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On a 2-dimensional graph,....

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On a 2-dimensional graph,....

by factor26 » Sat Oct 22, 2011 10:05 am
On a 2-dimensional graph, are the points (x, y) and (a, b) equidistant from the point (0,0), the origin of the graph?

(1) x - y = -1

(2) a = x + 1 and b = 1 - y

CREDITED ANSWER IS C I UNDERSTAND A IS INCORRECT AS WE HAVE NO INFORMATION ON POINTS A,B. BUT CAN SOMEONE EXPLAIN THE LOGIC/UNDERSTANDING BEHIND THE QUESTION?

THANKS!
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by Amiable Scholar » Sat Oct 22, 2011 10:35 am
Distance of point x,y from origin
=sqrt(x^2 + y ^2)

similarly distance of point a, b from origin
=sqrt(a^2 + b^2)

now first statement doesn't solve the question here


so using second statement in a, b distance
=sqrt(x^2+1+2*x + 1 + y^2-2*y)
=sqrt(x^2 + y^2 + 2 + 2*(x-Y))

no conclusion can be drawn from here
use first statement here
=sqrt(x^2 + y^2)

tells the question answer. so both statements are compulsory to solve
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by neelgandham » Sat Oct 22, 2011 1:04 pm
Distance between two points (p,q) and (m,n) on a two dimensional plane is Square root(((p-m)^2) + ((q-n)^2))

From the above equation,
Distance from origin (0,0) to point (x,y) =
Square root(((x-0)^2) + ((y-0)^2)) = Square root((x^2) + (y^2)) -(1)
Distance from origin (0,0)to point (a,b) =
Square root(((a-0)^2) + ((b-0)^2)) = Square root((a^2) + (b^2)) -(2)

Since the question posed is 'Is (x, y) and (a, b) equidistant from the point (0,0) ?', equating (1) and (2) above, the question can be rephrased to

Is Square root((x^2) + (y^2)) = Square root((a^2) + (b^2)) ?
or
Is (x^2) + (y^2) = (a^2) + (b^2) ?

SOLUTION

1) x - y = -1 , Insufficient !
2) a = x + 1 and b = 1 - y
Squaring and adding the above you get, (a^2)+(b^2) = (x^2)+(y^2)+2(1+x-y) Insufficient !

Using 1 and 2, you get,

1) x-y = -1 => x-y+1 = 0
2)(a^2)+(b^2) = (x^2)+(y^2)+2(1+x-y)

Replacing the value of 1+x-y = 0 in the second equation you get (a^2)+(b^2) = (x^2)+(y^2, which answers the question !
Anil Gandham
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