Question 1043743: http://i.imgur.com/XGZpcvN.jpg In the figure above. AB=12. AC=15. and CD=6. Both AB and CD are perpendicular to AC. If the area of triangle ABE is p and the area of triangle CDE is q, then what is the value of p-q?
Answer by rothauserc(4718)   (Show Source):
You can put this solution on YOUR website! we are going to determine the equations of AD and BC, this will allow us to determine the altitudes of triangles ABE and CDE
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locate the vertex at the origin (0,0) then the figure is in quadrant 1 (x>0, y>0)
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slope of AD = 6/15 = 2/5 and equation of AD is y = 2x/5
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slope of BC = 12/-15 = -4/5 and equation of BC is y = -4x/5 + 12
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The coordinates for point E is determined by setting the above equations = to each other
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2x/5 = -4x/5 + 12
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2x = -4x + 60
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x = 10 then
y = 2(10)/5 = 4
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point E is located at (10,4)
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we can use the formula for the perpendicular distance from a point to a line
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d = |Am + Bn +C| / square root(A^2 + B^2) where (m,n) is the point (10,4) and
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Ax +By + C = 0
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consider line segment AB, its equation is x = 0
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d for E to AB = 10
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p = (1/2) * 12 * 10 = 60
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consider line segment CD, its equation is x -15 = 0
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d for E to CD = 5
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q = (1/2) * 5 * 6 = 15
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p - q = 60 - 15 = 45
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