Chapter 14.3, Problem 16WE

a.

To determine

To graph: $\Delta POQ$

Answer to Problem 16WE

  $\Delta POQ$ is graphed.

Explanation of Solution

Given:

  $P(0,3),O(0,0)\&Q(6,0)$

Concept Used:

Graph is plot based on points (x-axis, y-axis) coordinates available. Using 2-dimension cartesian plane, this can be easily done.

Calculation:

As per the given problem

  $P(0,3),O(0,0)\&Q(6,0)$

On graph above, using points $P(0,3),O(0,0)\&Q(6,0)$

  $\Delta POQ$ is plot.

Hence,

  $\Delta POQ$ is graphed.

b.

To determine

To graph: $\Delta P\text{'}O\text{'}Q\text{'}\&\Delta P"O"Q"$

Answer to Problem 16WE

  $\Delta P\text{'}O\text{'}Q\text{'}\&\Delta P"O"Q"$ is graphed.

Explanation of Solution

Given:

  $T_1:(x,y)\to (x+2,y-4)\&T_2:(x,y)\to (x+5,y+6)$

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

  $T_1:(x,y)\to (x+2,y-4)\&T_2:(x,y)\to (x+5,y+6)$

  $\begin{array}{l}{T}_1:\Delta POQ\to \Delta P\text{'}O\text{'}Q\text{'}\\ T_2:\Delta P\text{'}O\text{'}Q\text{'}\to \Delta P\text{'}\text{'}O\text{'}\text{'}Q\text{'}\text{'}\end{array}$

  $P(0,3),O(0,0)\&Q(6,0)$

  $\begin{array}{l}{T}_1:P(0,3)\to (0+2,3-4)=(2,-1)=P\text{'}\\ T_1:O(0,0)\to (0+2,0-4)=(2,-4)=O\text{'}\\ T_1:Q(6,0)\to (6+2,0-4)=(8,-4)=Q\text{'}\end{array}$

  $\begin{array}{l}{T}_2:P\text{'}(2,-1)\to (2+5,-1+6)=(7,5)=P\text{'}\text{'}\\ T_2:O\text{'}(2,-4)\to (2+5,-4+6)=(7,2)=O\text{'}\text{'}\\ T_2:Q\text{'}(8,-4)\to (8+5,-4+6)=(13,2)=Q\text{'}\text{'}\end{array}$

Hence,

  $\Delta P\text{'}O\text{'}Q\text{'}\&\Delta P"O"Q"$ is graphed.

c.

To determine

To find: $T_3$ translation maps $\Delta POQ$ directly to $\Delta P"O"Q"$

Answer to Problem 16WE

$T_3$ translation maps $\Delta POQ$ directly to $\Delta P"O"Q"$ is $T_3:(x,y)\to (x+7,y+2)$ .

Explanation of Solution

Given:

  $T_1:(x,y)\to (x+2,y-4)\&T_2:(x,y)\to (x+5,y+6)$

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

  $T_1:(x,y)\to (x+2,y-4)\&T_2:(x,y)\to (x+5,y+6)$

  $\begin{array}{l}{T}_3(x,y)=T_2(T_1(x,y))=T_2(x+2,y-4)=((x+2)+5,(y-4)+6))\\ \Rightarrow (x+7,y+2)\\ \therefore T_3:(x,y)\to (x+7,y+2)\end{array}$

Hence,

$T_3$ translation maps $\Delta POQ$ directly to $\Delta P"O"Q"$ is $T_3:(x,y)\to (x+7,y+2)$ .

d.

To determine

To find: $T_2\&T_3$ in vector form & find the relation between $T_1,T_2\&T_3$ .

Answer to Problem 16WE

Vector form are $\stackrel{\rightharpoonup }{T_2}=(5,6)$ & $\stackrel{\rightharpoonup }{T_3}=(7,2)$ and also relation between vectors is $\stackrel{\rightharpoonup }{T_3}=\stackrel{\rightharpoonup }{T_1}+\stackrel{\rightharpoonup }{T_2}$ .

Explanation of Solution

Given:

  $\stackrel{\rightharpoonup }{T_1}=(2,-4)$ is describe in vector form.

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

  $T_1:(x,y)\to (x+2,y-4)\&T_2:(x,y)\to (x+5,y+6)$

From above sub-part we found,

  $T_3:(x,y)\to (x+7,y+2)$

$T_1$ glides all points 2 units right & 4 units down describe by vector $\stackrel{\rightharpoonup }{T_1}=(2,-4)$ .

Similarly,

$T_2$ glides all points 5 units right &6 units up describe by vector $\stackrel{\rightharpoonup }{T_2}=(5,6)$ .

$T_3$ glides all points 7 units right &2 units up describe by vector $\stackrel{\rightharpoonup }{T_3}=(7,2)$ .

Vector relation: -

  $\stackrel{\rightharpoonup }{T_3}=\stackrel{\rightharpoonup }{T_1}+\stackrel{\rightharpoonup }{T_2}$

Hence,

Vector form are $\stackrel{\rightharpoonup }{T_2}=(5,6)$ & $\stackrel{\rightharpoonup }{T_3}=(7,2)$ and also relation between vectors is $\stackrel{\rightharpoonup }{T_3}=\stackrel{\rightharpoonup }{T_1}+\stackrel{\rightharpoonup }{T_2}$ .