ABCD is a trapezium in which AC is parallel to BD and AC
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In this question ABCD is a trapezium in which AC is parallel to BD and AC is 4/5 times
the length of BD....See attached file for full problem description.
Find Image of C C1
A1C1 // B1D1
A1C1 must have x-coordinate = 1
B1D1 = 1
A1C1 =
1 =
Therefore, C1 = (1, )
Find Image of M M1
M = intersection of AC and BD
M1 = intersection of A1C1 and B1D1
Equation of A1B1 y = 0
y = mx + c
m =
=
=
c = 1
y = x + 1
Substitute y = 0 into the equation y = x + 1
0 = x + 1
- 1 = x
x = 5
Therefore, M1 = (5, 0)
Find Image of N N1
N = intersection of AD and BC
N1 = intersection of A1D1 and B1C1
Equation of A1D1 y = mx + c
m = - 1 and c = 1
hence, equation of A1D1 is: - y = - x + 1
2 points: - (0, 0) and (1, )
Equation of B1C1 y = mx + c
m =
=
=
c = 0
y = x
y = - x + 1
y = x
x = - x + 1
x =
y = x
=
=
Therefore, N1 = ( , )
Find Image of P P1
P = intersection of MN and AC
P1 = intersection of M1N1 and A1C1
Equation of A1C1 x = 1
2 points: - (5, 0) and ( , )
y = mx + c
m =
=
=
y = x + c
Substitute (5, 0) into the equation y = x + c
y = (5) + c
c =
y = x +
Substitute x = 1 into the above equation: - y = x +
y = (1) +
=
Therefore, P1 = (1, )
Find Image of Q Q1
Q = intersection of MN and BD
Q1 = intersection of M1N1 and B1D1
y - = x
Equation of B1D1 x = 0
y - = (0)
y =
Therefore, Q1 = (0, )
________________________________________________________________________
b)
A¬¬1 = (1, 0); C1 = (1, ); P1 = (1, )
=
=
=
Therefore, AP: PC = 1: 1
M¬¬1 = (5, 0); P1 = (1, ); Q1 = (0, )
=
=
Therefore, MP: PQ = 4: 1
________________________________________________________________________
c)
B B1 = (0, 0) B1 = (0, 0) B = (2, 0) =
D D1 = (0, 1) A1 = (1, 0) A = (2, 1)
A A1 = (1, 0) D1 = (0, 1) D = (3, 2)
t-1 = A +
=
A =
=
=
coordinate of M:-
t-1 (M1) = +
= + =
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