## shivam_bhalla 3 years ago 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)

1. Ishaan94

The hardest part of this question is the language :/

2. shivam_bhalla

I swear :P

3. 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 ...

4. dumbcow

hmm not a fan of absolute value functions

5. Ishaan94

|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

6. Ishaan94

Is this right?

7. shivam_bhalla

yes. It is right. But I am facing the problem in understanding this question.

8. shivam_bhalla

@Ishaan94 and @dumbcow , how would you proceed further??

9. Ishaan94

Hmm they (the question) have defined a new distance formula.

10. Ishaan94

|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

11. dumbcow

correct

12. shivam_bhalla

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)

13. shivam_bhalla

*distance

14. Ishaan94

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

15. shivam_bhalla

It is a question from IITJEE .

16. dumbcow

it seems that no points outside (3,2) are equidistant to O and A

17. Ishaan94

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.

18. shivam_bhalla

The question is correct. I have referred from 2 books. It is a math question from IITJEE year 2000 (main)

19. Ishaan94

Let me check.

20. shivam_bhalla

See here. Question 4 -> http://www.askiitians.com/iit-papers/IIT-JEE-2000-mathematics-mains.aspx

21. Ishaan94

Strange the question is right http://www.askiitians.com/iit-papers/IIT-JEE-2000-mathematics-mains.aspx

22. Ishaan94

I have the 30 years, I will be back in a moment.

23. 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

24. Ishaan94

ohh yeah

25. Ishaan94

:-/ and I was trying to blame the question

26. shivam_bhalla

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

27. Ishaan94

lol, with enough practice you won't need wolfram. goodluck. :-)

28. shivam_bhalla

Thanks :D

29. eliassaab

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.

30. shivam_bhalla

@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 :)