Co-ordinate Geometry Query (1 Viewer)

[shaon0]

I find this topic the hardest in the Prelim MX1 course...i don't get the internal and external ratio concept. Could someone explain it to me?

 

[Aplus]

If point A and point B are divided by a ratio of m:n , they may be considered to be internally divided, or externally divided.

Formula For Point Of Internal Division
P = ( [mx₂ + nx₁] / [m + n] ) , ( [my₂ + ny₁] / [m + n] )

Formula For Point of External Division
The same concept is used, but where the ratio is m:n, you have to make either m or n negative. I'd prefer to go with the 2nd number, as it has a more natural feel about it.

Hence:

P = ( [mx₂ - nx₁] / [m + n] ) , ( [my₂ - ny₁] / [m + n] )

 

[shaon0]

Ok thanks

 

[kaz1]

For P = ( [mx₂ + nx₁ / [m + n] ) if the ratio is external you just change it to -m:n . Or you can use the external ratio formula.

 

[Aplus]

I'd rather have m - n than -m + n

Looks more natural and less ugly

 

[u-borat]

seriously, doing it by inspection is so much easier.

 

[lolokay]

dividing in ratio m:n from point A to B;
you're adding a ratio of m/AB to A, where AB is the distance from A to B represented either by m+n or n-m. draw a diagram to make this bit more clear

dividing internally
A + m/(m+n) * (B - A)

divide externally
A + m/(n-m) * (B - A)

just do it by inspection though. the formulas are terrible