Chapter 11.5, Problem 39E

To determine

To check: Whether the distance between any two points is invariant under rotation or not prove it by comparing the distance between the points and images.

Answer to Problem 39E

Yes, the distance between any two points is invariant under rotation.

Explanation of Solution

Given information:

The two points are $P$ and $Q$ . The images after rotation are $P^{\prime }$ and $Q^{\prime }$ respectively.

Assume that the coordinates of the points $P\left(x_1,y_1\right)$ and $Q\left(x_2,y_2\right)$ after rotation are $P^{\prime }\left({x^{\prime }}_1,{y^{\prime }}_1\right)$ and $Q^{\prime }\left({x^{\prime }}_2,{y^{\prime }}_2\right)$ respectively.

By the rotation the coordinates becomes.

  $\begin{array}{l}{x}_1={x^{\prime }}_1\cos \varphi -{y^{\prime }}_1\sin \varphi \\ y_1={x^{\prime }}_1\sin \varphi +{y^{\prime }}_1\cos \varphi \end{array}$

And,

  $\begin{array}{l}{x}_2={x^{\prime }}_2\cos \varphi -{y^{\prime }}_2\sin \varphi \\ y_2={x^{\prime }}_2\sin \varphi +{y^{\prime }}_2\cos \varphi \end{array}$

Now check the distance between the two points $P\left(x_1,y_1\right)$ and $Q\left(x_2,y_2\right)$ .

$d\left(P,Q\right)=\sqrt{{\left(y_2-y_1\right)}^2+{\left(x_2-x_1\right)}^2}$$=\sqrt{{\left({x^{\prime }}_2\sin \varphi +{y^{\prime }}_2\cos \varphi -\left({x^{\prime }}_1\sin \varphi +{y^{\prime }}_1\cos \varphi \right)\right)}^2+{\left({x^{\prime }}_2\cos \varphi -{y^{\prime }}_2\sin \varphi -\left({x^{\prime }}_1\cos \varphi -{y^{\prime }}_1\sin \varphi \right)\right)}^2}$$=\sqrt{{\left(\left({x^{\prime }}_2-{x^{\prime }}_1\right)\sin \varphi +\left({y^{\prime }}_2-{y^{\prime }}_1\right)\cos \varphi \right)}^2+{\left(\left({x^{\prime }}_2-{x^{\prime }}_1\right)\cos \varphi -\left({y^{\prime }}_2-{y^{\prime }}_1\right)\cos \varphi \right)}^2}$$=\sqrt{{\left({x^{\prime }}_2-{x^{\prime }}_1\right)}^2{\sin }^2\varphi +{\left({y^{\prime }}_2-{y^{\prime }}_1\right)}^2{\cos }^2\varphi +2\sin \varphi \cos \varphi \left({x^{\prime }}_2-{x^{\prime }}_1\right)\left({y^{\prime }}_2-{y^{\prime }}_1\right)+{\left({x^{\prime }}_2-{x^{\prime }}_1\right)}^2{\cos }^2\varphi +{\left({y^{\prime }}_2-{y^{\prime }}_1\right)}^2{\sin }^2\varphi -2\sin \varphi \cos \varphi \left({x^{\prime }}_2-{x^{\prime }}_1\right)\left({y^{\prime }}_2-{y^{\prime }}_1\right)}$

Further simplify,

  $\begin{array}{c}d\left(P,Q\right)=\sqrt{{\left({x^{\prime }}_2-{x^{\prime }}_1\right)}^2{\sin }^2\varphi +{\left({y^{\prime }}_2-{y^{\prime }}_1\right)}^2{\cos }^2\varphi +2\sin \varphi \cos \varphi \left({x^{\prime }}_2-{x^{\prime }}_1\right)\left({y^{\prime }}_2-{y^{\prime }}_1\right)+{\left({x^{\prime }}_2-{x^{\prime }}_1\right)}^2{\cos }^2\varphi +{\left({y^{\prime }}_2-{y^{\prime }}_1\right)}^2{\sin }^2\varphi -2\sin \varphi \cos \varphi \left({x^{\prime }}_2-{x^{\prime }}_1\right)\left({y^{\prime }}_2-{y^{\prime }}_1\right)}\\ =\sqrt{{\left({x^{\prime }}_2-{x^{\prime }}_1\right)}^2\left({\sin }^2\varphi +{\cos }^2\varphi \right)+{\left({y^{\prime }}_2-{y^{\prime }}_1\right)}^2\left({\sin }^2\varphi +{\cos }^2\varphi \right)}\\ =\sqrt{{\left({x^{\prime }}_2-{x^{\prime }}_1\right)}^2+{\left({y^{\prime }}_2-{y^{\prime }}_1\right)}^2}\\ =d\left(P^{\prime },Q^{\prime }\right)\end{array}$

As the distance $d\left(P,Q\right)$ is equal to $d\left(P^{\prime },Q^{\prime }\right)$ .

Therefore, the distance between any two points is invariant under rotation.