J13

Solution
In triangle OAB, OA + AB = OB
a + AB = b
AB = b – a
Along AB, AP = 7 AB
10
= 7 ( b
10
a)
OP = OA + AP = a + 7 ( b a )
Example
10
3 a
7
10 10
= b
In the figure below, OA = a and OB = b.
(a) E xpress BA in terms of a and b.
(b) If X is the mid-point of BA, show that OX = 1 2
(a + b).
(c) Given that OC = 3a, express BC in terms of a and b.
(d) Given that BY = mBC, express OY in terms of a, b, m.
(e) If OY = nOX use the results of (b) and (d) to evaluate m and n.
Solution
(a)
In t riangle OAP,
OB + BA = OA
b+ BA = a
BA = a – b
(b)
(c)
(d)
In triangle OBX, BX = 2 1 BA
= 2 1 (a – b)
OX = OB + BX = b + 2 1 (a – b)
= 2 1 (a + b)
In OBC, BC = BO + OC
= -b + 3a = 3a – b
In OBY, OY = OB + BY
= b + m(3a – b)
= 3ma + (1 – m)b (i)
(e) OY = nOX = n 2 1 (a + b) = 2 1 na + 2 1 nb (ii)