If the straight line 2x+3y−1=0,x+2y−1=0 and ax+by−1=0  from a triangle with origin as orthocenter, then (a,b)  is given by

Equation of AO is 2x+3y−1+λ(x+2y−1)=0 Where λ=−1,  since the line passes through the origin. (i.e.) x+y=0                       Since AO is perpendicular to BC, (−1)(−ab)=−1 ∴a=−b Similarly, (2x+3y−1)+μ(ax−ay−1)=0  will be the equation of BO for μ=−1.  BO is perpendicular to AC⇒{−(2−a)3+a}(−12)=−1.    ∴a=−8,b=8 .