The distance of the point (2,3) from the line 2x-3y+28=0, measured parallel to the line 3x-y+1=0, is equal to
- 3+4 2
- 6 3
- 4+6 3
- 4 2
Answer
(3) 4+6 3
Solution
[figure]
∵ Slope of 3x-y+1=0 is 3 ⇒ PQ makes 60 with the x-axis.
Parametric form: Q = (2+r 60 , 3+r 60 )
Q = (2+ r2, 3+ 3r2 )
Q lies on 2x-3y+28=0:
4+r-9- 3 32r+28=0 r ( 3 3-22 )=23
r= 463 3-2= 46(3 3+2)27-4=2(3 3+2)
PQ =4+6 3
Answer: (3)