SOLUTION: Find the area bounded by the equation x/-5 + y/2=1on the first quadrant and the line x = 10.

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Question 1186100: Find the area bounded by the equation x/-5 + y/2=1on the first quadrant and the line x = 10.
Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!
given:
x%2F-5+%2B+y%2F2=1
x+=+10
--------------------
x%2F-5+%2B+y%2F2=1 ....multiply by 10
-2x%2B5y+=+10
5y+=+2x%2B10
y+=+%282%2F5%29x%2B2
intersection point is at(+10,y)
find y
y+=+%282%2F5%2910%2B2
y+=+6 => intersection point is at(+10,6)

distance between (+10,0) and (+10,6) is 6 units



the area bounded by the lines on the first quadrant is the area of a trapezoid which is equal to half the product of the height and the sum of the two bases

in your case, the sum of parallel sides is 2%2B6=8 and perpendicular distance between the _parallel sides is 10
so, the area is
Area+=+%281%2F2%29%288%29%2A%2810%29
Area+=+40 square units