In the diagram, OABC is a parallelogram. X is a point on OB such that OX = 3 XB. P is on OA such that OA = ⅓ OP. It is given that OA = 2a, and OC = 4c.
(a) Express as simply as possible in terms of a and/or c
(i) OB
(ii) CX
(iii) CP
(b) Prove that CXP is a straight line.
(c) Find the ratio of CX:XP
(d) Given that Y is a point on AB such that CY = h CX and BY = k BA, write down a vector equation, relating vector CX, BA and CB. Hence solve for h and k.
(e) State the ratio of XY:YP.
*************************
Answer:
(a)
(i) AB = OC
Hence, OB = OA + AB = 2a + 4c
(ii) OX = 3XB
Hence, OX = ¾ OB
CX = CO + OX
= -4c + ¾ (2a + 4c)
= ½ (3a - 2c)
(iii) OA = ⅓OP
OP = 3 OA = 6a
CP = CO + OP
= -4c + 6a
(b) CP = 2 (3a - 2c)
Thus, CP = 4 CX
Hence, CXP is a straight line.
(c) CP = 4 CX
Thus, CX : XP = 1 : 3
(d) CB = CY + YB
CB = CY - BY
CB = h CX - k BA
Since CB = OA and BA = CO,
2a = h(½ (3a - 2c)) - k (-4c)
2a = (3h/2) a + (4k - h) c
Comparing coefficients of a,
3h/2 = 2
h = 4/3
Comparing coefficients of c,
4k - h = 0
k = 1/3
(e) CY = 4/3 CX
CX : XY = 3 : 1
Also, CX : XP = 1 : 3 = 3 : 9
Thus, XY : XP = 1 : 9
XY : YP = 1 : 8
