Let two straight lines drawn from the origin O intersect the line 3x + 4y = 12 at the points P and Q such that OPQ is an isosceles triangle and POQ = 90 . If l = OP2 + PQ2 + QO2, then the greatest integer less than or equal to l is :
- 42
- 46
- 48
- 44
Answer
(2) 46
Solution
[figure]
Since OP = OQ = r and = 90 , take P(r , r ) and Q(-r , r ).
Both points lie on 3x + 4y = 12, so
r(3 + 4 ) = 12 and r(4 - 3 ) = 12
Squaring and adding: 25r2 = 288 r2 = 28825
Also PQ2 = r2 + r2 = 2r2, hence
l = r2 + 2r2 + r2 = 4r2 = 115225 = 46.08 [l] = 46
Answer: (2)