Question

find the vertices of the polygon determined by the set of equations.
$y = -x + 2$
$y = -dfrac{3}{2}x - 3$
$y = dfrac{2}{3}x - 3$
$y = 3x + 6$

the vertices of the polygon are \boxed{}.
(type ordered pairs. use a comma to separate answers as needed.)

Explanation:

Step1: Find intersection of \( y = -x + 2 \) and \( y = 3x + 6 \)

Set \( -x + 2 = 3x + 6 \)
\( -x - 3x = 6 - 2 \)
\( -4x = 4 \)
\( x = -1 \)
Substitute \( x = -1 \) into \( y = -x + 2 \): \( y = -(-1) + 2 = 3 \)
So, intersection point: \( (-1, 3) \)

Step2: Find intersection of \( y = 3x + 6 \) and \( y = -\frac{3}{2}x - 3 \)

Set \( 3x + 6 = -\frac{3}{2}x - 3 \)
\( 3x + \frac{3}{2}x = -3 - 6 \)
\( \frac{6x + 3x}{2} = -9 \)
\( \frac{9x}{2} = -9 \)
\( x = -2 \)
Substitute \( x = -2 \) into \( y = 3x + 6 \): \( y = 3(-2) + 6 = 0 \)
Wait, error here. Let's recalculate:
\( 3x + 6 = -\frac{3}{2}x - 3 \)
Multiply both sides by 2: \( 6x + 12 = -3x - 6 \)
\( 6x + 3x = -6 - 12 \)
\( 9x = -18 \)
\( x = -2 \)
\( y = 3(-2) + 6 = 0 \)? Wait, no, \( y = -\frac{3}{2}(-2) - 3 = 3 - 3 = 0 \). So point: \( (-2, 0) \)? Wait, no, let's check with \( y = 3x + 6 \): \( 3*(-2)+6 = 0 \). Correct.

Step3: Find intersection of \( y = -\frac{3}{2}x - 3 \) and \( y = \frac{2}{3}x - 3 \)

Set \( -\frac{3}{2}x - 3 = \frac{2}{3}x - 3 \)
\( -\frac{3}{2}x - \frac{2}{3}x = -3 + 3 \)
\( -\frac{9x + 4x}{6} = 0 \)
\( -\frac{13x}{6} = 0 \)
\( x = 0 \)
Substitute \( x = 0 \) into \( y = \frac{2}{3}x - 3 \): \( y = -3 \)
So, intersection point: \( (0, -3) \)

Step4: Find intersection of \( y = \frac{2}{3}x - 3 \) and \( y = -x + 2 \)

Set \( \frac{2}{3}x - 3 = -x + 2 \)
\( \frac{2}{3}x + x = 2 + 3 \)
\( \frac{2x + 3x}{3} = 5 \)
\( \frac{5x}{3} = 5 \)
\( x = 3 \)
Substitute \( x = 3 \) into \( y = -x + 2 \): \( y = -3 + 2 = -1 \)
So, intersection point: \( (3, -1) \)

Wait, but we have four lines, so the polygon is a quadrilateral? Wait, maybe I missed some intersections. Wait, let's list all pairs:

1. \( y = -x + 2 \) & \( y = 3x + 6 \): \( (-1, 3) \)
2. \( y = 3x + 6 \) & \( y = -\frac{3}{2}x - 3 \): Let's recalculate correctly.
\( 3x + 6 = -\frac{3}{2}x - 3 \)
\( 3x + \frac{3}{2}x = -3 - 6 \)
\( \frac{6x + 3x}{2} = -9 \)
\( \frac{9x}{2} = -9 \)
\( x = -2 \)
\( y = 3*(-2) + 6 = 0 \). Correct. So \( (-2, 0) \)
3. \( y = -\frac{3}{2}x - 3 \) & \( y = \frac{2}{3}x - 3 \): \( (0, -3) \) (as above)
4. \( y = \frac{2}{3}x - 3 \) & \( y = -x + 2 \): \( (3, -1) \) (as above)
5. \( y = -x + 2 \) & \( y = \frac{2}{3}x - 3 \): Wait, we did that, got (3, -1)
6. \( y = 3x + 6 \) & \( y = \frac{2}{3}x - 3 \): Let's check.
Set \( 3x + 6 = \frac{2}{3}x - 3 \)
\( 3x - \frac{2}{3}x = -3 - 6 \)
\( \frac{9x - 2x}{3} = -9 \)
\( \frac{7x}{3} = -9 \)
\( x = -\frac{27}{7} \), which is not a nice number. So maybe the polygon is formed by the intersections of adjacent lines? Wait, the four lines: let's see their slopes.

Slope of \( y = -x + 2 \): -1

Slope of \( y = 3x + 6 \): 3

Slope of \( y = -\frac{3}{2}x - 3 \): -3/2

Slope of \( y = \frac{2}{3}x - 3 \): 2/3

Notice that \( -\frac{3}{2} \) and \( \frac{2}{3} \) are negative reciprocals? No, \( -\frac{3}{2} * \frac{2}{3} = -1 \), so they are perpendicular.

Wait, maybe the correct intersections are:

1. \( y = -x + 2 \) and \( y = 3x + 6 \): \( (-1, 3) \)
2. \( y = 3x + 6 \) and \( y = -\frac{3}{2}x - 3 \): Let's solve again.
\( 3x + 6 = -\frac{3}{2}x - 3 \)
\( 3x + \frac{3}{2}x = -9 \)
\( \frac{9x}{2} = -9 \)
\( x = -2 \)
\( y = 3*(-2) + 6 = 0 \). So (-2, 0)
3. \( y = -\frac{3}{2}x - 3 \) and \( y = \frac{2}{3}x - 3 \): \( (0, -3) \) (since when x=0, both y=-3)
4. \( y = \frac{2}{3}x - 3 \) and \( y = -x + 2 \): \( (3, -1) \) (solved earlier)
5. \( y = -x + 2 \) and \( y = \frac{2}{3}x - 3 \): Wait, we did that, got (3, -1)
6. \( y = 3x + 6 \) and \( y = \frac{2}{3}x - 3 \): As above, not a vertex of the polygon.

Wait, maybe the polygon is a quadrilateral with vertices at (-2, 0), (0, -3), (3, -1), (-1, 3)? Wait, no, let's check the intersection of \( y = -x + 2 \) and \( y = -\frac{3}{2}x - 3 \):

Set \( -x + 2 = -\frac{3}{2}x - 3 \)
\( -x + \frac{3}{2}x = -3 - 2 \)
\( \frac{-2x + 3x}{2} = -5 \)
\( \frac{x}{2} = -5 \)
\( x = -10 \)
\( y = -(-10) + 2 = 12 \). That's a different point. So maybe I made a mistake in identifying adjacent lines.

Wait, the problem says "the polygon determined by the set of equations". So we need to find all intersection points of the lines, then determine which form the polygon.

Let's list all pairs of lines and find their intersections:

1. Line 1: \( y = -x + 2 \) (L1)
Line 2: \( y = -\frac{3}{2}x - 3 \) (L2)
Intersection: \( -x + 2 = -\frac{3}{2}x - 3 \)
\( \frac{x}{2} = -5 \)
\( x = -10 \)
\( y = -(-10) + 2 = 12 \) → (-10, 12)

2. L1 and Line 3: \( y = \frac{2}{3}x - 3 \) (L3)
\( -x + 2 = \frac{2}{3}x - 3 \)
\( -x - \frac{2}{3}x = -5 \)
\( -\frac{5}{3}x = -5 \)
\( x = 3 \)
\( y = -3 + 2 = -1 \) → (3, -1)

3. L1 and Line 4: \( y = 3x + 6 \) (L4)
\( -x + 2 = 3x + 6 \)
\( -4x = 4 \)
\( x = -1 \)
\( y = 3 \) → (-1, 3)

4. L2 and L3:
\( -\frac{3}{2}x - 3 = \frac{2}{3}x - 3 \)
\( -\frac{3}{2}x - \frac{2}{3}x = 0 \)
\( -\frac{13}{6}x = 0 \)
\( x = 0 \)
\( y = -3 \) → (0, -3)

5. L2 and L4:
\( -\frac{3}{2}x - 3 = 3x + 6 \)
\( -\frac{3}{2}x - 3x = 9 \)
\( -\frac{9}{2}x = 9 \)
\( x = -2 \)
\( y = 3(-2) + 6 = 0 \) → (-2, 0)

6. L3 and L4:
\( \frac{2}{3}x - 3 = 3x + 6 \)
\( \frac{2}{3}x - 3x = 9 \)
\( -\frac{7}{3}x = 9 \)
\( x = -\frac{27}{7} \)
\( y = 3(-\frac{27}{7}) + 6 = -\frac{81}{7} + \frac{42}{7} = -\frac{39}{7} \) → (-\frac{27}{7}, -\frac{39}{7})

Now, to find the polygon, we need to see which intersection points are vertices. Let's analyze the slopes:

L1: slope -1

L2: slope -3/2

L3: slope 2/3

L4: slope 3

Notice that L2 and L3 are perpendicular (slope product -1), L4 and L1: slope product 3*(-1) = -3 ≠ -1, L2 and L4: slope product (-3/2)*3 = -9/2 ≠ -1, L3 and L1: slope product (2/3)*(-1) = -2/3 ≠ -1.

Wait, maybe the polygon is a quadrilateral formed by the intersections of L1-L4, L4-L2, L2-L3, L3-L1. Let's check the order:

- L1-L4: (-1, 3)

- L4-L2: (-2, 0)

- L2-L3: (0, -3)

- L3-L1: (3, -1)

Now, let's check if these points form a polygon. Let's plot mentally:

(-1, 3) is above, (-2, 0) to the left, (0, -3) below, (3, -1) to the right. Connecting these, it's a quadrilateral.

Wait, but earlier when we did L1-L2, we got (-10, 12), which is far left, so probably not part of the polygon. So the vertices are the intersections of consecutive lines (in terms of slope order) that form the bounded region.

So the correct vertices are:

(-1, 3) (intersection of L1 and L4),

(-2, 0) (intersection of L4 and L2),

(0, -3) (intersection of L2 and L3),

(3, -1) (intersection of L3 and L1)

Wait, let's verify with another pair: L3 and L1 is (3, -1), L1 and L4 is (-1, 3), L4 and L2 is (-2, 0), L2 and L3 is (0, -3). Yes, these four points form a quadrilateral.

Answer:

\((-1, 3)\), \((-2, 0)\), \((0, -3)\), \((3, -1)\)