Praneetha reddy
Last Activity: 11 Years ago
Given a point P(1,3,4)
Let Q be the image of the point.
the dr''s of PQ is (2,-1,1)
Eqn of line PQ= x-1/2 = y-3/-1 = z-4/1
Any point on line PQ is (2r+1, -r+3, r+4)
If this point is on the plane , then
2(2r+1)-(-r+3)+(r+4)+3=0
4r+2+r-3+r+4+3=0
6r+6=0
r=-1
Therefore coordinates of foot of the perpendicular (say S) = (-1,4,2)
We know that S is the middle point of PQ
=> -1=x1+1/2, 4=y1+3/2, 3=4+z1/2
=> x1=-2, y1=5, z1=2
Therefore , Image of the point P is Q(-2,5,2).