question_answer

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

A) (6, 4)    

B) (-3, 3)

C) (-8, 8)  

D) (0, 7)

Correct Answer: C

Solution :

[c] The equation of AO is \[2x+3y-1+\lambda (x+2y-1)=0\], where \[\lambda =-1\], sine the line passes through the origin, so, \[x+y=0.\]since AO is perpendicular to BC, we have \[(-1)\left( -\frac{a}{b} \right)=-1\] \[\therefore a=-b\] 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, \[\left\{ -\frac{(2-a)}{3+a} \right\}\left( -\frac{1}{2} \right)=-1\] \[\therefore a=-8,b=8\]