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Image of a line under a transformation (matrices)

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    The plane transformation V is represented by the matrix \begin{pmatrix} 1 & 4 \\2 & -1 \end{pmatrix} .

    L_1 is the line with equation y = 0.5x + k , and L_2 is the image of L_1 under V.

    Find, in the form y = mx + c , the cartesian equation for L_2 .


    So far I have considered the general point of y = 0.5x + k to be \begin{pmatrix} t  , & 0.5t + k \end{pmatrix}.

    So \begin{pmatrix} 1 & 4 \\2 & -1 \end{pmatrix}\begin{pmatrix} t \\0.5t + k \end{pmatrix}  = \begin{pmatrix} 3t + 4k \\1.5t - k \end{pmatrix}

    From here on I get a little stuck so could do with some assistance, thanks
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    (Original post by bonana567)
    The plane transformation V is represented by the matrix \begin{pmatrix} 1 & 4 \\2 & -1 \end{pmatrix} .

    L_1 is the line with equation y = 0.5x + k , and L_2 is the image of L_1 under V.

    Find, in the form y = mx + c , the cartesian equation for L_2 .


    So far I have considered the general point of y = 0.5x + k to be \begin{pmatrix} t  , & 0.5t + k \end{pmatrix}.

    So \begin{pmatrix} 1 & 4 \\2 & -1 \end{pmatrix}\begin{pmatrix} t \\0.5t + k \end{pmatrix}  = \begin{pmatrix} 3t + 4k \\1.5t - k \end{pmatrix}

    From here on I get a little stuck so could do with some assistance, thanks
    So you equation
    \begin{pmatrix}x\\y\end{pmatrix}  =\begin{pmatrix}4k\\-k\end{pmatrix}+t\begin{pmatrix}3  \\1.5\end{pmatrix}
    Eliminate t.
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    (Original post by ztibor)
    So you equation
    \begin{pmatrix}x\\y\end{pmatrix}  =\begin{pmatrix}4k\\-k\end{pmatrix}+t\begin{pmatrix}3  \\1.5\end{pmatrix}
    Eliminate t.
    Gaaaaa so easy thanks :L

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Updated: May 17, 2012
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