Chapter 14.3, Problem 2WE

a.

To determine

To graph: $\Delta ABC$ and its image $\Delta A\text{'}B\text{'}C\text{'}$ and check is $\Delta ABC$ $\widetilde{=}$ $\Delta A\text{'}B\text{'}C\text{'}$

Answer to Problem 2WE

  $\Delta ABC$ $\widetilde{=}$ $\Delta A\text{'}B\text{'}C\text{'}$ i.e. they are congruent and similar.

Explanation of Solution

Given:

  $\begin{array}{l}T:(x,y)\to (x-3,y-6)\\ A(3,6),B(-3,6),C(-1,-2)\end{array}$

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

Translation $T:(x,y)\to (x-3,y-6)$

  $A(3,6),B(-3,6),C(-1,-2)$

$x$ moves left 3 units & $y$ moves down 6 units.

  $\begin{array}{l}A\to A\text{'}\\ B\to B\text{'}\\ C\to C\text{'}\end{array}$

  $\begin{array}{l}T(3,6)\to (3-3,6-6)=(0,0)\\ T(-3,6)\to (-3-3,6-6)=(-6,0)\\ T(-1,-2)\to (-1-3,-2-6)=(-4,-8)\end{array}$

  $\begin{array}{l}A(3,6)\to A\text{'}(0,0)\\ B(-3,6)\to B\text{'}(-6,0)\\ C(-1,-2)\to C\text{'}(-4,-8)\end{array}$

On graph above triangle, $\Delta A\text{'}B\text{'}C\text{'}$ and below $\Delta ABC$ .

We can refer from the graph that all angles & corresponding angles are same.

  $\begin{array}{l}AB=A\text{'}B\text{'}\\ BC=B\text{'}C\text{'}\\ CA=C\text{'}A\text{'}\\ \angle A=\angle A\text{'}\\ \angle B=\angle B\text{'}\\ \angle C=\angle C\text{'}\end{array}$

Hence,

  $\Delta ABC$ $\widetilde{=}$ $\Delta A\text{'}B\text{'}C\text{'}$ i.e. they are congruent and similar.

b.

To determine

To draw: Arrows from $A\to A\text{'}$ , $B\to B\text{'}$ and $C\to C\text{'}$ .

Answer to Problem 2WE

The arrows are drawn from $A\to A\text{'}$ , $B\to B\text{'}$ and $C\to C\text{'}$ .

Explanation of Solution

Given:

  $\begin{array}{l}T:(x,y)\to (x-3,y-6)\\ A(3,6),B(-3,6),C(-1,-2)\end{array}$

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

Translation $T:(x,y)\to (x-3,y-6)$

  $A(3,6),B(-3,6),C(-1,-2)$

  $\begin{array}{l}A\to A\text{'}\\ B\to B\text{'}\\ C\to C\text{'}\end{array}$

  $\begin{array}{l}T(3,6)\to (3-3,6-6)=(0,0)\\ T(-3,6)\to (-3-3,6-6)=(-6,0)\\ T(-1,-2)\to (-1-3,-2-6)=(-4,-8)\end{array}$

  $\begin{array}{l}A(3,6)\to A\text{'}(0,0)\\ B(-3,6)\to B\text{'}(-6,0)\\ C(-1,-2)\to C\text{'}(-4,-8)\end{array}$

Hence,

The arrows are drawn from $A\to A\text{'}$ , $B\to B\text{'}$ and $C\to C\text{'}$ .

c.

To determine

To find:The arrows are drawn from $A\to A\text{'}$ , $B\to B\text{'}$ and $C\to C\text{'}$ are of the same length and parallel?

Answer to Problem 2WE

The arrows are drawn from $A\to A\text{'}$ , $B\to B\text{'}$ and $C\to C\text{'}$ are of the same length and are parallel.

Explanation of Solution

Given:

  $\begin{array}{l}T:(x,y)\to (x-3,y-6)\\ A(3,6),B(-3,6),C(-1,-2)\end{array}$

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

Translation $T:(x,y)\to (x-3,y-6)$

  $A(3,6),B(-3,6),C(-1,-2)$

  $\begin{array}{l}A\to A\text{'}\\ B\to B\text{'}\\ C\to C\text{'}\end{array}$

  $\begin{array}{l}T(3,6)\to (3-3,6-6)=(0,0)\\ T(-3,6)\to (-3-3,6-6)=(-6,0)\\ T(-1,-2)\to (-1-3,-2-6)=(-4,-8)\end{array}$

  $\begin{array}{l}A(3,6)\to A\text{'}(0,0)\\ B(-3,6)\to B\text{'}(-6,0)\\ C(-1,-2)\to C\text{'}(-4,-8)\end{array}$

From the graph it was found each arrow length = $6.7$ & slope of $AA\text{'},BB\text{'},CC\text{'}$ i.e. $\angle m=$ ${63.4}^{°}$

Hence,

The arrows are drawn from $A\to A\text{'}$ , $B\to B\text{'}$ and $C\to C\text{'}$ are of the same length.