Divide these intervals internally/externally (1 Viewer)

[Verify]

Hello everyone,

I'm currently stuck on these types on questions in my workbook. It would be much appreciated if I got a reply showing the formula (if there is one) and how to work it out.

A question dividing intervals internally is:
(-1,5) and (0,-4) in the ratio of 2:3

A question dividing intervals externally is:
(-2,3) and (6,1) in the ratio of 1:5.

Again, any reply is much appreciated!

 

[D94]

The formula is:

x = (nx₁+mx₂)/(m+n)

where x₁ is the first x coordinate, x₂ is the second x coordinate, and the ratio is m:n. So if the ratio was 1:5, then m = 1, and n = 5.

Likewise for y.

If it's an internal ratio, the ratio is in the form m:n, if it's external, the ratio is in the form -m:n or m:-n, where you change one of the parts to be negative. If the ratio was 2:3 externally, then m = -2, and n = 3, OR, m = 2, n = -3.

 

[Fawun]

Hello everyone,

I'm currently stuck on these types on questions in my workbook. It would be much appreciated if I got a reply showing the formula (if there is one) and how to work it out.

A question dividing intervals internally is:
(-1,5) and (0,-4) in the ratio of 2:3

A question dividing intervals externally is:
(-2,3) and (6,1) in the ratio of 1:5.

Again, any reply is much appreciated!

Here's my working out because i can't be bothered latexing the whole thing lol

I can't be bothered doing the external one lol but in the formula, m and n are the ratio 2:3 (m:n)

EDIT: I think I got it wrong lol my bad

 

[CrackerMo]

When external, sub the smaller number of the ratio into the formula as a negative

 

[kunal96]

I'm a bit rusty but anyways.. :/

I use two diff formulas for internal and external division. I find it easier to remember two formulas rather than trying to work it out in the exam..

Internal Div: (m x x2+n x x1)/(m+n) , (m x y2+n x y1)/(m+n)

External Div: (m x x2-n x x1)/(m-n) , (m x y2-n x y1)/(m-n)

Where for example in this question: (2:3)=(M:N)

AND (x1,y1)=(-1,5)
AND (x2, y2)=(0,-4)

Then sub in the values into above internal division formula... (2 x 0+3 x-1)/(2+3) AND (2 x -4 + 3 x 5)/ (2+3)=-3/5 and 7/5

For the external divison also sub in the values like you did for internal division except this time use the external division formula.

 

[kunal96]

answer to external division question should be (1, 5/4)