Question 9:
Diagram 2 shows a triangle EFG.
It is given ER:RF=1:2, FG:TG=3:1, −−→ER=4x˜ and −−→EG=6y˜.
a) Express in terms of x˜ and y˜: (i) −−→GR (ii) −−→GT
[3 marks]
(b) If line GR is extended to point K such that −−→GK=h−−→GR and −−→EK=6x˜−3y˜, find the value of h.
[3 marks]
Solution:
(a)(i)
−−→GR=−−→GE+−−→ER=−6y˜+4x˜
(a)(ii)
−−→GT=13−−→GF=13(−−→GR+−−→RF)=13(−6y˜+4x˜+8x˜)←−−→RF=2−−→ER=−2y˜+4x˜
(b)
−−→GK=h−−→GR−−→GE+−−→EK=h−−→GR−6y˜+(6x˜−3y˜)=h(−6y˜+4x˜)−9y˜+6x˜=−6hy˜+4hx˜Compare,−6h=−9h=32