Question

If origin and $(3,2)$ are contained in the same angle of the lines $2x+y-a=0,x-3y+a=0$, then $a$ must lie in the interval-

Answer 1

$L_1:2x+y-a=0$

$L_2:x-3y+a=0$

$L_1(V,U)=-a,L_2(V,U)=a$

$L_1(3,2)=8-a,L_2(3,2)=a-3$

$L_1(U,U)$ & $L_{(}3,2)=a-3$

$L_1(U,U)$ & $L_1(3,2)$ must have sure sign

$L_2(U,U)$ & $L_2(3,2)$ must have sure sign

If, $a>U,-a<0$

$a-3>0,8-a<0$

$\Rightarrow a>3,\therefore a>8$

$a\in (8,\infty )$......... (1)

if, $a<U,-a>0$

$a-3<U,-a>0$

$a<3,a<8$

$a\in (-\infty ,3)$.....(2)

$\therefore a\in (-\infty ,3)\cup (8,\infty )$

Hence, this is the answer.