All the points lying on or inside the triangle formed by the points $A(1,3), B(5,6), C(-1,2)$ satisfy
(A) $3x+2y\geq 0$
(B) $2x+y+1\geq 0$
(C) $2x+3y-12 \geq 0$
(D) $2x+11 \geq 11$
My Approach:
Equation of lines formed by given points are $3x-4y+9=0, x-2y+5=0, 2x-3y+8=0$ .
And Now let say point inside triangle be $P(\alpha, \beta)$.
Point $P$ lies inside the triangle formed by given lines
If $P$ and $A$ lies on same side of line $2x-3y+8=0$. i.e. $\; 2\alpha -3\beta +8\geq 0$
Also point $P$ and $B$ lies on same side of $x-2y+5=0$ $\;$i.e.$\; \alpha-2\beta+5\leq0$
And Point $P$ and $C$ lies on same side of $3x-4y+9=0$ i.e. $\; 3\alpha-4\beta+9\leq 0$
Now I don't know how to proceed further.
Hint:
Let the line $L_1(x,y)=3x+2y-0$, then $L_1(1,3)=9>0, L_1(5,6)=15+12=27>0, L_3(-1,2)=-3+4=1>0$. This implies that all points inside ABC will satisfy $3x+2y \ge 0$
Similarly, you can check for other given lines.