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Find the affine transformation that sends the line $ε:3x+2y+4=0$ of $\mathbb{R^2}$ to the line $x=0$

I am having some problems here, and I am getting confused, if anyone could help I would appreciate it.

The first issue that I have is that an affine transformation is define to be a function $f:\mathbb{R}^2 \rightarrow \mathbb{R}^2 $ such that $f(x)=Ax+k$ , $x\in \mathbb{R}^2$ $,A\in Gl(n), k\in \mathbb{R}^2$ how we "feed" a line equation in the function?

Then i thought to use the fundamental theorem of affine geometry and the points $(0,0)$ , $(0,1)$ and $(1,0)$ didn't belong to the line $ε$ so I peek $3$ points on the line $a,b,c$ to qualify the equations $f(a)=(0,0)$ etc but i couldent find $A , k$

領域展開
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  • Almost by definition, there are infinitely many affine transformations that map the line $3x+2y+4=0$ to the line $x=0$. Are there perhaps some additional constraints to be satisfied? – hardmath Oct 12 '21 at 22:19
  • @hardmath This doesn't help me, since I don't exactly know how these transformations map a line to another line. – 領域展開 Oct 12 '21 at 22:25
  • Focus on how these transformations map a point to another point. Pick two distinct points on the line $3x+2y+4=0$ and devise an affine map that send them to two distinct points on $x=0$ (also known as the $y$-axis). But my Comment was aimed at how you open the body of your post. You ask to "Find the affine transformation...", and I'm remarking that there is no unique "the" transformation without imposing additional conditions. – hardmath Oct 12 '21 at 22:31

1 Answers1

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Mapping a line (or other object) to another means that it maps every point from that line to a point on the other bijectively. So we want to take the set $\varepsilon = \{(x, y) \in \mathbb{R} : 3x + 2y + 4 = 0\}$ and for each point $p \in \varepsilon$ map it to a point $q \in l = \{(x, y) : x = 0\}$. And, in this case, we want to do so using an affine transform.

The easiest way to do this would be to pick a few points from $\varepsilon$ and assign points on $l$ to send them to, and see what that means for the transformation. Since it takes two points to uniquely define a line, that seems like a reasonable starting point - maybe take the points where $x = 0$ and $y = 0$, and go from there.

ConMan
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  • By using the method I describe in my question, i get $\begin{pmatrix}
    a &b \ c&d \end{pmatrix} \binom{0}{-2} +\binom{m}{n}=\binom{0}{0}$ , $ \begin{pmatrix}
    a &b \ c&d \end{pmatrix} \binom{-4/3}{0} +\binom{m}{n}=\binom{0}{1}$ etc for the 3rd point, but I can't solve the system     – 領域展開 Oct 12 '21 at 22:44
  • oh but wait, the system i get is $b=2/3a$ and $-4/3c+2d=1$ if i consider the matrix $\begin{pmatrix} 3/2 &1 \ -3/4&0
    \end{pmatrix}$ and $\binom{m}{n}=\binom{2}{0}$ I think I get what I want, right?     – 領域展開 Oct 12 '21 at 22:55