Question

Find the angle from the first line to the second.
3x = 4 , 3x + 2y = -8

Answer 1

can you help me please i will give medal :D

Answer 2

The angle between two lines is defined by:
\[Cos \alpha =\frac{ \left| \vec{u}. \vec{v} \right| }{ \left| \vec{u} \right|\left| \vec{v} \right| }\]
Quite the thing huh?
but u and v, represent the direction vectors of the line in question, the first ti the second, that's the order.
But let's rewrite that equation because it is a little messy that way (I will do it in such way that i will be easier to replace), so let's see the following:
\[\left| \vec{a}. \vec{b} \right|=(x _{a}x _{b})+ (y _{a}y _{b})\]
And we can calculate the module of a vector, in a ortonormated reference point as:
\[\left| \vec{b} \right|=\sqrt{x _{b }^{2}+y _{b}^{2}}\]
So, replacing:
\[Cos \alpha = \frac{ (x_u x_v + y_u y_v ) }{ \sqrt{x _{u}^{2} +y _{u}^{2} }\sqrt{x _{v}^{2}+y _{v}^{2}} }\]

Answer 3

So, let's find the directional vectors of those lines, let's call them "u" and "v":
We can find the directional vector of any line by the following equation:
\[ax +by +c=0\]
\[ \vec{w}(-b,a) \]
so, we can get the directional vector with the (-b,a) form, you can also use (b,-a) but I personally prefer the first one.
So, taking the lines:
\[3x=4 \rightarrow 3x-4=0\]
so therefore, the vector "u" is:
\[ \vec{u}(0,3)\]
And the second line:
\[3x + 2y=8 \rightarrow 3x +2y-8=0\]
The second vector "v" is:
\[ \vec{v}(-2,3)\]
Now, it's just a matter of applying the equation I stated above.

Answer 4

i think its ( 2, -3)

Answer 5

what form do you use to find the directional vecotr of a line?

Answer 6

A slope

Answer 7

(0,3) , (2, -3) isnt it?

Answer 8

Sure, we can go with that.
Try placing it in the equation I wrote in the first post.

Answer 9

i will use the slope formula to get the angle :D

Answer 10

m = y2-y1/ x2-x1

Answer 11

Okay, let me see your work when you're finished.

Answer 12

then m = tan data :D

Answer 13

i got a negative angle

Answer 14

-71.57 degrees

Answer 15

is this right?

Answer 16

Well, I believe you're using:
\[m=\tan \alpha\]
But that alpha there, is the angle the line has referential to the x-axis.
Like this:

Answer 17

so what you calculated is that little angle, you did well, but it's not the angle yo uare asked for.

Answer 18

how can we get it ? D:

Answer 19

I gave you the formula, in the first post I made.

Answer 20

ok, i will solve for it

Answer 21

i got 33.68 degrees

Answer 22

If you did the maths correctly, that should be correct.

Answer 23

:D , can you help me another run

Answer 24

sure!