We have the points $A = (2, - 11)$, $B = (10, - 5)$ and $C = (0,0)$.
(a) Determine the equation of the line $AB$ in Hesse normal form and determine the distance between $C$ and $AB$. Give also the line reflection $\sigma$ at the line $AB$ in analytic form and calculate $\sigma (C)$.
(b) Show that there is a rotation $\delta$ with center $C$ that maps $A$ to $B$. Give $\delta$ in analytic form and determine an equation of theline $\delta (g)$ for $g:=AB$.
(c) Is $\delta \circ \sigma$ a rotation, translation, line reflection or glide reflection?
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For (a):
We have that $\vec{AB}=\begin{pmatrix}8 \\ 6\end{pmatrix}$.
A normal vector to $\vec{AB}$ is for example $\vec{n}=\begin{pmatrix}6 \\ -8\end{pmatrix}$.
We check the direction of $\vec{n}$ :\begin{equation*}\vec{n}\circ \vec{a}=\begin{pmatrix}6 \\ -8\end{pmatrix}\circ \begin{pmatrix}2 \\ -11\end{pmatrix}=12+88=100>0\end{equation*} The direction of $\vec{n}$ is therefore correct.
The absolute value of $\vec{n}$ is \begin{equation*}|\vec{n}|=\sqrt{6^2+(-8)^2}=\sqrt{36+64}=\sqrt{100}=10\end{equation*} So the Hesse normal form of the equation of the line $AB$ is \begin{equation*}\frac{1}{10}\cdot \begin{pmatrix}6 \\ -8\end{pmatrix}\circ \left [\vec{x}- \begin{pmatrix}2 \\ -11\end{pmatrix}\right ]=0\end{equation*}
To calculate the distance between $C$ and $AB$ we substitute $C$ in the Hesse normal form of $AB$. \begin{equation*}d_C=\frac{1}{10}\cdot \begin{pmatrix}6 \\ -8\end{pmatrix}\circ \left [\begin{pmatrix}0 \\ 0\end{pmatrix}- \begin{pmatrix}2 \\ -11\end{pmatrix}\right ]=\frac{1}{10}\cdot \begin{pmatrix}6 \\ -8\end{pmatrix}\circ \begin{pmatrix}-2 \\ 11\end{pmatrix}=\frac{1}{10}\cdot \left (-12-88\right ) =\frac{1}{10}\cdot \left (-100\right )=-10 \end{equation*} Since $d_C < 0$ the point $C$ is on the same side of the line as the origin.
Is that correct so far?
Could you give me a hint how to calculate $\sigma$ ?