1. Tardigrade
  2. Question
  3. Mathematics
  4. If it is possible to draw a line which belongs to all the given family of lines y-2 x+1+λ1(2 y-x-1)=0,3 y-x-6 +λ2(y-3 x+6)=0, a x+y-2+λ3(6 x+a y-a)=0, then

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Q. If it is possible to draw a line which belongs to all the given family of lines $y-2 x+1+\lambda_{1}(2 y-x-1)=0,3 y-x-6$ $+\lambda_{2}(y-3 x+6)=0, a x+y-2+\lambda_{3}(6 x+a y-a)=0$, then

Straight Lines

Solution:

The first two families of lines pass through $(1,1)$
and $(3,3)$, respectively. The point of intersection of
the lines belonging to the third family of lines will lie on line $y=x$. Hence,
$a x+x-2=0 \text { and } 6 x+a x-a=0$
or $ \frac{2}{a+1}=\frac{a}{6+a}$
or $a^{2}-a-12=0$ or $(a-4)(a+3)=0$