SOLUTION: Graph the following ewuation and compute the area it encloses: abs(2x -10) + abs (5y

SOLUTION: Graph the following ewuation and compute the area it encloses: abs(2x -10) + abs (5y - 10) = 20

Question 41947: Graph the following ewuation and compute the area it encloses:
abs(2x -10) + abs (5y - 10) = 20
Answer by psbhowmick(878) (Show Source): You can put this solution on YOUR website!
Graph the following ewuation and compute the area it encloses:
abs(2x -10) + abs (5y - 10) = 20

When x < 5, 2x < 10 or |2x - 10| = 10 - 2x
i) When y < 2, 5y < 10 or |5y - 10| = 10 - 5y. So the given equation becomes 2x + 5y = 0 (brown).
ii) When y > 2, 5y > 10 or |5y - 10| = 5y - 10. So the given equation becomes 2x - 5y + 20 = 0 (green).

When x > 5, 2x > 10 or |2x - 10| = 2x - 10
i) When y < 2, 5y < 10 or |5y - 10| = 10 - 5y. So the given equation becomes 2x - 5y = 20 (blue).
ii) When y > 2, 5y > 10 or |5y - 10| = 5y - 10. So the given equation becomes 2x + 5y = 40 (violet).

From the graph find the coordinates of the points of intersection of the straight lines.
The coordinates of the points of intersection of the brown and green lines are (-5,2).
The coordinates of the points of intersection of the brown and blue lines are (5,-2).
The coordinates of the points of intersection of the blue and violet lines are (15,2).
The coordinates of the points of intersection of the violet and green lines are (5,6).
So, clearly, the quadrilateral formed by the intersection of the straight lines is a rhombus.
The diagonals of this rhombus are 8 and 20 units respectively.
Hence the reqd. area = area of the rhombus = x 8 x 20 = 80 sq units.
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