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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.
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 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
Dear student,
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
C divides AB externally.
From the similar triangles, CAM and CBN, we have
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
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
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