Let A(a, b), B(3, 4) and C(-6,-8) respectively denote the centroid, circumcentre and orthocentre of a triangle. Then, the distance of the point P(2a+3, 7b+5) from the line 2x+3y-4=0 measured parallel to the line x-2y-1=0 is
- 17 56
- 15 57
- 17 57
- 517
Answer
(3) 17 57
Solution
Orthocentre, centroid, circumcentre are collinear and the centroid divides them in 2:1
[figure]
A= ( 2(3)+(-6)3, 2(4)+(-8)3 )=(0, 0) P(3, 5)
Slope of x-2y-1=0 is 12 = 2 5, = 1 5
Point at distance r from P along it: (3+ 2r 5, 5+ r 5 ) lies on 2x+3y-4=0
6+ 4r 5+15+ 3r 5-4=0 7r 5=-17
|r|= 17 57
Answer: (3)