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In the rectangular coordinate system, what is the area of the region d [#permalink]
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Good question.


The tricky part was analyzing XY> 0

The region covered by this equation is the entire quadrant I and quadrant III where X and Y have the same sign.


0 < Y < 2

The line given by Y = 0 is actually the Horizontal X Axis

The line Give by Y = 2 will be parallel to the X axis and be the Horizontal Line that passes through Y Coordinate of 2 on the Y axis

The region encompassed will be between these 2 parallel lines.


Lastly, X < Y + 1

Rearranging;

Y > X - 1

To analyze this region, first graph Y = X - 1

Y = X will be the line with a 45 degree positive slope that passes through the Origin


Adding (-1) outside the f(x) will shift the Graph of the Line Y = X ——-> 1 unit down on the Y Axis - passes through (0 , -1) Y Intercept.

The region given by Y > X - 1 will be the region to the Upper Left above the line.

The Y axis
The X axis
0 < Y < 2
And
Y > X - 1

All intersect to create a trapezoid in Quadrant 1

The height will be 2 and the 2 parallel sides will be 3 units (along line Y = 2) and 1 unit (along Y = 0)

Area of Trapezoid = (1/2) * (1 + 3) * 2 =

4

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In the rectangular coordinate system, what is the area of the region d [#permalink]
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