Cartesian Plane

[JSmith]

If a point on the Cartesian plane lies at (4, 2) what is the angle (in radians) made with the line containing the point and the origin, and the negative y-axis?

Help... so it is asking for the angle that is created from the y-axis and the line going from (0,0) out to the point?

 

[lookagain]

If a point on the Cartesian plane lies at (4, 2) what is the angle (in radians)
made with the line containing the point and the origin, and the negative y-axis? The question is not specific enough.

Help... so it is asking for the angle that is created from the y-axis and the line going from (0,0) out to the point?

Jsmith,

it is not just the y-axis. Your question mentions the negative y-axis. The question is flawed, because the
negative y-axis and the line containing (0, 0) and the origin form both an obtuse angle and an acute angle.


Jsmith,

it is not known which angle is meant from your question.

If the intent was to know the the angle on the right-hand side
of the y-axis, a couple of fixes of the question might be either
of the following:


"If a point on the Cartesian plane lies at (4, 2) what is the angle (in radians)
made with the line segment containing the point (4, 2) and the origin (0, 0),
and the negative y-axis?"


"If a point on the Cartesian plane lies at (4, 2) what is the angle (in radians)
made with the ray containing the point (4, 2), the origin (0, 0) as an endpoint
of the ray, and the negative y-axis?"

 

[dsk2]

Jsmith,

Note that the point is in the first quadrant.

slope = tan(theta) where theta is angle made with the positive x-axis. tan(theta)=(2/4).

Angle = atan(2/4) radians or atan(2/4)*180/pi degrees (use google search bar or your calculator) gives the angle that the line makes with the positive x-axis.

Here 'atan' means inverse of tan or \(\displaystyle tan^{-1}\)

Plot it out and find the answer.

Note: The angle that the line makes with the positive y-axis is different from the angle that the line makes with the negative x-axis. Hence the question specifically mentions negative y-axis. When you draw it, it will become clear to you.

Cheers,
Sai.

Edited 1:09 pm.

 

[JSmith]

Ok. So I determined the number of degrees, divided by 360. I ended up with 4.22 degrees. Is this correct??

 

[JSmith]

ok so the angle going out from the origin to the point (4,2) is 0.463 radians. Now what is my next step? I thought I would add pi/2, however that doesnt provide me with an answer that matches one of the options

 

[pka]

ok so the angle going out from the origin to the point (4,2) is 0.463 radians. Now what is my next step? I thought I would add pi/2, however that doesnt provide me with an answer that matches one of the options

What are the options?

Is one of them 1.107?

 

[lookagain]

No, please read my post. I pointed out that there are two angles. It doesn't make sense to speak
about one angle, because there are two of them even with the negative portion of the y-axis
given in the problem as a side of the angle.

The acute angle is located in the third quadrant, and the obtuse angle is located partially
in the first quadrant and partially in the fourth quadrant.

These two angles are supplementary to each other.

 

[Deleted member 4993]

If a point on the Cartesian plane lies at (4, 2) what is the angle (in radians) made with the line containing the point and the origin, and the negative y-axis?

Help... so it is asking for the angle that is created from the y-axis and the line going from (0,0) out to the point?

Have you been taught vector scalar (dot) product?

 

[soroban]

Hello, JSmith!

If a point on the Cartesian plane lies at (4, 2), what is the angle (in radians)
made with the line containing the point and the origin, and the negative y-axis?

The graph looks like this:

          |
          |               A
         2+ - - - - - - - * (4,2)
          |           *   :
          |       *       :
          |   * θ         :
      ----+---------------+----
         O| π/2           4
          |
          |
          B

We want \(\displaystyle \angle AOB \:=\:\theta + \frac{\pi}{2}\)

\(\displaystyle \tan\theta \:=\:\frac{2}{4} \quad\Rightarrow\quad \theta \:=\:\tan^{-1}(0.5) \;\approx\;0.464 \)

Therefore: .\(\displaystyle \angle AOB \;=\;0.464 + \frac{\pi}{2} \;\approx\;2.034\) radians.

 

[dsk2]

lookagain,

I think I know what you are saying. The two points naturally led me to picking only one case. But, you are right that the other angle that the line makes with the -ve y-axis is the acute angle. I was only giving a pointer to the solution, I wasn't being comprehensive.

I was confused when you referred to (0,0) as the end point of the ray. I believe this case would be the case where the origin of the ray is at (0,0) and the ray is shooting into the third quadrant, since the ray doesn't end at the origin.

I hope I am following your argument, please clarify if I am not.

Thanks for pointing out.

In summary, if \(\displaystyle \alpha\) is the angle (positive) that the line makes in the first quadrant with the positive x-axis, then the angles that the line makes with the negative y-axis are, \(\displaystyle 90+\alpha\) degrees and \(\displaystyle 90-\alpha\) degrees.

Cheers,
Sai.

P.S: There was a typo in my earlier post. It should have been 'y-axis' not 'x'-axis

 

[lookagain]

lookagain,

I was confused when you referred to (0,0) as the end point of the ray.

There is a ray (or the line segment will do) that has (0, 0) as an endpoint and passes through (4, 2).
That situation limits that to being one of the sides to the obtuse angle. The other side of the obtuse
angle is the negative y-axis (which should be a ray with an open endpoint at (0, 0)).

I believe this case would be the case where the origin of the ray is at (0,0)

and the ray is shooting into the third quadrant, since the ray doesn't end at the origin.

The open end-pointed ray in question is the negative y-axis. Actually the ray "shoots"
between the third and fourth quadrants (or 270 degrees).

...