OABC is a rectangle in which OC = 2x, CB = 2y, and M is the midpoint of AB. OA is produced to D such that 3OA = 2OD and OM is produced to meet BD at N.
(a) Express the following vectors in terms of x and y, giving each answer in its simplest form.
(i) OB,
(ii) OM,
(iii) BD
(b) If BN = α BD and ON = β OM, form an equation connecting α, β, x and y and find the values of α and β.
(c) Hence, find the value of
(i) OM/ON
(ii) area of ΔMDN / area of ΔMDO
(iii) area of ΔOBD / area of trapezium OCBD
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Answer:
(a)
(i) OB = OC + CB = 2x + 2y
(ii) AB = OC = 2x
OA = CB = 2y
OM = OA + ½AB
= 2y + x
(iii) 3OA = 2 OD
OD = 3/2 OA = 3y
BD = BO + OD
= OD - OB
= 3y - (2x + 2y)
= y - 2x
(b) BN = α BD
BO + ON = α( y - 2x)
-(OB) + β OM = α( y - 2x)
-(2x + 2y) + β(2y + x) = α( y - 2x)
Comparing coefficients of x,
-2 + β = -2α ----------------------(1)
Comparing coefficients of y,
-2 + 2β = α -----------------------(2)
Sub (2) into (1)
-2 + β = -2(-2 + 2β)
-2 + β = 4 - 4β
5β = 6
β = 6/5
α = -2 + 2β
α = -2 + 2(6/5)
α= 2/5
(c)
(i) ON = 6/5 OM
OM / ON = 5/6
(ii) area of ΔMDN / area of ΔMDO
= MN / OM (since they have the same common height)
= (ON - OM) / OM
= ON / OM - OM / OM
= 6/5 - 1
= 1/5
(iii) Area of ΔOBC / Area of ΔOBD
= Area of ΔOAB / Area of ΔOBD (since area of ΔOBC = area of ΔOAB)
= OA / OD (since they have the same common height)
= 2/3
area of trapezium OCBD / area of ΔOBD (find opposite first)
= (area of ΔOBD + area of ΔOBC) / area of ΔOBD
= 1 + Area of ΔOBC / Area of ΔOBD
= 1 + 2/3
= 5/3
Thus,
area of ΔOBD / area of trapezium OCBD = 3/5 (flip over)
