HOW TO DETERMINE THE EQUATION USING ARC TAN???

A

Ana.stasia

Junior Member
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The question is:
Determine the equation of the line containing the point (-1, -4) and with the positive part of the abscissa form the angle arc tan (-2)

My biggest problem is arc tan (-2). What am I supposed to do with it???

IMG_20210205_170710.jpg
 
Does the problem say "positive part of the abscissa"? What does that mean? The abscissa is a number and does not have positive and negative parts.

If the arc tangent is - 2, what does that tell you about the sine and cosine?
 
JeffM said:
Does the problem say "positive part of the abscissa"? What does that mean? The abscissa is a number and does not have positive and negative parts.

If the arc tangent is - 2, what does that tell you about the sine and cosine?
Click to expand...

The problem does say "the positive part of abscissa", however, that's my straight up translation of it so It's possible there is a language barrier as I don't know how It should be translated. In the equation y=kx +n "the positive part of abscissa is supposed to be that" n"

The best I got is that since arc tan is tan^-1 (-2) tan is sine (-2) divided by cosine (-2).
 
Your sketch shows a line w/ a positive slope ... I don't think this is the case.

The slope of any non-vertical line is [MATH]m = \dfrac{\Delta y}{\Delta x} = \tan{\theta}[/MATH], where [MATH]\theta[/MATH] is the angle the line forms relative to the direction of the positive x-axis.

[MATH]\theta = \arctan(-2) \implies \tan{\theta} = -2 \implies m = -2[/MATH]
use that slope and the point to determine the desired equation
 
Ana.stasia said:
The problem does say "the positive part of abscissa", however, that's my straight up translation of it so It's possible there is a language barrier as I don't know how It should be translated. In the equation y=kx +n "the positive part of abscissa is supposed to be that" n"

The best I got is that since arc tan is tan^-1 (-2) tan is sine (-2) divided by cosine (-2).
Click to expand...
Can you show us the exact wording in the original language, so we can try to work out a better translation?

It sounds like you are saying that "the positive part of abscissa" is used to mean what we call the y-intercept (taken as a number rather than a point); but your picture shows an angle at the x-intercept. And since the angle shown is acute, its tangent is positive. So there is a lot here that is confusing.
 
Dr.Peterson said:
Can you show us the exact wording in the original language, so we can try to work out a better translation?

It sounds like you are saying that "the positive part of abscissa" is used to mean what we call the y-intercept (taken as a number rather than a point); but your picture shows an angle at the x-intercept. And since the angle shown is acute, its tangent is positive. So there is a lot here that is confusing.
Click to expand...

It's possible I drew it wrong. I am supposed to figure out y=kx +n and then use that for Ax+By+C
 
Ana.stasia said:
It's possible I drew it wrong. I am supposed to figure out y=kx +n and then use that for Ax+By+C
Click to expand...
Then we really need to see the original problem, as the difficulty seems to be entirely in deciding what it means.
 
Dr.Peterson said:
Then we really need to see the original problem, as the difficulty seems to be entirely in deciding what it means.
Click to expand...

I'll try to explain it to the best of my abilities.
The goal is to determine the equation of a line. (this question is taken from a lecture about this form of equation: y= kx +n)
That line contains a point (-1,-4)
And as far as I understand, that line forms an angle arc tg (-2) with the positive side of the x "spectrum" (however you call it)

I know that k = tg a, and that a (alpha) is the angle mentioned in the text.
 
Ana.stasia said:
Must be a mistake in the book.
Either way could you explain how you got #10
Click to expand...

point-slope form of a linear equation is ...

[MATH]y - y_1 = m(x - x_1)[/MATH], where [MATH](x_1,y_1)[/MATH] is a point on the line and [MATH]m[/MATH] is the slope

in this problem, [MATH]x_1 = -1 \text{ and } y_1 = -4[/MATH].
slope is [MATH]m = -2[/MATH]
 
skeeter said:
point-slope form of a linear equation is ...

[MATH]y - y_1 = m(x - x_1)[/MATH], where [MATH](x_1,y_1)[/MATH] is a point on the line and [MATH]m[/MATH] is the slope

in this problem, [MATH]x_1 = -1 \text{ and } y_1 = -4[/MATH].
slope is [MATH]m = -2[/MATH]
Click to expand...

That helped me solve it. Thank you
 
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