The equations y=-2X+12, y=6-X, and X=0 enclose a region in quadrant I. when that region is reflected over the line y=6-X, a portion of the reflection is in quadrant II. Find the area of that resulting region.
To find the area of the resulting region, we first need to find the coordinates of the points where the given equations intersect. We can do this by setting the equations equal to each other and solving for the values of X and Y.
1. For y = -2X + 12 and y = 6 - X:
-2X + 12 = 6 - X
Simplify:
-2X + X = 6 - 12
-X = -6
Solve for X:
X = -6 / -1
X = 6
Substitute X = 6 into one of the equations to find Y:
y = -2(6) + 12
y = -12 + 12
y = 0
Therefore, the point of intersection is (6, 0).
2. For y = 6 - X and X = 0:
Substituting X = 0 into y = 6 - X:
y = 6 - 0
y = 6
Therefore, the point of intersection is (0, 6).
Now, we have the two points of intersection: (6, 0) and (0, 6). Let's draw a sketch to visualize the region enclosed by the given equations and the line y = 6 - X:
```
(6,0)
+
|
|
|
|
|
+
(0,6)
```
The line y = 6 - X divides the region into two parts. We are interested in the portion of the reflection in quadrant II. From the sketch, it's clear that the area of that region is a triangle.
To find the area of the triangle, we need to calculate the base and the height. The base is the difference in X-coordinates, and the height is the difference in Y-coordinates.
Base = 6 - 0 = 6
Height = 6 - 0 = 6
Now, we can calculate the area of the triangle using the formula:
Area = (1/2) * base * height
Area = (1/2) * 6 * 6
Area = 3 * 6
Area = 18
Therefore, the area of the resulting region after reflecting over the line y = 6 - X in quadrant II is 18 square units.