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Transformation geometry

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Feb 15, 2012
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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MHB Math Scholar
Feb 15, 2012
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:


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

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


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

so your transformation is:

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

i do not know how you arrived at your answer, but after a lengthy calculation i got:

(4a-3b)x + (b-2a)y + (-8a+9b-2c) = 0

which appears to be your equation multiplied by -2 (-2*0 is still 0, so it shouldn't matter).

my calculations depended on b being non-zero. performing a second calculation when b = 0, i got:

2ax - ay + (4a-c) = 0, which leads me to believe we're both right.