proof of section formula in straight line , where a point devides line sigment in m:n ratio

Consider ? ABB’  

Since BB’||PQ and AP:PB = m:n (see figure given below)

AQ/AB' = PQ/BB' = m/(m+n) (= AP/AB) 

? x – x1/x2 – x1 = m/(m+n) 

? x = (nx1 + mx2)/(m + n) and y = (ny1 + my2)/(m + n) 

If P is outside AB (less assume it is at P’) 

We have (x – x1)/(x2 – x1) = m/(m + n) 

? x = (nx1 + mx2)/(m + n) and y = (ny1 + my2)/(m + n) 

Similarly if P is at P” then 

x = (–mx2 + nx1)/(n-m), y = (– my2 + ny1)/(n-m)

Note:  

m:n can be written as m/n or ?:1. So any point on line joining A and B will be P{(?x2 + x1)/(? + 1), (?y2 + y1)/(? +1)}. It is useful to assume ?:1 because it involves only one variable. 

Regards

Arun (askIITians forum expert)