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Question Number 42315 by mondodotto@gmail.com last updated on 23/Aug/18

$$\mathbf{the}\mathbf{point}(2,-1)\mathbf{is}\mathbf{reflected}$$$$\mathbf{in}\mathbf{the}\mathbf{line}\mathit{x}=4\mathbf{find}\mathbf{the}$$$$\mathbf{image}\mathbf{point}.$$

Answered by tanmay.chaudhury50@gmail.com last updated on 23/Aug/18

Answered by Rio Michael last updated on 23/Aug/18

$$Reflectioninthelinex=afor(x,y)is$$$$\left(\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right)=\left(\begin{array}{cc}-1& 0\\ 0& 1\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)+\left(\begin{array}{c}2a\\ 0\end{array}\right)StandardFormular$$$$atlinex=4andpoint(2,-1)$$$$\left(\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right)=\left(\begin{array}{cc}-1& 0\\ 0& 1\end{array}\right)\left(\begin{array}{c}2\\ -1\end{array}\right)+\left(\begin{array}{c}8\\ 0\end{array}\right)\left\{wemultiplyROW\times COLUMN\right\}$$$$=\left(\begin{array}{c}-2\\ -1\end{array}\right)+\left(\begin{array}{c}8\\ 0\end{array}\right)$$$$\left(\begin{array}{c}{x}^{\prime }\\ y^{\prime }\end{array}\right)=\left(\begin{array}{c}6\\ -1\end{array}\right)$$

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