I know that the section formula can be derived by using the similarity of triangles concept.But can it also be be derived using the distance between two points concept, i.e., for line AB with point p dividing it in the ratio m : n; here A = (x1,y1), B = (x2,y2) and p = (x,y).(distance of AP) / (distance of PB) = m/nI tried doing it but got ended up getting a very scary looking expression.Any help in this regard will be really appreciated.-Neel.

Dear student,

You are on the right track and this is the way of solving section formula....

Please feel free to ask your queries here. We are all IITians and here to help you in your IIT JEE preparation.

All the best.

Win exciting gifts by answering the questions on Discussion Forum. So help discuss any query on askiitians forum and become an Elite Expert League askiitian.

Now you score 5+15 POINTS by uploading your Pic and Downloading the Askiitians Toolbar  respectively : Click here to download the toolbar..

Askiitians Expert

Sagar Singh

B.Tech, IIT Delhi

sagarsingh24.iitd@gmail.com

section formula

if a company wants to insert a tower between two points in a way such that building of  darshanam vatika is 5 times nearer to tower than the building of gordhan park.

darshanam vatika(34km,28km);gordan park(83km,56km)

position of tower=[5*34+83]/6 , 5*28+56/6]

                       = [253/6,196/6]

                       = [42.16km,32.66km]

the distance between dv and gp is divided in such a way that the tower is five times as close to dv than gp.Thuscoordinates of dv is taken five times than coordinates of gp.

you can check this:

difference of tower from dv(x coordinate)=42.16-34=8.16

difference of tower from gp(x coordinate)=42.16-83=-40.84km 0r distance=40.84km

therefore;40.84/8.16=5