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 -

• A. $(-\infty ,0)\cup (8,\infty )$
• B. $(-\infty ,0)\cup (3,\infty )$
• C. $(0,3)$
• D. $(3,8)$

Answer 1

Answer: (A)

$(-\infty ,0)\cup (8,\infty )$

$L_1(x,y)=2x+y-a=0$ .... (1)

$L_2(x,y)=x-3y+a=0$ ....(2)

at $(0,0),L_1(0,0)=-a$

$L_2(0,a)=a$

at $(3,2)$

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

$L_2(3,2)=-3+a$

how , if origin and $(3,2)$ are can tained in the same angled, than $L_1(0,0)\&L_1(3,2)$ must have same sign & $L_2(0,0)\&L_2(3,2)$ must have same sign.

case 1, if $a>0\Rightarrow -a<0$

$8-a<0\Rightarrow a>8$ &

$-3+a>0\Rightarrow a>3$

thus $a\in (8,\infty )$

Case 2, if $a<0,-a>0$

$8-a>0,\Rightarrow a<8$

$-3+a<0,a<3$

thus $a\in (-\infty ,0)$

Thus $a\in (-\infty ,0)\cup (8,\infty )$