Chapter 14.3, Problem 1WE

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 1WE

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

Explanation of Solution

Given:

  $\begin{array}{l}T:(x,y)\to (x-2,y+6)\\ A(-2,0),B(0,4),C(3,-1)\end{array}$

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 $T:(x,y)\to (x-2,y+6)$

  $A(-2,0),B(0,4),C(3,-1)$

$x$ moves left 2 units & $y$ moves up 6 units.

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

  $\begin{array}{l}T(-2,0)\to (-2-2,0+6)=(-4,6)\\ T(0,4)\to (0-2,4+6)=(-2,10)\\ T(3,-1)\to (3-2,-1+6)=(1,5)\end{array}$

  $\begin{array}{l}A(-2,0)\to A\text{'}(-4,6)\\ B(0,4)\to B\text{'}(-2,10)\\ C(3,-1)\to C\text{'}(1,5)\end{array}$

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 1WE

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

Explanation of Solution

Given:

  $\begin{array}{l}T:(x,y)\to (x-2,y+6)\\ A(-2,0),B(0,4),C(3,-1)\end{array}$

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 $T:(x,y)\to (x-2,y+6)$

  $A(-2,0),B(0,4),C(3,-1)$

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

  $\begin{array}{l}T(-2,0)\to (-2-2,0+6)=(-4,6)\\ T(0,4)\to (0-2,4+6)=(-2,10)\\ T(3,-1)\to (3-2,-1+6)=(1,5)\end{array}$

  $\begin{array}{l}A(-2,0)\to A\text{'}(-4,6)\\ B(0,4)\to B\text{'}(-2,10)\\ C(3,-1)\to C\text{'}(1,5)\end{array}$

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?

Answer to Problem 1WE

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

Explanation of Solution

Given:

  $\begin{array}{l}T:(x,y)\to (x-2,y+6)\\ A(-2,0),B(0,4),C(3,-1)\end{array}$

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 $T:(x,y)\to (x-2,y+6)$

  $A(-2,0),B(0,4),C(3,-1)$

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

  $\begin{array}{l}T(-2,0)\to (-2-2,0+6)=(-4,6)\\ T(0,4)\to (0-2,4+6)=(-2,10)\\ T(3,-1)\to (3-2,-1+6)=(1,5)\end{array}$

  $\begin{array}{l}A(-2,0)\to A\text{'}(-4,6)\\ B(0,4)\to B\text{'}(-2,10)\\ C(3,-1)\to C\text{'}(1,5)\end{array}$

From the graph it was found each arrow length = $6.32$ .

Hence,

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