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Question Number 139639 by bemath last updated on 30/Apr/21
What is the reflection of the point (2,2) in the line x+2y = 4?
$$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{reflection}\:\mathrm{of}\:\mathrm{the}\: \\ $$$$\mathrm{point}\:\left(\mathrm{2},\mathrm{2}\right)\:\mathrm{in}\:\mathrm{the}\:\mathrm{line}\:\mathrm{x}+\mathrm{2y}\:=\:\mathrm{4}? \\ $$
Commented by bramlexs22 last updated on 30/Apr/21
Let P(2,2) & P′(a,b) is the reflectional image of P in L. let Q(((a+2)/2),((b+2)/2)) is the midpoint PP′ and PQ perpendicular to L (•) m_(PQ) = ((b−2)/(a−2)) ; m_L = −(1/2) ⇒m_(PQ) × m_L =−1 ⇒((b−2)/(a−2)) = 2 ⇒b−2=2a−4 ; 2a−b=2 (••) line L passes through point Q ⇒((a+2)/2) +2(((b+2)/2))=4 ⇒a+2+2b+4=8 ; a+2b=2 solving for a & b ⇒a+2b=2...(×2)⇒2a+4b=4 2a−b=2 { ((5b=2 ; b=(2/5))),((a=2−2b=2−(4/5)=(6/5))) :} therefore P′((6/5),(2/5)).
$$\mathrm{Let}\:\mathrm{P}\left(\mathrm{2},\mathrm{2}\right)\:\&\:\mathrm{P}'\left(\mathrm{a},\mathrm{b}\right)\:\mathrm{is}\:\mathrm{the}\:\mathrm{reflectional} \\ $$$$\mathrm{image}\:\mathrm{of}\:\mathrm{P}\:\mathrm{in}\:\mathrm{L}. \\ $$$$\mathrm{let}\:\mathrm{Q}\left(\frac{\mathrm{a}+\mathrm{2}}{\mathrm{2}},\frac{\mathrm{b}+\mathrm{2}}{\mathrm{2}}\right)\:\mathrm{is}\:\mathrm{the}\:\mathrm{midpoint} \\ $$$$\mathrm{PP}'\:\mathrm{and}\:\mathrm{PQ}\:\mathrm{perpendicular}\:\mathrm{to}\:\mathrm{L} \\ $$$$\left(\bullet\right)\:\mathrm{m}_{\mathrm{PQ}} \:=\:\frac{\mathrm{b}−\mathrm{2}}{\mathrm{a}−\mathrm{2}}\:;\:\mathrm{m}_{\mathrm{L}} \:=\:−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{m}_{\mathrm{PQ}} \:×\:\mathrm{m}_{\mathrm{L}} \:=−\mathrm{1} \\ $$$$\Rightarrow\frac{\mathrm{b}−\mathrm{2}}{\mathrm{a}−\mathrm{2}}\:=\:\mathrm{2}\:\Rightarrow\mathrm{b}−\mathrm{2}=\mathrm{2a}−\mathrm{4}\:;\:\mathrm{2a}−\mathrm{b}=\mathrm{2} \\ $$$$\left(\bullet\bullet\right)\:\mathrm{line}\:\mathrm{L}\:\mathrm{passes}\:\mathrm{through}\:\mathrm{point}\:\mathrm{Q} \\ $$$$\Rightarrow\frac{\mathrm{a}+\mathrm{2}}{\mathrm{2}}\:+\mathrm{2}\left(\frac{\mathrm{b}+\mathrm{2}}{\mathrm{2}}\right)=\mathrm{4} \\ $$$$\Rightarrow\mathrm{a}+\mathrm{2}+\mathrm{2b}+\mathrm{4}=\mathrm{8}\:;\:\mathrm{a}+\mathrm{2b}=\mathrm{2} \\ $$$$\mathrm{solving}\:\mathrm{for}\:\mathrm{a}\:\&\:\mathrm{b} \\ $$$$\Rightarrow\mathrm{a}+\mathrm{2b}=\mathrm{2}…\left(×\mathrm{2}\right)\Rightarrow\mathrm{2a}+\mathrm{4b}=\mathrm{4} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{2a}−\mathrm{b}=\mathrm{2} \\ $$$$\begin{cases}{\mathrm{5b}=\mathrm{2}\:;\:\mathrm{b}=\frac{\mathrm{2}}{\mathrm{5}}}\\{\mathrm{a}=\mathrm{2}−\mathrm{2b}=\mathrm{2}−\frac{\mathrm{4}}{\mathrm{5}}=\frac{\mathrm{6}}{\mathrm{5}}}\end{cases} \\ $$$$\mathrm{therefore}\:\mathrm{P}'\left(\frac{\mathrm{6}}{\mathrm{5}},\frac{\mathrm{2}}{\mathrm{5}}\right). \\ $$
Answered by mr W last updated on 30/Apr/21
the line from (2,2) and perpendicular to line x+2y=4 is ((x−2)/1)=((y−2)/2)=t or x=2+t, y=2+2t intersection point with the line: (2+t)+2(2+2t)=4 ⇒t=−(2/5) the image point is of double distance, i.e. t=2×(−(2/5))=−(4/5) ⇒x=2−(4/5)=(6/5) ⇒y=2−2×(4/5)=(2/5) image of (2,2) in line is ((6/5),(2/5)).
$${the}\:{line}\:{from}\:\left(\mathrm{2},\mathrm{2}\right)\:{and}\:{perpendicular} \\ $$$${to}\:{line}\:{x}+\mathrm{2}{y}=\mathrm{4}\:{is} \\ $$$$\frac{{x}−\mathrm{2}}{\mathrm{1}}=\frac{{y}−\mathrm{2}}{\mathrm{2}}={t} \\ $$$${or}\:{x}=\mathrm{2}+{t},\:{y}=\mathrm{2}+\mathrm{2}{t} \\ $$$${intersection}\:{point}\:{with}\:{the}\:{line}: \\ $$$$\left(\mathrm{2}+{t}\right)+\mathrm{2}\left(\mathrm{2}+\mathrm{2}{t}\right)=\mathrm{4} \\ $$$$\Rightarrow{t}=−\frac{\mathrm{2}}{\mathrm{5}} \\ $$$${the}\:{image}\:{point}\:{is}\:{of}\:{double}\:{distance}, \\ $$$${i}.{e}.\:{t}=\mathrm{2}×\left(−\frac{\mathrm{2}}{\mathrm{5}}\right)=−\frac{\mathrm{4}}{\mathrm{5}} \\ $$$$\Rightarrow{x}=\mathrm{2}−\frac{\mathrm{4}}{\mathrm{5}}=\frac{\mathrm{6}}{\mathrm{5}} \\ $$$$\Rightarrow{y}=\mathrm{2}−\mathrm{2}×\frac{\mathrm{4}}{\mathrm{5}}=\frac{\mathrm{2}}{\mathrm{5}} \\ $$$${image}\:{of}\:\left(\mathrm{2},\mathrm{2}\right)\:{in}\:{line}\:{is}\:\left(\frac{\mathrm{6}}{\mathrm{5}},\frac{\mathrm{2}}{\mathrm{5}}\right). \\ $$

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