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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. Can anyone supply me the proof with the above mentioned method. Thank you . -Neel.


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.


Can anyone supply me the proof with the above mentioned method.


Thank you.


-Neel.


Grade:12

1 Answers

SAGAR SINGH - IIT DELHI
879 Points
10 years ago

Dear student,

Case (i)

C divides AB internally.

Let A (x1, y1) and B (x2, y2) be the two points joined by line segment AB. Let C (x, y) be the point on the line segment such that 

(In this case, AC and CB are real in the same direction on the line AB.)

Draw AP, CR and BQ perpendicular to x-axis.

AM perpendicular to CR and CM perpendicular to BQ.

AM = PR = x-x1

CN = RQ = x2-x

CM = y-y1

BN = y2-y

From the similar triangles, CAM and BCN, we have

Case (ii)

C divides AB externally.

From the similar triangles, CAM and CBN, we have

 

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Sagar Singh

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