Chapter 14.3, Problem 17WE

To determine

To locate:Midpoints of $\overline{AA\text{'}},\overline{BB\text{'}},\overline{CC\text{'}}$ .

Answer to Problem 17WE

The midpoints of segments $\overline{AA\text{'}},\overline{BB\text{'}},\overline{CC\text{'}}$ lie on the reflecting line.

Explanation of Solution

Given information :

A glide reflection maps $\Delta ABC$ to $\Delta A\text{'}B\text{'}C\text{'}$ .

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, A glide reflection maps $\Delta ABC$ to $\Delta A\text{'}B\text{'}C\text{'}$ .

Below diagram is provided in problem.

To get midpoints of $\overline{AA\text{'}},\overline{BB\text{'}},\overline{CC\text{'}}$ we will join $A$ to $A\text{'}$ , $B$ to $B\text{'}$ & $C$ to $C\text{'}$ .

It seems that the midpoints of segments $\overline{AA\text{'}},\overline{BB\text{'}},\overline{CC\text{'}}$ lie on the reflecting line.

For example

  $A(4,2),B(2,2)\&C(2,4)$

  $T:(x,y)\to (x+4,y+6)$

After translation,

  $\begin{array}{l}T:A(4,2)\to (4+4,2+6)=(8,8)\\ T:B(2,2)\to (2+4,2+6)=(6,8)\\ T:C(2,4)\to (2+4,4+6)=(6,10)\end{array}$

Say reflection line

  $R_m\to y=x$

The point $(x,y)$ is mapped to $(y,x)$ .

  $\begin{array}{l}\therefore A\text{'}=(8,8)\\ B\text{'}(8,6)\\ C\text{'}(10,6)\end{array}$

Midpoints of segments $\overline{AA\text{'}},\overline{BB\text{'}},\overline{CC\text{'}}$

  $(\frac{4+8}{2},\frac{2+8}{2});(\frac{2+8}{2},\frac{2+6}{2});(\frac{2+10}{2},\frac{4+6}{2})=(6,5);(5,4);(6,5)$

Almost lie on the reflection line.

Hence,

The midpoints of segments $\overline{AA\text{'}},\overline{BB\text{'}},\overline{CC\text{'}}$ lie on the reflecting line.