Question
Wed February 27, 2013 By:

Prove that the points (a,0), (b,0) and (1,1) are collinear if 1/a+1/b=1.

Expert Reply
Wed February 27, 2013
Sometimes there can be printing mistakes in the book of sample papers also. If you will try to solve it using the section method and try to find the ratio, you will find that the ratio would come out to be 0 if you consider the points that you provided, which is not possible for collinear points. So, yeah, there can be mistakes everywhere. Sometimes NCERT books (even Board exam papers) also have mistakes, so, its alright. 
 
begin mathsize 16px style If space the space question space as space straight I space said space would space have space been space " Prove space that space the space points
left parenthesis straight a comma space 0 right parenthesis comma space left parenthesis 0 comma space straight b right parenthesis space and space left parenthesis 1 comma space 1 right parenthesis space are space collinear space if space 1 over straight a plus 1 over straight b equals 1 " space then space you space can space solve space it space as
Let space straight A left parenthesis straight a comma 0 right parenthesis comma space straight B left parenthesis 0 comma straight b right parenthesis space and space straight C left parenthesis 1 comma 1 right parenthesis space be space the space 3 space collinear space points.
Let space straight B space divide space the space line space segment space joining space AC space in space the space ratio space of space straight k colon 1.
Then comma space by space section space formula space the space coordinates space of space straight B space would space be space given space by space open parentheses fraction numerator straight k plus straight a over denominator straight k plus 1 end fraction comma fraction numerator straight k over denominator straight k plus 1 end fraction close parentheses
However comma space we space already space know space that space the space coordinates space of space straight B space is space left parenthesis 0 comma space straight b right parenthesis
So comma space equating space fraction numerator straight k plus straight a over denominator straight k plus 1 end fraction equals 0
straight k plus straight a equals 0
straight k equals negative straight a
Also comma space straight b space equals fraction numerator straight k over denominator straight k plus 1 end fraction
Substituting space straight k equals negative straight a space we space get
straight b equals fraction numerator negative straight a over denominator negative straight a plus 1 end fraction space
1 over straight b equals fraction numerator straight a minus 1 over denominator straight a end fraction
1 over straight b equals 1 minus 1 over straight a
1 over straight a plus 1 over straight b equals 1
Hence comma space proved.
end style
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