Chapter 14.3, Problem 15WE

a.

To determine

To graph:Points $A(-1,2),B(0,6),A\text{'}\&B\text{'}$

Answer to Problem 15WE

Points $A,B,A\text{'}\&B\text{'}$ are graphed.

Explanation of Solution

Given:

  $\begin{array}{l}T(A)=A\text{'}\\ T(B)=B\text{'}\end{array}$

Translation T glides all point 3 units right & 4 units down.

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.

Calculation:

As per the given problem

  $\begin{array}{l}T(A)=A\text{'}\\ T(B)=B\text{'}\end{array}$

Translation T glides all point 3 units right & 4 units down.

  $\Rightarrow T:(x,y)\to (x+3,y-4)$

  $\begin{array}{l}T(A):(-1,2)\to (-1+3,2-4)=(2,-2)=A\text{'}\\ T(B):(0,6)\to (0+3,6-4)=(3,2)=B\text{'}\end{array}$

  $\therefore A\text{'}=(2,-2)\&B\text{'}=(3,2)$

On graph above, points $A,B,A\text{'}\&B\text{'}$ are plot.

Hence,

Points $A,B,A\text{'}\&B\text{'}$ are graphed.

b.

To determine

To find: Kind of figure $AA\text{'}BB\text{'}$ &it’s perimeter.

Answer to Problem 15WE

The figure of $AA\text{'}BB\text{'}$ is parallelogram & perimeter is $6\sqrt{17}$ .

Explanation of Solution

Given:

  $\begin{array}{l}T(A)=A\text{'}\\ T(B)=B\text{'}\end{array}$

Translation T glides all point 3 units right & 4 units down.

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.

Calculation:

As per the given problem

  $A=(-1,2),B=(0,6),A\text{'}=(2,-2)\&B\text{'}=(3,2)$

As we refer diagram in above sub-part, figure of $AA\text{'}BB\text{'}$ is parallelogram.

Perimeter = $AA\text{'}+A\text{'}B\text{'}+B\text{'}B+BA$

Using the distance formula,

  $\begin{array}{l}P(x,y)\&Q(x\text{'},y\text{'})\\ PQ=\sqrt{{(x\text{'}-x)}^2+{(y\text{'}-y)}^2}\end{array}$

  $\begin{array}{l}AB=\sqrt{{(-1-0)}^2+{(2-6)}^2}=\sqrt{17}\\ A\text{'}B=\sqrt{{(2-0)}^2+{(-2-6)}^2}=2\sqrt{17}\end{array}$

Since, $AA\text{'}BB\text{'}$ is parallelogram

  $\begin{array}{l}AB=A\text{'}B\text{'}\\ A\text{'}B=BA\text{'}\end{array}$

Perimeter = $\sqrt{17}+\sqrt{17}+2\sqrt{17}+2\sqrt{17}=6\sqrt{17}$

Hence,

The figure of $AA\text{'}BB\text{'}$ is parallelogram & perimeter is $6\sqrt{17}$ .