The line of:

Celia Horne
2022-01-31
Answered

How can you find a reflection matrix about a given line, using matrix multiplication and the idea of composition of transformations?

The line of:$y=-\frac{2x}{3}$ , all in $\mathbb{R}}^{2$

The line of:

You can still ask an expert for help

dikgetse3u

Answered 2022-02-01
Author has **10** answers

Step 1

So you have a vector$({x}_{1},{y}_{1})}^{T$ , and you want to find a matrix M such that $({x}_{2},{y}_{2})}^{T}=M{({x}_{1},{y}_{1})}^{T$ is the reflection across $y=-\frac{2x}{3}$ .

So what does the reflection means? It means that the middle of the two points is on the line of reflection, and the line between those points is perpendicular to the reflection line. The first condition can be written as

$\frac{{y}_{1}+{y}_{2}}{2}=-\frac{23}{\frac{{x}_{1}+{x}_{2}}{2}}$

The second condition means that the slope of the line between the two points is$-\frac{1}{m}$ where m is the slope of the reflection line:

$\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}=-\frac{1}{-\frac{2}{3}}=\frac{32}{}$

You now write$x}_{2$ and $y}_{2$ in terms of $x}_{1$ and $y}_{1$ :

$x}_{2}={m}_{11}{x}_{1}+{m}_{12}{y}_{1$

$y}_{2}={m}_{21}{x}_{1}+{m}_{22}{y}_{1$

Then the matrix you are looking for has these coefficients.

$M=\left(\begin{array}{cc}{m}_{11}& {m}_{12}\\ {m}_{21}& {m}_{22}\end{array}\right)$

So you have a vector

So what does the reflection means? It means that the middle of the two points is on the line of reflection, and the line between those points is perpendicular to the reflection line. The first condition can be written as

The second condition means that the slope of the line between the two points is

You now write

Then the matrix you are looking for has these coefficients.

Karly Logan

Answered 2022-02-02
Author has **11** answers

Step 1

One strategy is to rotate the plane so that the line becomes$y=0$ , apply the reflection in the x-axis, and then rotate back.

Let$\mathrm{tan}\theta =\frac{2}{3}$ . Then the required matrix is

$\left(\begin{array}{cc}\mathrm{cos}\theta & \mathrm{sin}\theta \\ -\mathrm{sin}\theta & \mathrm{cos}\theta \end{array}\right)\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right)\left(\begin{array}{cc}\mathrm{cos}\theta & -\mathrm{sin}\theta \\ \mathrm{sin}\theta & \mathrm{cos}\theta \end{array}\right)$

$=\left(\begin{array}{cc}\mathrm{cos}2\theta & -\mathrm{sin}2\theta \\ -\mathrm{sin}2\theta & -\mathrm{cos}2\theta \end{array}\right)$

One strategy is to rotate the plane so that the line becomes

Let

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