Straight Line Question 148
If the straight lines $ 2x+3y-1=0,x+2y-1=0,~y+2x-1=0,andax+by-1=0 $ form a triangle with the origin as orthocenter, then (a, b) is given by
Options:
A) (6, 4)
B) (-3, 3)
C) (-8, 8)
D) (0, 7)
Correct Answer: C The 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, we have $ (-1)( -\frac{a}{b} )=-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, $ { -\frac{(2-a)}{3+a} }( -\frac{1}{2} )=-1 $ $ \therefore a=-8,b=8 $
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BETA
