Chapter 14, Problem 6CR

b.

Yes the distance is invariant under $T$ .

To determine

To determine whether the distance is invariant under $T:(x,y)\to (x+2,y-4)$ or not.

Answer to Problem 6CR

Yes the distance is invariant under $T$ .

Explanation of Solution

Given information:

Given that $T:(5,5)\to (7,1)$ .

Calculation:

Taking another point $(7,7)$ then,

  $T:(7,7)\to (7+2,7-4)=(9,3)$

And thus we get,

  $\begin{array}{l}(5,5)\to (7,1),\text{ Distance=}\sqrt{{(1-5)}^2+{(7-5)}^2}=\sqrt{20}\\ (7,7)\to (9,3),\text{ Distance=}\sqrt{{(1-5)}^2+{(7-5)}^2}=\sqrt{20}\mathrm{...........}(i)\end{array}$

Which implies the function $T(x,y)$ is preserve length.

c.

Yes the angle measure is invariant under $T$ .

To determine

To determine wheather the mapping $T(x,y)$ is angle preserving or not.

Answer to Problem 6CR

Yes the angle measure is invariant under $T$ .

Explanation of Solution

Given information:

Given that $T:(5,5)\to (7,1)$ .

Calculation:

In part (b) we see that is the function $T(x,y)$ maps every segment to congruent segment. Therefore it is called isometry. And every isometry preserves lengths, area and angles.

d.

Yes the area is invariant under $T$ .

To determine

To determine wheather the mapping $T(x,y)$ is preserving area or not.

Answer to Problem 6CR

Yes the area is invariant under $T$ .

Explanation of Solution

Given information:

Given that $T:(5,5)\to (7,1)$ .

Calculation:

From part (c) we see that function $T(x,y)$ maps every segment to congruent segment. Therefore it is called isometry. And every isometry preserves lengths, area and angles.