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If a coordinate system is devised so that the positive y-axis makes an angle of 60 degrees with the positive x-axis, what is the distance between the points with coordinates (4,-3) and (5,1)?

I'm sure you guys can get it without the multiple choice answers.

Keep in mind the college board expects 18 and 17 year old kids to be able to answer this in 62 seconds on average.

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My question is whether you can at least attempt this – Henry Nov 11 '13 at 8:28
I think I just figured out my own question. It's sqrt(21). You just make a few 30 60 90 triangles one with hypotenuse 3 and one with hypotenuse 1. 3/sin90 = 1.5/sin30 = 1.5sqrt(3)/sin60 – MathematicalAnomaly Nov 11 '13 at 9:00
up vote 0 down vote accepted

One brute force way to solve this is to note that the change in $x$ values is $1$ and the change in $y$ values is $4$. By considering the triangle formed by the two given points and the point $(5,-3)$, we form a Side-Angle-Side triangle with side $1$, angle $120^\circ$, and side $4$. Applying Cosine Law, we obtain: $$ c^2 = 1^2 + 4^2 - 2\cdot 1 \cdot 4 \cdot \cos(120^\circ) = 17 - 8(-1/2) = 21 \implies c = \sqrt{21} $$

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If you design the standard coordinate system and the devised one (lets say for general with angle f, instead of $60^{\circ} $) you will see that the new coordinates are:

(x', y')=(x+cos(f)*y, sin(f)*y)

So the points A(4,-3) and B(5,1), with the standard coordinates are $A'(\frac52, -\frac{3\sqrt3}2), B'(\frac{11}2, \frac{\sqrt3}2)$

So you now get the distance A'B' with a result of $\sqrt{21}$

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