Angle between two lines given the slopes

[@jc.nogueira]

Hi, I believe it would be easier just to get both slopes (m1, m2) from the equations. Since m1=tan(alpha1) and m2=tan(alpha2), we get both angles. Then we can calculate the angle between the slopes. I really enjoy your channel. Greetings from Uruguay, South America.

[@Dominus_Potatus]

I thought a really conplicated workings and forget about the simple working. This is why I love your videoes, really reminds me how math is'nt supposed to be complicated.

[@Mehrdad_Basiry-fj4rl]

Nice question... learn alot...❤❤❤.

[@mattsgamingstuff5867]

I did it by defining unit vectors along the lines and using the dot product. a (dot) b = ab cos theta. Since I made a and b unit I just took the arccos of the dot product and it gave me an angle. It ended up being the larger angle but it was not hard to find the smaller angle since they add up to 180 degrees.

[@davidgagen9856]

Great explanation as always!

[@emmanuelo.oketch3250]

Lovely presentation! 👏👏👏

[@JSSTyger]

You're a great teacher.

[@PrimeNewtons]

I appreciate that!

[@danobro]

114 views and no comment? - let me fix that! (love your vids btw)

[@Maths_3.1415]

There was a comment before you and prime Newton removed it

[@wes9627]

y=mx+b where m is the slope and b is an offset. We can ignore b in each equation. For y=x/2-1, m1=1/2. For y=-x+2, m2=-1 Let u1=[1, 1/2]/(√5/2) represent a unit vector parallel to the first line. Let u2=[1, -1]/√2 represent a unit vector parallel to the second line. Let cosθ=u1•u2=√10/5(1-1/2)=√10/10; θ=acos(√10/10)=71.565° The symbol • represents the inner product between two vectors.

[@leonardobarrera2816]

you are so cool

[@samarcando75]

Nice exercise, but I'm worried, this kind of exercises do not explained, only mechanical processes, no analysis, thus the knowledge is limited, you make the maths so pleasant...hugs from Colombia!

[@Vengeance-yb8lm]

Could vectors be used to solve this problem too ?

[@holyshit922]

Let l_{1} : y=a_{1}x+b_{1} l_{2} : y = a_{2}x+b_{2} Slope is tangent of angle between line and abscissa a_{1} = tan(theta_{1}) a_{2} = tan(theta_{2}) Now angle between lines will be difference of angles theta_{2} and theta_{1} so we calculate tangent of the difference of angles and if denominator is zero we have perpendicular lines otherwise we calculate inverse tangent (arctan)

[@JourneyThroughMath]

Because I've never seen this type of problem before, I'll generalize a solution before I watch the video: i would find a lin perpendicular to on of the given lines at a point of my choosing. Either in this case because the slope of a perpendicular would be an integer. This gives me a right triangle. Then i find the lengths of two of the sides (this would require me knowing the intersection). Then use inverse trig to find the angle.

[@JourneyThroughMath]

I was way off!

[@menh6619]

Can you do a video about how to draw a graph of a 2 variables equation and explain it For example (x+2y)(x+3y+1)=2

[@AS-ix3qd]

you can simply use: theta=abs(atan(m1)-atan(m2)) in other words... theta=abs(atan(1/2)-atan(-1))

[@nikhilprabhakar7116]

This was interesting. However, a derivation or proof of the formula could be better as then there won't be any confusions regarding the signs. Now, its not clear why the signs are flipped. Also, I would really appreciate few more examples and edge cases.

[@jacobweinstein9136]

It's related to the dot product between vectors which represent the lines.

[@nanamacapagal8342]

There's two cases: If the two lines extend in different quadrants, then the answer is the sum of the two angles. However, one of these angles will be negative and the other positive, so you have to account for that by inserting a minus sign into the formula. If the two lines instead extend to the same quadrants, the answer is the difference of the two angles. The formula thus already has the minus sign built into it.

[@francescosermi9136]

Take the formula of sin(x+y) and cos(x+y). Then we have that tan(x+y) = [sen(x)cos(y) + cos(x)sen(y)]/[cos(x)cos(y) - sen(x)sen(y)]. We can factor cos(x)cos(y) and, finally, we get that (tan(x) - tan(y))/(1 - tan(x)tan(y)). By this proof, you see that the problem is not the signs of the tangents (because we haven't used any condition regarding their signs) but the geometric interpretation of the slope of a line: the angular coefficient is the tangent that the line creates with the x-axis and here, in the example brought in this video, the angle was -alpha or pi - alpha, not alpha P.S. sorry for the mistakes I might have written, but I'm not a native speaker

[@johnka5407]

Don’t you think arctan(3) would be better than tan^(-1)(3)?

[@Ron_DeForest]

You are good. Just for shiggles I checked to see what the arctan of 3 was in degrees was. It’s 71.5650512. Good guess.

[@jursamaj]

Guess? Like he hadn't working it out exactly before filming? LOL

[@waferlayout]

To draw y(_x curve y strain number, x independent variableand that is a ....

[@junkgum]

I thought arctan m1 - arctan m2 was the fastest answer.

[@EgoFang]

-3?

[@축복이-x6u]

asnwer=1/2

[@i_want_to_be_free]

Can we just say the tan(alpha mines beta)

[@golddddus]

Not good. So what is the slope of the line? By definition tg(φ)=k. φ is the angle between the straight line and positive part of the x-axis. The difference between two angles gives the angle between two straight lines. At the same time, it gives both a obtuse and a acute angle. In our case, the solution calls for a negative value. You can see from your picture that the angle in the first quadrant is obtuse. And solution is π-acctg(3).😎

[@PrimeNewtons]

I would be surprised if you claim you watched the video before writing this comment.

[@elie1117]

Chat am I cooked?