Questions
The image of the point ( -1,3,4) in the plane x-2y -0 is
a
−173,−193,4
b
(15,11,4)
c
−173,−193,1
d
95,−135,4
detailed solution
Correct option is D
Let P(α,β,γ) be the image of the point Q(−1,3,4). Midpoint of PQ lies on x−2y=0. Then, α−12−2β+32=0or α−1−2β−6=0 or α−2β=7----iAlso PQ rs perpendicular to the plane. Then, α+11=β−3−2=γ−40-----iiSolving (i) and (ii), we get α=95,β=−135,γ=4Therefore, image is 95,−135,4Alternate method: For image, α−(−1)1=β−3−2=γ−40=−2(−1−2(3))(1)2+(−2)2 or α=95,β=−135,γ=4Talk to our academic expert!
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