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```        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/n

I tried doing it but got ended up getting a very scary looking expression.
Any help in this regard will be really appreciated.
-Neel.
```
6 years ago

SAGAR SINGH - IIT DELHI
879 Points
```										Dear student,
You are on the right track and this is the way of solving section formula....

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.

Sagar Singh
B.Tech, IIT Delhi
sagarsingh24.iitd@gmail.com

```
6 years ago
sudeep mohanty
18 Points
```										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

```
5 years ago
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