Chapter 14.3, Problem 23WE

To determine

To proof:Points $P\&Q$ and their image points $P\text{'}\&Q\text{'}$ where $T$ is isometry i.e. $PQ=P\text{'}Q\text{'}$ .

Answer to Problem 23WE

$T$ is isometry where $PQ=P\text{'}Q\text{'}$ .

Explanation of Solution

Given information :

Under translation $T$ , points $P\&Q$ have their image points $P\text{'}\&Q\text{'}$ .

Concept used:

A translation moves ("slides") an object a fixed distance in a given direction without changing its size or shape, and without turning it or flipping it.The original object is called the pre-image, and the translation is called the image.

Under translation $T$ , points $P\&Q$ have their image points $P\text{'}\&Q\text{'}$ .

Below diagram is provided in problem.

Translation, $T:(x,y)\to (x+a,y+b)$

Where points $P(x_1,y_1)\&Q(x_2,y_2)$

Their image points will be:

  $\begin{array}{l}T:(P)=(x_1,y_1)\to (x_1+a,y_1+b)\\ T:(Q)=(x_2,y_2)\to (x_2+a,y_2+b)\end{array}$

We will use the distance formula to determine the isometry of $T$

For example

  $\begin{array}{l}M(a,b)\&N(c,d)\\ MN=\sqrt{{(c-a)}^2+{(d-b)}^2}\end{array}$

So using above formula,

  $P=(x_1,y_1),Q=(x_2,y_2),P\text{'}=(x_1+a,y_1+b),Q\text{'}=(x_2+a,y_2+b)$ ……………. (I)

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

And $\begin{array}{l}P\text{'}Q\text{'}=\sqrt{{((x_2+a)-(x_1+a))}^2+{((y_2+b)-(y_1+b))}^2}\\ \Rightarrow \sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}\end{array}$ ……………………... (II)

(Here when we open bracket $a$ & $b$ get cancelled)

Comparing I & II

  $PQ=P\text{'}Q\text{'}$

Hence,

$T$ is isometry where $PQ=P\text{'}Q\text{'}$ .