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

Options

A. (6,4)
B. (-3,3)
C. (-8,8)
D. (0,7)

Answer

C

Explanation

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)(-(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,
`{-((2-a))/(3+a)}(-(1)/(2))=-1`
`therefore a=-8, b=8`