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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?
Whatisthereflectionofthepoint(2,2)inthelinex+2y=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)).
LetP(2,2)&P(a,b)isthereflectionalimageofPinL.letQ(a+22,b+22)isthemidpointPPandPQperpendiculartoL()mPQ=b2a2;mL=12mPQ×mL=1b2a2=2b2=2a4;2ab=2()lineLpassesthroughpointQa+22+2(b+22)=4a+2+2b+4=8;a+2b=2solvingfora&ba+2b=2(×2)2a+4b=42ab=2{5b=2;b=25a=22b=245=65thereforeP(65,25).
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)).
thelinefrom(2,2)andperpendiculartolinex+2y=4isx21=y22=torx=2+t,y=2+2tintersectionpointwiththeline:(2+t)+2(2+2t)=4t=25theimagepointisofdoubledistance,i.e.t=2×(25)=45x=245=65y=22×45=25imageof(2,2)inlineis(65,25).

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