Question

If it is possible to draw a line which belongs to all the given family of lines $y-2x+1+{\lambda }_1(2y-x-1)=0,$
$3y-x-6+{\lambda }_2(y-3x+6)=0,$

$ax+y-2+{\lambda }_3(6x+ay-a)=0$, then

• A. $a=4$
• B. $a=3$
• C. $a=-2$
• D. $a=2$

Answer 1

Answer: (A)

$a=4$

$y-2x+1+{\lambda }_1(2y-x-1)=0$ always passes through intersection of $y-2x+1=0$ and $(2y-x-1)=0$
their intersection is $(1,1)\equiv Q$
Similarly $3y-x-6=0+{\lambda }_2(y-3x+6)=0$ passes through $(3,3)\equiv P$
Equation of $PQ$ is :$y=x$
According to given condition ,third family of equation should always pass through the point which lies on $y=x$
putting $y=x$ in $ax+y-2+{\lambda }_3(6x+ay-a)=0$
we get $ax+x-2=0$ and $6x+ay-a=0$
$⟹\frac{2}{a+1}=\frac{a}{6+a}$
$⟹12+2a=a^2+a⟹a^2-a-12=0⟹(a-4)(a+3)=0$
$\therefore a=4$