Added on - 01 Oct 2019

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MATHEMATICS FOR COMPUTING COURSEWORKQuestion 1(i)As per the information given in the question, one point A of the quadrilateral has the coordinates (1,1). Now we have to take the other three points whose x and y coordinates are greater than 1. So let usplot the points in such a manner that the quadrilateral looks like a parallelogram.The points are:B (4, 2)C (5, 5)D (2, 4)The figure is shown below:
Calculating the slope of the line AB:m1= yb– ya/ xb– xa= 2 – 1 / 4 – 1= 1/3.Calculating the slope of the line BC:m2= yc– yb/ xc– xb= 5 – 2 / 5 – 4= 3/1= 3.Calculating the slope of the line CD:m3= yd– yc/ xd– xc= 4 – 5 / 2 – 5= 1/3.Calculating the slope of the line DA:m4= yd– ya/ xd– xa= 4 – 1 / 2 – 1= 3.We see that the opposite lines have the same slope thus the opposite sides are parallel.We can also calculate the mid points of the diagonals Ac and BD which comes out to be M (3, 3) hence,we see that AM = MC = 2.828 and BM = MD = 2.828.In this case, the distances i.e. AM = BM but not always as it depends on selection of points.Thus, the diagonals bisect each other.Therefore, we can say thatthe Quadrilateral ABCD is a Parallelogram.
(ii)Scaling down the matrix M by 1/3 means that all the coordinates of the points will be scaled down by1/3.Therefore, the new points are:A’ (1/3, 1/3)B’ (4/3, 2/3)C’ (5/3, 5/3)D’ (2/3, 4/3)Now rotate the points about origin by an angle of 55oclockwise one by one using the transformation:x''y''=[cos55sin55sin55cos55]x'y'For A’’:x''y''=[cos55sin55sin55cos55]1/31/3=0.460.08For B’’:x''y''=[cos55sin55sin55cos55]4/32/3=1.30.7For C’’:x''y''=[cos55sin55sin55cos55]5/35/3=2.30.4For D’’:x''y''=[cos55sin55sin55cos55]2/34/3
=1.460.18Thus, the new points after rotation by 55oare:A” (0.46, -0.08)B” (1.3, -0.7)C” (2.3, -0.4)D” (1.46, 0.18)Reflection of the parallelogram about y axis gives us the new coordinates of the points as:A”’ (-0.46, -0.08)B”’ (-1.3, -0.7)
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