GMATPrep Q

First calculate the distance between two points (-sqrt3,1) & (0,0) by applying the formula of coordinate geometry. You will get a value of sqrt2. As you can see the sector made a 90 degrees angle with help of two radius, qox is 45 degrees. Let us imagine a Pythagorean triangle oqt, where qt is perpendicular on ox. Now you can see, oqt is a 45-45- 90 triangle. As the hypotenuse is sqrt2, then the value of the remaining legs will be 1 for each one. So the correct answer is 1.
Hope it helps.

Distance between P and (0,0) is 2 not sqrt2.
Why do you say qot is 45 degress? The question doesn't say so.

Please advise?
Thanks
M

PO=2 so sqrt(s²+t²)=2
and the distance between P and Q is
sqrt( (s+sqrt(3))² + (t-1)² )=sqrt( 8 ) given we have a right triangle with sides 2 and 2

Another approach:
op=oq=2
OP forms angle 30(from left side) with x axis 1:sqroot3:2..hence OQ forms 60(180-90-30) with x axis...hence s= 1 and t=sqroot3

pepeprepa,

You need to solve the eq. "sqrt( (s+sqrt(3))² + (t-1)² )=sqrt( 8 )" in order to determine s. How do you do that?

Thanks bha.

M

héhé that's long it's clear there are better methods.

sqrt(s²+t²)=2
s²+t²=4
s²=4-t²
s=sqrt( (2+t)(2-t) )

sqrt( (s+sqrt(3))² + (t-1)² )=sqrt( 8 )
(s+sqrt(3))² + (t-1)²=8
s² + 2*sqrt(3)*s + 3 + t²-2t+1=8
s² + 2*sqrt(3)*s + t² - 2t = 4
(2+t)(2-t) + 2*sqrt(3)*(2+t)(2-t) + t² -2t=4
4 -t² + 2*sqrt(3*(2+t)(2-t)) + t² - 2t=4
2*sqrt(3*(2+t)(2-t)) -2t=0
sqrt(3*(2+t)(2-t))=t
3*(2+t)(2-t)=t²
12 + -3t² =t²
t²=3
t=sqrt(3)

héhé that's the sucker way

bha wrote:Another approach:
op=oq=2
OP forms angle 30from left side with x axis 1:sqroot3:2..hence OQ forms 60(180-90-30) with x axis...hence s= 1 and t=sqroot3

how?