Chapter 14.3, Problem 7WE

To determine

To graph: $\Delta ABC,\Delta A\text{'}B\text{'}C\text{'}\&\Delta A"B"C"$ .

Answer to Problem 7WE

The graph of $\Delta ABC,\Delta A\text{'}B\text{'}C\text{'}\&\Delta A"B"C"$ is drawn on graph.

Explanation of Solution

Given information :

$\Delta A\text{'}B\text{'}C\text{'}$ is image under glide of $\Delta ABC$ move up 4 units and $\Delta A"B"C"$ is image under reflection of $\Delta A\text{'}B\text{'}C\text{'}$ in the $y-$ axis.

  $A(1,0),B(4,2)\&C(5,6)$

Concept used:

In 2-dimensional geometry, a glide reflection is a symmetry operation that consists of a reflection over a line and then translation along that line, combined into a single operation.

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.

As per the given problem

$\Delta A\text{'}B\text{'}C\text{'}$ is image under glide of $\Delta ABC$ move up 4 units and $\Delta A"B"C"$ is image under reflection of $\Delta A\text{'}B\text{'}C\text{'}$ in the $y-$ axis.

  $A(1,0),B(4,2)\&C(5,6)$

Since, $\Delta A\text{'}B\text{'}C\text{'}$ is image under glide of $\Delta ABC$ move up 4 units

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

$\Delta A"B"C"$ is image under reflection of $\Delta A\text{'}B\text{'}C\text{'}$ in the $y-$ axis.

Here, substitute x-coordinates with opposite (negative) numbers.

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

Hence,

The graph of $\Delta ABC,\Delta A\text{'}B\text{'}C\text{'}\&\Delta A"B"C"$ is drawn on graph.