Chapter 10.5, Problem 57AYU

Use the rotation formulas $\left(5\right)$ to show that distance is invariant under a rotation of axes. That is, show that the distance from $P_1=\left(x_1,y_1\right)$ to $P_2=\left(x_2,y_2\right)$ in the $xy$-plane equals the distance from $P_1=\left(x{\text{'}}_{\text{1}},y{\text{'}}_{\text{1}}\right)$ to $P_2=\left(x{\text{'}}_2,y{\text{'}}_{\text{2}}\right)$ in the $x\text{'},y\text{'}$-plane.

To determine

To prove: The distance $P_1=(x_1,y_1)$ to $P_2=(x_2,y_2)$ in $xy$ plane equals the distance from $P_1^{\text{'}}=({x^{\prime }}_1,y_1\text{'})$ to $P_2=({x^{\prime }}_2,y_2\text{'})$.

Answer to Problem 57AYU

Distance is invariant under rotation of axis.

Explanation of Solution

Given:

The Equations $A{x}^2+Bxy+C{y}^2+Dx+Ey+F=0$ and $Ax{\text{'}}^2+Bx\text{'}y\text{'}+Cy{\text{'}}^2+Dx\text{'}+Ey\text{'}+F\text{'}=0$.

Formula used:

$x=x’\cos \theta -y’\sin \theta ,y=x’\sin \theta +y’\cos \theta$

Proof:

The Equations $A{x}^2+Bxy+C{y}^2+Dx+Ey+F=0,B\ne 0$.

$d^2={\left(y_2-y_1\right)}^2+{\left(x_2-x_1\right)}^2$

$d^2={\left(x_2^{\text{'}}\sin \theta +y_2^{\text{'}}\cos \theta -x_1^{\text{'}}\sin \theta -y_1^{\text{'}}\cos \theta \right)}^2+{\left(x_2^{\text{'}}\cos \theta -y_2^{\text{'}}\sin \theta -x_1^{\text{'}}\cos \theta +y_1^{\text{'}}\sin \theta \right)}^2$

$d^2={\left((x_2^{\text{'}}-x_1^{\text{'}})\sin \theta +(y_2^{\text{'}}-y_1^{\text{'}})\cos \theta \right)}^2+{\left((x_2^{\text{'}}-x_1^{\text{'}})\cos \theta -\left(y_2^{\text{'}}-y_1^{\text{'}}\right)\sin \theta \right)}^2$

$d^2={\left({(x_2^{\text{'}}-x_1^{\text{'}})}^2{\sin }^2\theta +{(y_2^{\text{'}}-y_1^{\text{'}})}^2{\cos }^2\theta \right)}^2+\left({(x_2^{\text{'}}-x_1^{\text{'}})}^2{\cos }^2\theta +{\left(y_2^{\text{'}}-y_1^{\text{'}}\right)}^2{\sin }^2\theta \right)+2(x_2^{\text{'}}-x_1^{\text{'}})\left(y_2^{\text{'}}-y_1^{\text{'}}\right)\left(\sin \theta \cos \theta -\cos \theta \sin \theta \right)$

$d^2={\left((x_2^{\text{'}}-x_1^{\text{'}}\right)}^2+{\left(y_2^{\text{'}}-y_1^{\text{'}}\right)}^2)\left({\sin }^2\theta +{\cos }^2\theta \right)$

$d^2={\left(x_2^{\text{'}}-x_1^{\text{'}}\right)}^2+{\left(y_2^{\text{'}}-y_1^{\text{'}}\right)}^2$

Hence, distance is invariant under rotation of axis.