The image of  the point ( -1,3,4) in the plane x-2y -0 is

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=4