Let the image of the point P(0, -5, 0) in the line x-12= y1= z+1-2 be the point R and the image of the point Q (0, -12, 0 ) in the line x-1-1= y+94= z+11 be the point S. Then the square of the area of the parallelogram PQRS is .
Answer
162
Solution
Line L1: any point (1+2 , , -1-2 ), direction (2, 1, -2)
Foot F of P(0, -5, 0): PF (2, 1, -2)=0
2(1+2 )+( +5)-2(-1-2 )=0 9 +9=0 =-1
F(-1, -1, 1), R=2F-P=(-2, 3, 2)
Line L2: any point (1- , -9+4 , -1+ ), direction (-1, 4, 1)
Foot G of Q (0, - 12, 0 ): -(1- )+4 (4 - 172 )+( -1)=0
18 -36=0 =2 G(-1, -1, 1)
S=2G-Q= (-2, - 32, 2 )
Both feet are the same point (-1, -1, 1), the common mid point of the diagonals PR and QS, so PQRS is a parallelogram.
PQ= (0, 92, 0 ), PS= (-2, 72, 2 )
PQ PS=(9-0, 0-0, 0+9)=(9, 0, 9)
(Area)2= | PQ PS |2=81+81=162
Answer: 162