D_{O,k}:\left(x,y\right)\to \left(kx,ky\right)$ .

Proof:

A transformation related to similarity rather than congruence is called dilation. The dilation $D_{O,k}$ has center $O$ and nonzero scale factor $k$ .

The dilation $D_{O,k}$ maps at any point $P$ to a point $P^{\prime }$ can be determined as follows:

(1) If $k$ is greater than 0, $P^{\prime }$ lie on $\overline{OP}$ and $O{P}^{\prime }=k\cdot OP$ .

(2) If $k$ is less than 0, $P\text{'}$ lies on the ray opposite to $\overline{OP}$ and $O{P}^{\prime }=\left|k\right|\cdot OP$ .

(3) The center $O$ is its own image.

The dilation is called an expansion for $\left|k\right|>1$ and the dilation is called an contraction for $\left|k\right|<1$ .

Consider a point $P\left(x,y\right)$ . Now consider a dilation $D_{O,2}$ .

  $D_{O,k}:P\to P^{\prime }$

Thus,

  $O{P}^{\prime }=\left|k\right|\cdot OP$

And,

  $\begin{array}{c}OP=\sqrt{{\left(x-0\right)}^2+{\left(y-0\right)}^2}\\ =\sqrt{x^2+y^2}\end{array}$

And

  $\begin{array}{c}O{P}^{\prime }=\left|2\right|\cdot OP\\ =\left|2\right|\cdot \sqrt{x^2+y^2}\end{array}$

As under the dilation the triangle should be similar therefore, the point $P\text{'}$ will lie on the line $OP$ .

The equation of the line $OP$ is,

  $y_1=\frac{y}{x}{x}_1$

As the point $P^{\prime }$ lies on the line $OP$ . So,

  $\begin{array}{c}\sqrt{x_1^2+y_1^2}=\left|k\right|\sqrt{x^2+y^2}\\ \sqrt{x_1^2+{\left(\frac{y}{x}{x}_1\right)}^2}=\left|k\right|\sqrt{x^2+y^2}\\ x_1^2\left[1+\left(\frac{y}{x}\right)\right]=k^2\left(x^2+y^2\right)\\ x_1^2=\frac{k^2\left(x^2+y^2\right)}{\left[1+\left(\frac{y}{x}\right)\right]}\end{array}$

Simplify further.

  $\begin{array}{c}{x}_1^2=\frac{k^2\left(x^2+y^2\right)}{\left[\frac{x^2+y^2}{x^2}\right]}\\ =k^2{x}^2\\ =\pm kx\end{array}$

Put the value of $x_1$ in the equation of line $OP$ .

  $\begin{array}{c}{y}_1=\frac{y}{x}kx\\ =ky\end{array}$

So, the coordinates of the point $P^{\prime }$ are $\left(kx,ky\right)$ .

Hence, the Theorem $14-5$ is proved by the coordinate definition of a dilation $D_{O,k}:\left(x,y\right)\to \left(kx,ky\right)$ .