F3 Maths Problems

1.
(a)
In the figure, Point P(a, 0) should be A(2a, 0).

Slope of OC = (2c - 0) / (2b - 0) = c/b
Slope of AB = (2c - 0) / [(2a + 2b) - 2a] = c/b
Slope of OC = Slope of AB
Thus, OC // AB.

Slope of BC = (2c - 2c) / [(2a + 2b) - 2b) = 0
Slope of OA = (0 - 0) / (2a - 0) = 0
Slope of BC = Slope of OA
Thus, BC // OA.

OC // AB and BC // OA
Thus, OABC is a parallelogram.

(b)
Mid-point of OB = ((2b + 2c + 0)/2, (2c + 0)/2) = (a + b, c)
Mid-point of AC = ((1a + 2b)/2, (0 + 2c)/2) = (a + b, c)

OB and AC meet at their mid-points.
Thus, the two diagonals bisect each other.


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2.
(a)
Let O be the origin and OX be the x-axis.
Hence, O = (0,0) and let X = (a, 0)

ZX // OX. ZY is thus a horizontal line. The y-coordinates of Z and Y are equal.
Then, Let (b, d) be Z and (c, d) be Y.

M and N are the mid-points of OZ and XY respectively.
Coordinates of M = ((b + 0)/2, (d + 0)/2) = (b/2, d/2)
Coordinates of N = ((a + c)/2, (0 + d)/2) = ((a + c)/2, d/2)

Slope of MN = [(d/2) - (d/2)] / [(a + c)/2 - (b/2)] = 0
MN is a horizontal line.
OX and ZY are also horizontal lines.
Thus, MN // OX // ZY

(b)
MN
= [(a + c)/2] - (b/2)]
= (a - b + c)/2

(1/2)(OX + ZY)
= (1/2)[(a - 0) + (c - b)]
= (1/2)(a - b + c)

Thus, MN = (1/2)(OX + ZY)

(a)In the figure, Point P(a, 0) should be A(2a, 0).

Slope of OC = (2c - 0) / (2b - 0) = c/b
Slope of AB = (2c - 0) / [(2a + 2b) - 2a] = c/b
Slope of OC = Slope of AB
Thus, OC // AB.

Slope of BC = (2c - 2c) / [(2a + 2b) - 2b) = 0
Slope of OA = (0 - 0) / (2a - 0) = 0
Slope of BC = Slope of OA
Thus, BC // OA.