# Angle $\theta$ lies in quadrant II with point $A(-4, 6)$ on the terminal arm.

Angle $\theta$ lies in quadrant II with point $A(-4, 6)$ on the terminal arm. Point $P$ is the point of intersection of the terminal arm of $\theta$ and the unit circle centered at $(0,0)$. Determine the exact value of the $X$ coordinate of point $P$.

Not sure if I'm attempting it correctly... and not sure what to do next if I'm wrong. Could you explain and show the correct reasoning?

My Attempt: For the terminal arm, since one end is at the origin, the intercept value $b$ of the terminal arm's straight-line equation $$y = mx + b$$ must be $b = 0$. When $y = 6$ for this line, $x = -4$ so that the slope value $m$ must be such that $$6 = m\cdot(-4)$$ Thus $m = -3/2$ and $y = (-3/2)\cdot x$ for all points of the terminal arm's line. For the unit circle, $x^2 + y^2 = 1$. Substitute for $y$ from the straight-line equation into the circle's equation and get $x^2 + [(-3/2)x]^2 = 1$.

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The reasoning is correct. You wil then have to solve the final equation in $x$ that you got. There are two solutions. You will want to use the negative one, since the point $P$ is in the second quadrant.

I would prefer to argue as follows. Our point $P$ will have coordinates $(-4t,6t)$, where $t$ is positive. This point is on the unit circle, and therefore $$(-4t)^2 +(6t)^2=1^2.$$ We get $52t^2=1$, and therefore $t=\dfrac{1}{\sqrt{52}}$.

Thus the required $x$-coordinate is $-\dfrac{4}{\sqrt{52}}$. There are various ways to express this in different ways, such as $-\dfrac{2}{\sqrt{13}}$ or $-\dfrac{2\sqrt{13}}{13}$.

The argument above uses the same basic ideas as yours, but may feel a little simpler.

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An easy way to solve for this problem is to use similarity.

The points $A(-4,6), O(0,0) \text{and} B(-4,0)$ form a right triangle with length $$OA = 2\sqrt{13}, OB = 4, BA = 6$$.

On the unit circle, you can find a point, say, $A'$ which is the intersection with the linesegment $OA$.

The length of $OA'$ equals $1$, so using similarity,

$$\frac{OA}{OA'} = \frac{OB}{OB'}$$

$$2\sqrt{13} = {4 \over x}$$

Where $x$ is the distance from the origin. And then you can solve for the $x$-coordinate.

Does that help ?

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Note that $x$ here is just the positive horizontal distance from the point to the origin. Don't forget that the $x$-coordinate is actually negative. –  Adriano Jun 1 '13 at 23:26