If the straight lines 2 x+3 y-1=0 , x+2 y-1=0 and a x+b y-1=0 form a triangle with orthocentre at the origin, then (a, b)=

Step-by-step solution

Equation of line perpendicular to line a x+b y-1=0 and passes through the origin is b x-a y=0 Now, altitude (i) passes through the intersection of lines 2 x+3 y-1=0 , and x+2 y-1=0 So, aligned -b-a & =0 a+b & =0 aligned Due to relation (ii), the line a x+b y-1=0 , becomes x-y- 1 a =0 Now, the altitude perpendicular to the line 2 x+3 y-1=0 , and passes through origin is 3 x=2 y The point of intersection line (iii) and x+2 y-1=0 is ( 1 3 (1+ 2 a ), 1 3 (1- 1 a ) ) satisfy the line So, a=-8 and b=8 (a, b)=(-8,8) .