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For all real values of \mathrm{a} and \mathrm{b} lines \mathrm{(2 a+b) x+(a+3 b) y+(b-3 a)=0} and \mathrm{m x+2 y+6=0} are concurrent, then \mathrm{\mathrm{m}} is equal to

Option: 1

-2


Option: 2

-3


Option: 3

-4


Option: 4

-5


Answers (1)

best_answer

Given equations are concurrent for all real values of \mathrm{a} and \mathrm{b}.

\mathrm{Then \, \frac{2 a+b}{m}=\frac{a+3 b}{2}=\frac{b-3 a}{6}}

From last two ratios

\mathrm{\mathrm{b}=\frac{-3}{4} \mathrm{a}}
And from first two ratios
\mathrm{\mathrm{m}=-2}

Posted by

avinash.dongre

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