Finding the angle between two line equations

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I need to find the angle between two lines in $y = mx + b$ form and they are:

\begin{equation*} y = 4x + 2~\text{and}~y = -x + 3 \end{equation*}

I have no idea how to solve this and if you could please consider that I'm in Grade 12 Calculus and Vectors. So my knowledge consist of adding vectors, derivatives, etc.

Somehow the answer is suppose to be 59 degrees but I don't know how to get that

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8 Answers

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1. Find the slope of each line.
2. Find the angle of inclination of each line, using $\theta=\tan^{-1}m$. (Here, $\theta$ is the angle of inclination, $m$ is the slope.)
3. Subtract the two angles.
4. Handle the case where this difference is not an acute angle. (If you get a negative angle, take its absolute value. Also, when two lines intersect, they form two pairs of equal angles. Unless the lines are perpendicular, one pair will be acute and the other obtuse. You want to find the acute pair, so if you calculated the obtuse pair just subtract the value from pi radians or $180°$ to get the acute values.)

There are other ways, such as finding vectors on the lines and using their dot products. But the way I showed uses simple, basic trig methods.

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The final answer is

$$\pi-\left|\tan^{-1}(4)-\tan^{-1}(-1)\right|$$

converted to degrees, if the calculator is in radians mode, or

$$180°-\left|\tan^{-1}(4)-\tan^{-1}(-1)\right|$$

if the calculator is in degrees mode.

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You should be knowing tan and arctan...

Angle between two lines of slope $m_1$ and $m_2$ is $ \tan^{-1}\dfrac{m_1 - m_2}{1+m_1 m_2} $

This is derived on basis of tangent of angular difference of given lines.

In your case $ m_1 = 4 $ and $ m_2 =-1$.

One angle the first line makes to x-axis $ \phi_1$ is in the first and the other $\phi_2$ is in the second quadrants, so you can take the difference angle in anticlockwise direction. Angle between them is $ 45^{\circ} +75.96^{\circ}=$

$ \tan^{-1}\dfrac{4 - (-1)}{1+(4 *-1)}=\tan^{-1}\dfrac{5}{-3}\approx 120.96^{\circ} $

all anti clockwise direction as shown in Geogebra construction.

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the first vector that parallel to $y=4x+2$ $$v_1=i+4j$$ the second vector is $$v_2=-i+j$$

then use the following formula $$\theta =\cos^{-1}(\frac{v_1.v_2}{|v_1||v_2|})$$ $$\theta =\cos^{-1}(\frac{(1\times-1+4\times1)}{\sqrt{17}\sqrt{2}})=59.03624347$$

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By elementary trigonometry, the angle formed by a line with the horizontal axis is given by $$\tan\theta=m,$$ or $$\theta=\arctan(m)$$ (if you increase $x$ by $1$, $y$ increases by $m$). So the angle between your two lines is obtained by difference

$$\arctan(4)-\arctan(-1)=75.963\cdots°-(-45°)=120.963\cdots°$$

But actually there are two angles that can be considered, supplementary to each other, and the other is

$$180°-120.963\cdots°=59.036\cdots°.$$

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The man at the top got it right, but failed to remember that when using the tan function on a calculator it only returns angles in the first and fourth quadrant. $\tan (-1)$ is an example of this and to get the correct angle you must subtract $180^{\circ}$ or $\pi$ to get the inclination angle. If you follow this step of $180 - | (180+\tan (-1)) - \tan(4)|$ then you will get the angle of $59.03624347$.

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Here is a way to do it with the cosine rule. It's very long but if you are not getting the right answer you could try this method.

First, find the intercept of the two equations:

$4x+2=−x+3, x=0.2, y=4(0.2)+2, y=2.8, (0.2, 2.8)$

Now, sub in any y value into both the equations and find the coordinates. I will be using 10 as an example.

Eqn 1: $10=4x+2, x=2, (2,10)$

Eqn 2: $10=-x+3, x=-7, (-7,10)$

Then, find the distance between all three of the points. These three distances will be used subbed in the cosine rule so make sure that the distance that does not connect to the intercept point is $a$

$a$: $2-(-7)=9$

$b$: $\sqrt{(10-2.8)^2+(2-0.2)^2}=7.42$

$c$: $\sqrt{(10-2.8)^2+(-7-0.2)^2}=10.18$

Sub the values into the formula $cos(\theta)=\frac{b^2+c^2-a^2}{2bc}$

$\theta=cos^{-1}(\frac{(7.42)^2+(10.18)^2-(9)^2}{2(7.42)(10.18)})=59°$(approximate answer)

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You can use the unit circle. If you know the slope you can set up a system of equations.

Y = mx +b

X^2 + y^2 = 1

(Unit circle equation)

If you solve for x do arccos(x) If you solve for y do arcsin(y)

This will give you the angle of a line to positive part of the x axis. If you have two lines simply do this twice and find the difference between them.

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We have tangents and cosines above, but one can also use sines. Take the vector cross product of v1=i+4j and v2=-i+j. Then |v1 x v2|/|v1||v2| = Sin(angle between them).

On HP Prime vector operations are hidden under matrices. Go into the MATH toolbox, then select MATRIX/VECTOR/CROSS PRODUCT. Poke your two vectors in and you will find that the cross product is 5. In the same menu as cross product, use L2norm to calculate the norms of [1,4,0] = sqrt(17) and [-1,1,0] = sqrt(2). Then sin(angle)=5/sqrt(34), so angle = asin(5/sqrt(34)) = 59.0362.... degrees if your machine is set to the degrees mode.

The TInspire CAS machine has similar capability.

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