Question

The line $3x+2y=6$ will divide the quadrilateral formed by the lines $x+y=5,y-2x=8,3y+2x=0$ & $4y-x=0$ in

• A. two quadrilaterals
• B. one pentagon and one triangle
• C. two triangles
• D. one triangle and one quadrilateral

Answer 1

The correct option is A two quadrilaterals

$L_1:x+y=5$

$L_2:y-2x=8$

$L_3:3y+2x=0$

$L_4:4y-x=0$

$L_5:(3x+3y=6$

By solving we get, Vertices of quadrilateral

$O(0,0),A(4,1),B(-1,6),C(-3,2)$

Now,

$L_5\left(O\right)=-6<0$

$L_5\left(A\right)=12+2-6=8>0$

$L_5\left(B\right)=-3+12-6=3>0$

$L_5\left(C\right)=-9+4-6=-11<0$

$O$ and $C$ points are the same side

$A$ and $B$ points are other same side w.r.t. to $L_5$

So, $L_5$ divides the quadrilateral in two quadrilateral.