Let two straight lines drawn from the origin O intersect the line 3 x+4 y=12 at the points P and Q such that OPQ is an isosceles triangle and POQ =90^ . If l= OP ^2+ PQ ^2+ QO ^2 , then the greatest integer less than or equal to l is :

Step-by-step solution

aligned & 3 x +4 y =12 & 3( r )+4( r )=12 & r (3 +4 )=12 (1) & 3(- r )+4( r )=12 & r (-3 +4 )=12 (2) & ( 12 r )^2+ ( 12 r )^2=(3 +4 )^2+(-3 +4 )^2 & 2 ( 12 r )^2=9+16 & 2 144 r ^2 =25 288=25 r ^2 & 288 25 = r ^2 & 2 ( 12 5 )= r & = OP ^2+ PQ ^2+ QO ^2 & = r ^2+ r ^2 + r ^2( + )^2+ r ^2( + )^2 & =2 r ^2+ r ^2(1+ 2 +1-2 2 ) & =2 r ^2+2 r ^2 & =4 r ^2 & =4 ( 288 25 )= 1152 25 =46.08 & [ ]=46 aligned