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The angle of line

  • Thread starter KrazyFire
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  • #1
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Homework Statement


The angle of the line (x+ SquaredRoot of 3y=12 makes with the positive direction of the x- axis is?


Homework Equations





The Attempt at a Solution


y=mx+c
Make x=0 or y=0
but dont get what to do with it :/...
 

Answers and Replies

  • #2
berkeman
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Homework Statement


The angle of the line (x+ SquaredRoot of 3y=12 makes with the positive direction of the x- axis is?


Homework Equations





The Attempt at a Solution


y=mx+c
Make x=0 or y=0
but dont get what to do with it :/...
Welcome to the PF. Is this your equation?

[tex]x + \sqrt{3y} = 12[/tex]

If so, first put it in the form of y = f(x), that is, y is a function of x. Isolate y all by itself on the left side of the = sign.

Then I think you will see that it is not in the form of a straight line, like y = mx + b

If it's not a straight line, how do you find the slope of the curve at some point?
 
  • #3
7
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Hey and thx :)

So the x- axis would be [tex]1/\sqrt[]{}3[/tex]X

Althought what would be the angle of the equation exactly

Its a Multiple Choice:
A) 30 Deg.
B) 120 Deg.
C) 150 Deg.
D) 60 Deg.
 
  • #4
berkeman
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57,304
7,282
Hey and thx :)

So the x- axis would be [tex]1/\sqrt[]{}3[/tex]X

Althought what would be the angle of the equation exactly

Its a Multiple Choice:
A) 30 Deg.
B) 120 Deg.
C) 150 Deg.
D) 60 Deg.
Could you please show your work step-by-step? You need to end up with an equation y = f(x)

Have you had calculus yet? If not, it may be easiest to just graph the function y= = f(x) and estimate the angle. The choices you are given are far enough apart, so you should be able to estimate the answer pretty well.
 
  • #5
7
0
Well y=-(1/squared root 3)x+12/squared root 3

and f(x)=-[tex]1/\sqrt{}3x[/tex]+[tex]12/\sqrt{}3[/tex]

and by using the 1,2 and[tex]\sqrt{}3[/tex] triangle

I got 30 Deg. :/?
 
  • #6
berkeman
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57,304
7,282
Well y=-(1/squared root 3)x+12/squared root 3

and f(x)=-[tex]1/\sqrt{}3x[/tex]+[tex]12/\sqrt{}3[/tex]

and by using the 1,2 and[tex]\sqrt{}3[/tex] triangle

I got 30 Deg. :/?
That does not look right. When you start with this:

[tex]x + \sqrt{3y} = 12[/tex]


What is the first step to start isolating y all by itself on the left of the = sign? And then what do you have to do to BOTH sides to get the SQRT off of the y? When you do that to both sides, you will not end up with x or SQRT(x) on the right. What do you end up with?
 
  • #7
7
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I am struggling with much demand of help in need :/
and as i have exam tomorrow i really need to brush off all these kinds of exact value questions...
Pls can you help me and answer it so i can see where to get the answer :)?
 
  • #8
berkeman
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57,304
7,282
I am struggling with much demand of help in need :/
and as i have exam tomorrow i really need to brush off all these kinds of exact value questions...
Pls can you help me and answer it so i can see where to get the answer :)?
No, sorry. We do not give answers here on the PF.

This is just an algebra problem. How do you isolate the y here? Just do it step by step. The first step is to get the x term onto the righthand side (RHS) of the equation. Show us what that looks like.

Then what do you do to both sides of the equation to turn

[tex]\sqrt{3y}[/tex]

into just

[tex]3y[/tex]
 
  • #9
7
0
Okay so what I have done is:

1) Make everything to y= which is:

[tex]\sqrt{}3[/tex]y= -x+12

2) take the Squared root over to make it:

3y=x(squared)-24x+144
 
  • #10
berkeman
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57,304
7,282
Okay so what I have done is:

1) Make everything to y= which is:

[tex]\sqrt{}3[/tex]y= -x+12

2) take the Squared root over to make it:

3y=x(squared)-24x+144
Great! You're almost there...
 
  • #11
7
0
Yeah, okay, but how do we get the angle from that equation though?
 
  • #12
berkeman
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57,304
7,282
Yeah, okay, but how do we get the angle from that equation though?
If you have taken calculus, you would differentiate the equation to get dy/dx, and set the x and y values to where the curve y = f(x) crosses the x-axis.

Without calculus, you need to plot the curve, to see what the angle is when it crosses the x-axis.

What is the value of y when the curve crosses the x-axis?
 
  • #13
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dy/dx of the equation= 2x-24
 
  • #14
berkeman
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57,304
7,282
dy/dx of the equation= 2x-24
You forgot to take the 3 across the = sign from the LHS to the RHS.

But you're on the right track. Once you get the correct dy/dx, you need to plug in the number for where the curve crosses the positive x-axis. To find that, I think you will need to factor the equation y = f(x) to find where the curve crosses the x-axis. It may cross it more than once, so be sure to check that you find a solution for the positive x-axis.
 

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