Section Formula Doubt (Practice Assignment 2)

[Lokesh Anand]

How do we decide what should be taken as (x1,y1) and (x2,y2) in the section formula? 

In Practice Assignment 2, I took the reverse points as the solution and got 1,1 as an answer. 

My calculation below: 

x1,y1 = 1,-1

x2,y2 = -2,2

m:n = 2:1

xP = [2 X (-2) + 1 X 1]/3 = 1

yP = [2 X 2 + 1 X (-1)]/3 = 1

I saw the solution is the other way around, but how is it determined which coordinate should be taken and x1,y1 and x2,y2 respectively?

Where am I going wrong?

[Anand Iyer]

there's a calculation mistake in the x coordinate.  it's (-4 + 1)/3 = -1

[Debajyoti Biswas]

Hi Lokesh,

When you take a point to be (x1,y1), the other (opposite) co-ordinate should be cut at the ratio m.   I have attached an image, if it helps.