700 level distance DS

[vkb16]

In the rectangular coordinate system, are the points (a, b) and (c, d) equidistant from the origin?

(1) a/b = c/d

(2) a^2 + b^2 = c^2 + d^2

oa is C

I read the mgmat explantion, but didnt quite agree with it!
I think its E

[fleurdelisse]

The distance between any two points in the rectangular coordinate system is given by the square root of the sum of difference between the coordinates of the points.

That is, the coordinates of the origin O are (0,0)

So the distance from O to point (a,b) is:
sqrt((a-0)^2+(b-0)^2))=sqrt(a^2+b^2)
More simply put, you can just draw a rectangle triangle with 0 and point (a,b) as vertexes and apply Pythagorean theorem

The same goes for the distance from the origin to point (c,d), it's:
sqrt(c^2+d^2)

Now for both distances to be equal, we need to have:
sqrt(a^2+b^2)=sqrt(c^2+d^2)
=> a^2+b^2 = c^2+d^2

SO (2) would be sufficient, I don't understand why the answer is C??

Anyone?

[vkb16]

oh my bad! sorry!

the second option is sqrt(a^2) + sqrt(b^2) = sqrt(c^2) + sqrt(d^2)

sorry..

[fleurdelisse]

In this case, 2 is not sufficient since it only tells you that a + b = c+ d

Taking both (1) and (2) together, you would get:

from (2) square both sides:

(a+b)^2 = (c+d)^2
a^2 + 2ab + b^2 = c^2 + 2cd + d^2
but from (1): ab = cd, so you can eliminate them on both sides

This results in:

a^2 + b^2 = c^2 + d^2
exactly what the question is asking for

They are equidistant, so C is your answer

[vkb16]

but from (1): ab = cd, so you can eliminate them on both sides

but from 1 we have aD = bC[/quote]

[fleurdelisse]

right! sorry about that .

Ok so, let's go back:

we have: a+b = c+d from (2)

multiply both sides of equation by d:

ad + bd = cd + d^2

since ad = bc:

bc + bd = cd + d^2
b(c+d) = d(c+d)

so b=d

and from (1), a = b

The two points are one and the same, so obviously they're equidistant to the origin