Oh no ! Im stuck again.!! (1 Viewer)

Peek-a-boo!!!

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I have a question to ask again.Doh!!!. Someone will help me right? Thank-you!
the question is" THE interval AB is divided internally and externally in the ration 3:2. If A and B are the points(-1,3) and (5,8). Find the coordinates of the points of division." There! WHat a question! Help me!! :confused::eek:
 
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pLuvia

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Internal Division
Pt=([my1+nx1/(m+n)],[my2+nx2]/[m+n])
External Division
Pt=([my1-nx1/(m-n)],[my2-nx2]/[m-n])

For your question
Internal Pt
C=([3x3+2x-1]/[3+2],[3x8+2x5]/[3+2])=(7/5,34/5)
Exterrnal Pt
C=([3x3-2x-1]/[3-2],[3x8-2x5]/[3-2])=(11,14)
 

NT-social

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internally, u need to make one of the numbers in the ratio negative(usually the smaller number)
i.e.
internally
(-1,3) (5,8)
A : B
= 3 : -2


working out:
x= ((-1 x -2) + (5 x 3))/(3+ -2) = 17/1 =17
y= ((3 x -2) + (8 x 3))/(3+-2)= 18/1 = 18

therefore point= (17,18)

im assuming u know the formula, but thats the working out for the internally...
for externally it would be the same working out except with the ratio 3:2...ie no negative
answer=
point = (13/5,6)
 

Peek-a-boo!!!

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Wait i sort of understand the question but why the internal points have to be negative and why the external points have to be positive. ? Im just a liitle bit confused rite there!!
 

airie

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Because, (say the internal point of division is C and the external point of division id D) AC and BC are two intervals in opposite directions whereas AD and BD are in the same direction. See the diagram attached (note: not to scale :p)
 

Riviet

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You make one negative for finding an external ratio and leave the ratio positive for an interval ratio.
 

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