OABC is a parallelogram whose diagonals intersect at E; the diagonal AB produced to D and OD and CD are joined.
Given that OA = a, OB = b and DB = 2(a - b), express, as simply as possible, in terms of a and b
(i) AB
(ii) ED
(iii) OD
(iv) CD
Given also that X is the point on BD such that OX = ⅓ (8b - 5a), find the value of BX/BD.
If, in addition |a| = |b| = 3 and ∡AOB = 60°, find the numerical value of |b - a|.
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Answer:
(i) AB = AO + OB
= - OA + OB
= -a + b
(ii) EB = ½ AB (since E is midpoint of AB, properties of a parallelogram)
ED = EB + BD
= ½ AB - DB
= -½ a + ½ b - 2(a - b)
= -5/2 a + 5/2 b
(iii) OD = OB + BD
= OB - DB
= b - 2(a - b)
= -2a + 3b
(iv) OC = OA + AC
= OA + OB (AC = OB)
= a + b
CD = CO + OD
= - OC + OD
= -(a + b) + (-2a + 3b)
= -3a + 2b
BD = -DB = 2(b - a)
BX = BO + OX
= -b + ⅓(8b - 5a)
= ⅓(5b - 5a)
= (5/3)(b - a)
= (5/3) ½BD
= (5/6) BD
Hence, BX/BD = 5/6
From (i), AB = b - a
Hence, |b - a| = length AB
From |a| = |b| = 3, OA = OB = 3
AB² = OA² + OB² - 2 (OA)(OB) cos 60°
AB² = 3² + 3² - 2(3)(3)(½)
AB² = 9
AB = 3
Hence, |b - a| = 3
