SOLUTION: What is the area of the figure bounded by x = -3, x = 2, y + 4 = (-2/5)(x - 2), and 5y - x - 13 = 0 when graphed on a coordinate plane?

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Question 1152213: What is the area of the figure bounded by x = -3, x = 2, y + 4 = (-2/5)(x - 2), and 5y - x - 13 = 0 when graphed on a coordinate plane?
Found 2 solutions by MathLover1, MathTherapy:
Answer by MathLover1(20850)   (Show Source): You can put this solution on YOUR website!

What is the area of the figure bounded by
, ,
, ->
->




when graphed on a coordinate plane,the figure bounded by is a trapezoid

The area is the average of the two base lengths times the altitude:

from graph you see that (the distance from to )
we need to find coordinates of the ,,, and
vertex is intersection of the lines and
substitute in




=> is at (,)

vertex is intersection of the lines and
substitute in



=> is at (,)


vertex is intersection of the lines and
substitute in




=> is at (,)

vertex is intersection of the lines and



=> is at (,)

now find the length of bases and

...use coordinates of and

=> is at (,)
=> is at (,)







=> is at (,)
=> is at (,)





so, , , , and the area of the figure is:








Answer by MathTherapy(10552)   (Show Source): You can put this solution on YOUR website!

What is the area of the figure bounded by x = -3, x = 2, y + 4 = (-2/5)(x - 2), and 5y - x - 13 = 0 when graphed on a coordinate plane?
When plotted, the graphs of: 
a) and x = - 3 intersect at the point (- 3, 2)
b) and x = - 3 intersect at the point (- 3, - 2)
c) and x = 2 intersect at the point (2, 3)
d) and x = 2 intersect at the point (2, - 4)
These four points form a trapezoid, with a height of 5 (2 - - 3), as seen on the x-axis
With the trapezoid's shorter base being 4 [from (- 3, 2) to (- 3, - 2)] units, and longer base being 7 [from (2, 3) to (2, - 4)],
we then get the trapezoid's area or the area bounded by the four line graphs as:
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