question_answer

If the straight lines \[2x+3y-1=0,\] \[x+2y-1=0\]and \[ax+by-1=0\] form a triangle with origin as orthocentre, then (a, b) is given by

A)  \[(6,4)\]

B)  \[(-3,3)\]   

C)  \[(-8,8)\]

D)  \[(0,7)\]

Correct Answer: C

Solution :

Equation of AO is \[2x+3y-1+\lambda (x+2y-1)=0.\]

Where \[\lambda =-1\]since the line passes through the origin.

So, \[x+y=0.\]Since, AO is perpendicular to BC. So, \[(-1)\left( -\frac{a}{b} \right)=-1\]\[\Rightarrow \]\[a=-b\]

Similarly, \[(2x+3y-1)+\mu \,\,(ax-ay-1)=0\]will be the equation of BO for \[\mu =-1.\]

Now, BO is perpendicular to AC.

Hence,  \[\left\{ -\frac{(2-a)}{3+a} \right\}\left( -\frac{1}{2} \right)=-1\]

Hence,   \[a=-8,\]\[b=8\]