Question and Answers Forum

All Questions      Topic List

Relation and Functions Questions

Previous in All Question      Next in All Question      

Previous in Relation and Functions      Next in Relation and Functions      

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)).

$${the}\:{line}\:{from}\:\left(\mathrm{2},\mathrm{2}\right)\:{and}\:{perpendicular} \\ $$$${to}\:{line}\:{x}+\mathrm{2}{y}=\mathrm{4}\:{is} \\ $$$$\frac{{x}−\mathrm{2}}{\mathrm{1}}=\frac{{y}−\mathrm{2}}{\mathrm{2}}={t} \\ $$$${or}\:{x}=\mathrm{2}+{t},\:{y}=\mathrm{2}+\mathrm{2}{t} \\ $$$${intersection}\:{point}\:{with}\:{the}\:{line}: \\ $$$$\left(\mathrm{2}+{t}\right)+\mathrm{2}\left(\mathrm{2}+\mathrm{2}{t}\right)=\mathrm{4} \\ $$$$\Rightarrow{t}=−\frac{\mathrm{2}}{\mathrm{5}} \\ $$$${the}\:{image}\:{point}\:{is}\:{of}\:{double}\:{distance}, \\ $$$${i}.{e}.\:{t}=\mathrm{2}×\left(−\frac{\mathrm{2}}{\mathrm{5}}\right)=−\frac{\mathrm{4}}{\mathrm{5}} \\ $$$$\Rightarrow{x}=\mathrm{2}−\frac{\mathrm{4}}{\mathrm{5}}=\frac{\mathrm{6}}{\mathrm{5}} \\ $$$$\Rightarrow{y}=\mathrm{2}−\mathrm{2}×\frac{\mathrm{4}}{\mathrm{5}}=\frac{\mathrm{2}}{\mathrm{5}} \\ $$$${image}\:{of}\:\left(\mathrm{2},\mathrm{2}\right)\:{in}\:{line}\:{is}\:\left(\frac{\mathrm{6}}{\mathrm{5}},\frac{\mathrm{2}}{\mathrm{5}}\right). \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com