To draw the boundary lines, we will treat them as equations instead of inequalities. To graph the equations, we must put these equation in slope-intercept form. Notice that the third equation is already written in this form.
|c|l|l|l|
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-5pt & Inequality & Equation -2pt & Slope-Intercept Form
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-2pt i. -2pt & 2y≥ x-3 & 2y= x-3 -2pt & y= 12x-1.5 [0.5em]
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-2pt ii. -2pt & x-2y≥-7 & x-2y=-7 & y= 12x+3.5 -2pt [0.5em]
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-2pt iii. -2pt & y≤ -2x+6 & y=-2x+6 -2pt & y=-2x+6 [0.5em]
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-2pt iv. -2pt & -9 ≤ 2x+y & -9= 2x+y -2pt & y=-2x-9 [0.5em]
Now we can graph the equations by finding two points for each function and connecting them with a straight line. If we let the y-intercept be the first point, we can find a second point by using the line's slope. Notice that all of the inequalities are non-strict which means we keep the lines solid. Let's show this with the first graph.
If we repeat this for the remaining equations, we get the following diagram.
Shading
To complete the graphs we have to shade the correct side of each line. We can do that by testing a point that is not on any of the boundary lines. The easiest point we can choose is the origin.
|c|c|c|c|
-3pt & -3pt (x,y) -3pt & Inequality & Evaluate
-5pt i. -5pt & -3pt ( 0, 0) -3pt & 2( 0)? ≥ 0-3 & 0≥-3 ✓ [0.5em]
-3pt ii. -3pt & -3pt ( 0, 0) -3pt & 0-2( 0)? ≥-7 & -2 ≥-7 ✓ [0.5em]
-3pt iii. -3pt& -3pt ( 0, 0) -3pt & 0? ≤ -2( 0)+6 & 0 ≤6 ✓ [0.5em]
-3pt iv. -3pt & -3pt ( 0, 0) -3pt & -9 ? ≤ 2( 0)+ 0 & -9 ≤ 0 ✓ [0.5em]
Since the origin produces a true statement for all inequalities, we should shade the side of the lines that contains the origin. We end up with the following diagram.
The polygon resembles a rectangle. If it is, adjacent sides must be perpendicular. If they are, the product of the slopes equals -1. Let's multiply the slopes and organize our answers in a table to check if the product equals -1.
|c|c|c|c|
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Equations & -3pt m_1m_2 -3pt & -3pt = -3pt & -3pt Perpendicular? -3pt [0.2em]
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i. and iii. & 0.5(-2) & -1 & âś“ [0.2em]
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i. and iv. & 0.5(-2) & -1 & âś“ [0.2em]
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ii. and iii. & 0.5(-2) & -1 & âś“ [0.2em]
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ii. and iv. & 0.5(-2) & -1 & âś“ [0.2em]
All intersecting boundary lines are perpendicular. Therefore, we are certain that the polygon is a rectangle.
b To find the coordinates of the vertices we will study our coordinate plane from Part A.
The coordinates can identified as (1,4), (3,0), (-3,-3) and (-5,1).
