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Question Number 42315 by mondodotto@gmail.com last updated on 23/Aug/18
the point (2,−1) is reflected   in the line  x=4 find the  image point.
$$\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{point}}\:\left(\mathrm{2},−\mathrm{1}\right)\:\boldsymbol{\mathrm{is}}\:\boldsymbol{\mathrm{reflected}}\: \\ $$$$\boldsymbol{\mathrm{in}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{line}}\:\:\boldsymbol{{x}}=\mathrm{4}\:\boldsymbol{\mathrm{find}}\:\boldsymbol{\mathrm{the}} \\ $$$$\boldsymbol{\mathrm{image}}\:\boldsymbol{\mathrm{point}}. \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 23/Aug/18
Answered by Rio Michael last updated on 23/Aug/18
Reflection in the line x =a for (x,y) is   (((x′)),((y′)) ) =  (((−1),0),(0,1) ) ((x),(y) ) +  (((2a)),(0) ) Standard Formular  at line x=4 and point (2,−1)   (((x′)),((y′)) ) =  (((−1),0),(0,1) ) ((2),((−1)) ) +  ((8),(0) )   {we multiply ROW × COLUMN}             =  (((−2)),((−1)) ) +  ((8),(0) )             (((x′)),((y′)) ) =  ((6),((−1)) )
$${Reflection}\:{in}\:{the}\:{line}\:{x}\:={a}\:{for}\:\left({x},{y}\right)\:{is} \\ $$$$\begin{pmatrix}{{x}'}\\{{y}'}\end{pmatrix}\:=\:\begin{pmatrix}{−\mathrm{1}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{1}}\end{pmatrix}\begin{pmatrix}{{x}}\\{{y}}\end{pmatrix}\:+\:\begin{pmatrix}{\mathrm{2}{a}}\\{\mathrm{0}}\end{pmatrix}\:{Standard}\:{Formular} \\ $$$${at}\:{line}\:{x}=\mathrm{4}\:{and}\:{point}\:\left(\mathrm{2},−\mathrm{1}\right) \\ $$$$\begin{pmatrix}{{x}'}\\{{y}'}\end{pmatrix}\:=\:\begin{pmatrix}{−\mathrm{1}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{1}}\end{pmatrix}\begin{pmatrix}{\mathrm{2}}\\{−\mathrm{1}}\end{pmatrix}\:+\:\begin{pmatrix}{\mathrm{8}}\\{\mathrm{0}}\end{pmatrix}\:\:\:\left\{{we}\:{multiply}\:{ROW}\:×\:{COLUMN}\right\} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\:\begin{pmatrix}{−\mathrm{2}}\\{−\mathrm{1}}\end{pmatrix}\:+\:\begin{pmatrix}{\mathrm{8}}\\{\mathrm{0}}\end{pmatrix} \\ $$$$\:\:\:\:\:\:\:\:\:\:\begin{pmatrix}{{x}'}\\{{y}'}\end{pmatrix}\:=\:\begin{pmatrix}{\mathrm{6}}\\{−\mathrm{1}}\end{pmatrix} \\ $$

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