then image of the point `(-1,3,4)` in the plane `x-2y=0`

Options

A. `(-(17)/(3),(19)/(3),4)`
B. (15,11,4)
C. `(-(17)/(3),(19)/(3),1)`
D. `((9)/(5),-(13)/(5),4)`

Answer

d

Explanation

Let `P(alpha,beta,gamma)` be the image of the point Q(-1,3,4) Midpoint of PQ lies on x-2y=0. Then,
`(alpha-1)/(2)-2((beta-3)/(2))=0`
or `alpha-1-2beta-6=0oralpha-2beta=7" "(i)`
Also PQ is perpendicular to the plane. Then, `(alpha+1)/(1)=(beta-3)/(-2)=(gamma-4)/(0)" "(ii)`
Solving (i) and (ii) we get
`alpha=(9)/(5)beta,-(13)/(5),gamma=4`
Therefore, image is
`((9)/(5),(13)/(5),4)`
Alternate method:
For image,
`(alpha-(-1))/(1)=(beta-3)/(-2)=(gamma-4)/(0)=(-2(-1-2(3)))/((1)^(2)+(-2)^(2))`
or `alpha=(9)/(5),beta=-(13)/(5),gamma=4`