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)
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The hardest part of this question is the language :/
I swear :P
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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hmm not a fan of absolute value functions
|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
Is this right?
yes. It is right. But I am facing the problem in understanding this question.
@Ishaan94 and @dumbcow , how would you proceed further??
Hmm they (the question) have defined a new distance formula.
|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
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)
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
It is a question from IITJEE .
it seems that no points outside (3,2) are equidistant to O and A
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.
The question is correct. I have referred from 2 books. It is a math question from IITJEE year 2000 (main)
Let me check.
See here. Question 4
Strange the question is right
I have the 30 years, I will be back in a moment.
looks like we forgot the case where y>2 but x<3
--> x = 1/2
:-/ and I was trying to blame the question
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
lol, with enough practice you won't need wolfram. goodluck. :-)
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.
@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 :)