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If the straight lines 2x+3y-1=0,x+2y-1=0...

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

A

`(6,4)`

B

`(-3,3)`

C

`(-8,8)`

D

`(0,7)`

Text Solution

Verified by Experts

The correct Answer is:
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)(-(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`
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