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?

Ans : Two Quadrilaterals

Answer 1

Graphing it, we see the line divides the quadrilateral into two quadrilaterals.

Explanation:

Wolfram Alpha is always good for a picture. I drew on this one. There's actually a fair amount of calculating to verify that the line intersects at the sides, and not the vertices.

Let's find the vertices.

`x+y=5, y-2x=8, 3x=-3, x=-1, y=6, A(-1,6)`
`y-2x=8,3y+2x=0,4y=8,y=2,x=-3, B(-3,2)`
`3y+2x=0, 4y-x=0,` clearly `x=y=0, C(0,0)`
`4y-x=0, x+y=5, 5y=5, y=1, x=4, D(4,1)`

Side `AB` has parametric form:
`X=(1-t)A+tB = A + t(B-A)` for `0\le t le 1`
`AB=(-1-2t ,6 -4t )`
`BC=(-3 +3t ,2 - 2t)`
`CD=(4t,t)`
`DA= (4-5t,1+5t)`

Let's find the meets with `3x+2y=6`.
`AB: 3(-1-2t)+2(6-4t)=6, t=3/14`
Since `0 < t < 1,` that's a proper meet with side `AB,` not with vertex `A` or `B.`

`BC: 3(-3 + 3t) + 2(2-2t)=6. t=11/5` outside the segment
`CD: 3(4t)+2t=6, t=3/7` a proper meet.
`DA: 3(4-5t)+2(1+5t)=6, t=8/5` not a segment meet

For completeness,

`E= 11/14 A + 3/14 B = (- 10/7, 36/7 )`

`F=(4t,t)=(12/7, 3/7)`