The angle between two straight lines.

[adjacent]

Homework Statement

Find the angle between two straight lines,##y=x## and ##y=2x##.
Find the general formula for finding the angles.

Homework Equations

##f(x)=mx+c##
##\tan(\theta)=\frac{opposite}{adjacent}##

The Attempt at a Solution

First I assign a value for x and calculate the value for y.Using the tan rule,I can find the angle(β) between the line y=x and x-axis.
Again repeat the same with y=2x and find the angle(∑).The angle between y=2x and y=x is ∑-β.

First I plugged in a number into the function.
Then the resulting value becomes the opposite and the x becomes the adjacent .
So I get $$\tan^{-1} \left( \frac{f(n)}{n} \right) - \tan^{-1} \left( \frac{g(n)}{n} \right)$$.
Where n is any number.f is the function of the line with the greater slope.g is the function of the line with the lower slope.
Is this correct? I am really worried about the terminology.

 

[SteamKing]

Homework Statement

Find the angle between two straight lines,##y=x## and ##y=2x##.
Find the general formula for finding the angles.

Homework Equations

##f(x)=mx+c##
##\tan(\theta)=\frac{opposite}{adjacent}##

The Attempt at a Solution

First I assign a value for x and calculate the value for y.Using the tan rule,I can find the angle(β) between the line y=x and x-axis.
Again repeat the same with y=2x and find the angle(∑).The angle between y=2x and y=x is ∑-β.

First I plugged in a number into the function.
Then the resulting value becomes the opposite and the x becomes the adjacent .
So I get $$\tan^{-1} \left( \frac{f(n)}{n} \right) - \tan^{-1} \left( \frac{g(n)}{n} \right)$$.
Where n is any number.f is the function of the line with the greater slope.g is the function of the line with the lower slope.
Is this correct? I am really worried about the terminology.

The description of your method seems a bit round-about.

Look at it this way. Both of your lines have a y-intercept of zero. If you draw a simple sketch,
you can make a couple of right triangles. You know the slope of each line from its equation, and you should be able to find the acute angle by taking the arctan of the slope, since the slope is equal to the tangent of the acut angle.

The angle between the lines should be arctan (2) - arctan (1).

 

[adjacent]

The description of your method seems a bit round-about.

Did you understand my answer?
What I did was take a point in the x axis,for example, (4,0) and input the x value into the function to get the y value of the line at that point of x.Since we have got an opposite and an adjacent,we can use tan rule to find the angle.

Look at it this way. Both of your lines have a y-intercept of zero. If you draw a simple sketch,
you can make a couple of right triangles. You know the slope of each line from its equation, and you should be able to find the acute angle by taking the arctan of the slope, since the slope is equal to the tangent of the acut angle.

The angle between the lines should be arctan (2) - arctan (1).

Slope is equal to the tangent of the angle?
I think that method only works for lines passing through the origin.I was asked(Actually I am making it now lol) to find the general formula for all types of lines.
My method will also simplify to that if the lines passes through the origin.It also works for lines passing through any point in the y-axis.

 

[SteamKing]

The slope of a line is defined as the rise over the run, or the change in y over the change in x. Ergo, the tangent of the angle which the line makes with the x-axis is equal to the slope.

 

[HallsofIvy]

The simplest way to answer the question "what is the angle between a line with slope [itex]m_1[/itex] and [itex]m_2[/itex] (i.e. between the lines with equations [itex]y= m_1x+ b_1[/itex] and [itex]y= m_2x+ b_2[/itex].)?" is to use the "difference" formula:
[tex]tex(\alpha- \beta)= \frac{tan(\alpha)- tan(\beta)}{1+ tan(\alpha)tan(\beta)}[/tex]

Here, [itex]tan(\alpha)= m_1[/itex] and [itex]tan(\beta)= m_2[/itex] so that [itex]\alpha- \beta[/itex], the angle between the lines, satisfies [tex]tan(\alpha- \beta)= \frac{m_1- m_2}{1+ m_1m_2}[/tex]

 

[jbunniii]

Another approach is to use vectors and the dot product. A vector pointing in the direction of the first line is ##(1, m_1)##, and a vector pointing in the direction of the second line is ##(1, m_2)##. Then
$$\cos(\theta) = \frac{(1, m_1) \cdot (1, m_2)}{\sqrt{1 + m_1^2} \sqrt{1 + m_2^2}} = \frac{1 + m_1 m_2}{\sqrt{1 + m_1^2} \sqrt{1 + m_2^2}}$$
In this example, we have ##m_1 = 1## and ##m_2 = 2##, so ##\theta = \cos^{-1}(3 / \sqrt{10})##. It's not immediately obvious that this equals ##\tan^{-1}(2) - \tan^{-1}(1)##, but Matlab assures me that it does.