# Transformation geometry

#### Poirot

##### Banned
Consider the affine transformation $$f(P)=\begin{bmatrix}1 & 2 \\3 & 4\end{bmatrix}P+\begin{bmatrix}5\\6\end{bmatrix}$$.

Find the image of $$ax+by+c=0$$ under $$f$$.

My answer is $$\left(a-\frac{b}{2}\right)y+\left(\frac{3b}{2} -2a\right)x+4a-\frac{9b}{2}+c=0$$.

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#### Deveno

##### Well-known member
MHB Math Scholar
the command is $\text{\begin{pmatrix}...\end{pmatrix}}$ for a matrix with parentheses, and the same but {bmatrix} for a matrix delimited by brackets, and {vmatrix} for a mtrix delimited by vertical bars (like when you indicate you're taking the determinant).

the format for the rows is:

(entry)&(entry)&(entry)

a double back-slash indicates the start of a new row, so:

$\text{\begin{bmatrix} 2&1&0\\1&1&-3\\0&0&1 \end{bmatrix}}$

gives:

$\begin{bmatrix} 2&1&0\\1&1&-3\\0&0&1 \end{bmatrix}$

$f(P) = \begin{bmatrix}1&2\\3&4 \end{bmatrix}P + \begin{bmatrix}5\\6 \end{bmatrix}$