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ABCD is a trapezium in which AC is parallel to BD and AC

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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Solution Summary

The expert examines ABCD a trapezium in which AC is parallel to BD and AC are 4/5.

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