in $R^2$, let $L$ be the line $y = mx$ where $m$ is not zero. find an expression for $T(x,y)$ where $T$ is the reflection of $R^2$ about $L$

Question

i have no idea how to start this problem. there's an example in my book and i've tried to adapt it, but i haven't been able to get very far and i'm not even sure i'm on the right track.

i know i need an expression $T(a,b)$ for any $(a,b)$ in $R^2$.

i want to set the basis $A = \{(1,0), (0,1)\}$, but i'm unsure of how to calculate the values of $T(a,b)$ using these. i think $T(1,0)=(1,mx)$ and $T(0,1)=(0,0)$, but i don't think that this is right.

have i chosen the right basis $A$? if not, how can i find a better one? and how can i calculate the values of $T(a,b)$ using the correct basis? then how should i proceed? my book and lecture content are both unclear about this.

thank you in advance for any insight.

Answer 1

Sorry for the notations; I have switched $(x,y)$ and $(a,b)$.

Hint: $T(x,y)=(a,b)$ is characterized by two properties:

a) The line joining $(x,y)$ and $(a,b)$ is orthogonal to the given line. This gives $m\frac {y-b} {x-a}=-1$.

b) The mid-point of the line joining $(x,y)$ and $(a,b)$ lies on the given line. This gives $\frac {b+y} 2 =m \frac {x+a} 2$.

Now solve the two equations above for $a$ and $b$ in terms of $x$ and $y$ that will give you the formula for $T(x,y)$,

Answer 2

As you said, this linear transformation is determined completely by how it acts on two basis elements. But you don't have to check specifically for the standard basis! It would be wise to think, geometrically, which vectors would be best.

Note that a reflection about an axis keeps the axis in place, and so it would make things easy to choose one of the basis elements to lie on the axis of reflection (since we know that $T$ does not affect it). So we can already say that $T((1,m))=(1,m)$ (since this vector is on the axis of reflection).

Moreover, if $v$ is orthogonal to the axis of reflection, then $T$ will send it ot $-v$. So pick such a vector, such as $(-m,1)$. Now you can say that $T((-m,1))=(m,-1)$.

Since you know how $T$ acts on two basis elements, you can find a general formula for $T$. Let me know if you're having trouble with that.