anonymous
  • anonymous
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)
Mathematics
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
The hardest part of this question is the language :/
anonymous
  • anonymous
I swear :P
dumbcow
  • dumbcow
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 ...

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dumbcow
  • dumbcow
hmm not a fan of absolute value functions
anonymous
  • anonymous
|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
anonymous
  • anonymous
Is this right?
anonymous
  • anonymous
yes. It is right. But I am facing the problem in understanding this question.
anonymous
  • anonymous
@Ishaan94 and @dumbcow , how would you proceed further??
anonymous
  • anonymous
Hmm they (the question) have defined a new distance formula.
anonymous
  • anonymous
|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
dumbcow
  • dumbcow
correct
anonymous
  • anonymous
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)
anonymous
  • anonymous
*distance
anonymous
  • anonymous
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
anonymous
  • anonymous
It is a question from IITJEE .
dumbcow
  • dumbcow
it seems that no points outside (3,2) are equidistant to O and A
anonymous
  • anonymous
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.
anonymous
  • anonymous
The question is correct. I have referred from 2 books. It is a math question from IITJEE year 2000 (main)
anonymous
  • anonymous
Let me check.
anonymous
  • anonymous
See here. Question 4 ->http://www.askiitians.com/iit-papers/IIT-JEE-2000-mathematics-mains.aspx
anonymous
  • anonymous
Strange the question is right http://www.askiitians.com/iit-papers/IIT-JEE-2000-mathematics-mains.aspx
anonymous
  • anonymous
I have the 30 years, I will be back in a moment.
dumbcow
  • dumbcow
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
anonymous
  • anonymous
ohh yeah
anonymous
  • anonymous
:-/ and I was trying to blame the question
anonymous
  • anonymous
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
anonymous
  • anonymous
lol, with enough practice you won't need wolfram. goodluck. :-)
anonymous
  • anonymous
Thanks :D
anonymous
  • anonymous
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.
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anonymous
  • anonymous
@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 :)

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