Ana.stasia
Junior Member
- Joined
- Sep 28, 2020
- Messages
- 118
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?
Can you show us the exact wording in the original language, so we can try to work out a better translation?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).
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.
Then we really need to see the original problem, as the difficulty seems to be entirely in deciding what it means.It's possible I drew it wrong. I am supposed to figure out y=kx +n and then use that for Ax+By+C
Then we really need to see the original problem, as the difficulty seems to be entirely in deciding what it means.
The result is supposed to be 4x +2y +6=0
Y+4=-2x-2
y+4+2x+2=0
y+2x+6=0
the point (-1,-4) is not on the line 4x + 2y + 6 = 0 ...
[MATH]4(-1) + 2(-4) + 6 \ne 0[/MATH]
Must be a mistake in the book.
Either way could you explain how you got #10
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]