Question

If the straight line $x(a+2b)+y(a+3b)=a+b$ passes through a fixed point for different values of $a$ and $b$, then the fixed point is

• A. $(2,-1)$
• B. $(-2,1)$
• C. $(-1,2)$
• D. $(1,-2)$

Answer 1

The correct option is A $(2,-1)$

Given equation is
$x(a+2b)+y(a+3b)=a+b\Rightarrow a(x+y-1)+b(2x+3y-1)=0\cdots \left(1\right)$

${\lambda }_1{L}_1+{\lambda }_2{L}_2=0$ represents family of straight lines which will always pass through the point of intersection of lines $L_1=0,L_2=0$
i.e., $x+y-1=0$ and $2x+3y-1=0$,
Solving above equations, we get
$\Rightarrow y=-1,x=2$

Hence, the fixed point is $(2,-1)$.