A diagram showing a trapezium ABCD in which AD is paralel to BC and AB is perpendicular to BC and AD. The coordinates of A, B, C are (-2,5), (3,9),(7,4) respectively. AD cuts the x-axis at E. Given further that AE:ED is 2:3 and AE=BC, find the coordinates of D.

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A diagram showing a trapezium ABCD in which AD is paralel to BC and AB is perpendicular to BC and AD. The coordinates of A, B, C are (-2,5), (3,9),(7,4) respectively. AD cuts the x-axis at E. Given further that AE:ED is 2:3 and AE=BC, find the coordinates of D.

Mathematics
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D = (8,-7.5) slope of AB is 4/5, therefore the perpendicular slope of AD is -5/4 using point A (-2,5) the line connecting A to D: y = -5/4x + 5/2 the x-intercept can be found by plugging in 0 for y x = 2, therefore point E =(2,0) AE =BC = sqrt(41) using the given ratio: sqrt(41)/ED = 2/3 --> ED = 3sqrt(41)/2 Use the distance formula to find point D (x,y) sqrt[(x-2)^2 +y^2] = 3sqrt(41)/2 substitute in for y sqrt[(x-2)^2 +(-5/4x + 5/2)^2] = 3sqrt(41)/2 (x-2)^2 +(-5/4x + 5/2)^2 = 92.25 expand and simplify --> x^2 -4x -32 = 0 (x-8)(x+4) = 0 x = 8, x can't be -4 because of the direction of line AD plug it into equation of line to find y y = -5/4(8) +5/2 = -7.5

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