Question

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.

Answer 1

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