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The image of the point (-1,3,4) in the ...

The image of the point `(-1,3,4)` in the plane `x-2y=0` is a. `(-(17)/3,-(19)/3,4)` b. `(15 ,11 ,4)` c. `(-(17)/3,-(19)/3,1)` d. `(9/5,(-13)/5,4)`

A

`(-(17)/(3),(19)/(3),4)`

B

(15,11,4)

C

`(-(17)/(3),(19)/(3),1)`

D

`((9)/(5),-(13)/(5),4)`

Text Solution

Verified by Experts

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`
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