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shivam_bhalla Group Title

For Points P=(x_1,y_1) and Q(x_2,y_2) of the coordinate plane a new distance d(P,Q) is defined by d(P,Q) = |x_1 - x_2|+|y_1-y_2| Let O = (0,0) and A = (3,2). Prove that the set of points in the first quadrant which are equidistant (with respect to the new distance) from O and A consists of the union of a line segment of finite length and an infinite ray . Sketch this set in a labelled diagram. (10 marks)

  • 2 years ago
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  1. Ishaan94 Group Title
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    The hardest part of this question is the language :/

    • 2 years ago
  2. shivam_bhalla Group Title
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    I swear :P

    • 2 years ago
  3. dumbcow Group Title
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    using the new distance definition, then for some point (x,y) setting the distance equal to both O and A \[\rightarrow |x| +|y| = |x-3| +|y-2|\] now to show that is a union of line segment and ray ...

    • 2 years ago
  4. dumbcow Group Title
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    hmm not a fan of absolute value functions

    • 2 years ago
  5. Ishaan94 Group Title
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    |x| + |y| = |x-3| + |y-2| Since it's the first quadrant, x and y shall be positive. x + y = |x-3| + |y-2| Now up till (2,2) The line should be x+y = -x + 3 - y + 2 2x + 2y = 5

    • 2 years ago
  6. Ishaan94 Group Title
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    Is this right?

    • 2 years ago
  7. shivam_bhalla Group Title
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    yes. It is right. But I am facing the problem in understanding this question.

    • 2 years ago
  8. shivam_bhalla Group Title
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    @Ishaan94 and @dumbcow , how would you proceed further??

    • 2 years ago
  9. Ishaan94 Group Title
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    Hmm they (the question) have defined a new distance formula.

    • 2 years ago
  10. Ishaan94 Group Title
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    |dw:1335436017723:dw|And we are supposed to get the points which are equidistant from the points O and A, using the new distance formula. Is that right? @dumbcow

    • 2 years ago
  11. dumbcow Group Title
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    correct

    • 2 years ago
  12. shivam_bhalla Group Title
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    It should be the ditsnace individually. Suppose. See the dig below|dw:1335436193825:dw| Like distance between O and P (by new distance formula) should be equal to distance between A and Q (by new distance formula)

    • 2 years ago
  13. shivam_bhalla Group Title
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    *distance

    • 2 years ago
  14. Ishaan94 Group Title
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    I can't get the line after (3,2) x + y = x-3 + y-2 Hmm are you sure the equation is correct? @shivam_bhalla

    • 2 years ago
  15. shivam_bhalla Group Title
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    It is a question from IITJEE .

    • 2 years ago
  16. dumbcow Group Title
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    it seems that no points outside (3,2) are equidistant to O and A

    • 2 years ago
  17. Ishaan94 Group Title
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    Hmm but the result can't lie. Maybe you could check the equation again? because the question says an infinite ray and finite distance, and we didn't get the infinite one.

    • 2 years ago
  18. shivam_bhalla Group Title
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    The question is correct. I have referred from 2 books. It is a math question from IITJEE year 2000 (main)

    • 2 years ago
  19. Ishaan94 Group Title
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    Let me check.

    • 2 years ago
  20. shivam_bhalla Group Title
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    See here. Question 4 ->http://www.askiitians.com/iit-papers/IIT-JEE-2000-mathematics-mains.aspx

    • 2 years ago
  21. Ishaan94 Group Title
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    Strange the question is right http://www.askiitians.com/iit-papers/IIT-JEE-2000-mathematics-mains.aspx

    • 2 years ago
  22. Ishaan94 Group Title
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    I have the 30 years, I will be back in a moment.

    • 2 years ago
  23. dumbcow Group Title
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    http://www.wolframalpha.com/input/?i=abs%28x%29%2Babs%28y%29+%3D+abs%28x-3%29%2Babs%28y-2%29 looks like we forgot the case where y>2 but x<3 --> x = 1/2

    • 2 years ago
  24. Ishaan94 Group Title
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    ohh yeah

    • 2 years ago
  25. Ishaan94 Group Title
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    :-/ and I was trying to blame the question

    • 2 years ago
  26. shivam_bhalla Group Title
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    wish I have wolfram alpha with me in my exams. Thanks @Ishaan94 and @dumbcow for the help. I wish had 2 medals to give away

    • 2 years ago
  27. Ishaan94 Group Title
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    lol, with enough practice you won't need wolfram. goodluck. :-)

    • 2 years ago
  28. shivam_bhalla Group Title
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    Thanks :D

    • 2 years ago
  29. eliassaab Group Title
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    Study the 9 regions defined by the lines in the attacehd file, each one separtely. It would be easy to get rid the absloute sign. For eaxmple the bounded region we have \[ 0\le x \le 3\\ 0\le y \le 2 \] In this region we 3 - x + 2 -y = x + y 2 x + 2y = 5 y= -x + 5/2 Study the othe regions and see what you can get.

    • 2 years ago
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  30. shivam_bhalla Group Title
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    @eliassaab , Thanks for your help. I had initially problem in understanding the question. But I am comfortable with modulus function, just that my brain had gone blank at that time after reading the question :)

    • 2 years ago
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