Qualification > Math
IGCSE Maths doubt(paper 4)
Alpha:
--- Quote from: ~~~~shreycool~~~~ on October 31, 2009, 02:03:45 pm ---7 The position vectors of points A and B, relative to an origin O, are 2i + 4j and 6i + 10j respectively.
The position vector of C, relative to O, is ki + 25j, where k is a positive constant.
(i) Find the value of k for which the length of BC is 25 units. [3]
(ii) Find the value of k for which ABC is a straight line. [3]
l\just the last part!!!
rest is easy!!!
--- End quote ---
If ABC is a str. line,
AB = hBC, where h is a constant.
AB= OB- OA
BC = OC - OB
Continues in the attachment... I allways have problem explaining answers here... Formatting is confusing... Anyway, see the attachment.
nid404:
the 2nd part is right....even I got 16 using gradient formula......should the first 1 be found using th distance formula....i dunno...im askin cuz i didn't take add math for my igs...
Ghost Of Highbury:
--- Quote from: ~~~~shreycool~~~~ on October 31, 2009, 02:03:45 pm ---7 The position vectors of points A and B, relative to an origin O, are 2i + 4j and 6i + 10j respectively.
The position vector of C, relative to O, is ki + 25j, where k is a positive constant.
(i) Find the value of k for which the length of BC is 25 units. [3]
(ii) Find the value of k for which ABC is a straight line. [3]
l\just the last part!!!
rest is easy!!!
--- End quote ---
ok..
first. --> AB=OB - OA = 4i + 6j
--> BC = OC - OB = (k-6)i + 15j
BC = x(AB)
(k-6)i + 15j = 4xi + 6xj
k-6 = 4x
6x = 15
x = 2.5
--
k-6 = 4*2.5 = 10
therefore --> k =16
nid404:
why can't u use distance formula here ???
Ghost Of Highbury:
--- Quote from: ~Alpha on October 31, 2009, 02:23:55 pm ---
If ABC is a str. line,
AB = hBC, where h is a constant.
AB= OB- OA
BC = OC - OB
Continues in the attachment... I allways have problem explaining answers here... Formatting is confusing... Anyway, see the attachment.
--- End quote ---
@nid - show how did u use the distance formula?
-----
BC = k-6)i + 15j
therefore
(k-6)2 = 252 - 152 = 400
= 20
hence, k = 26
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