Question

find the angles between the lines: y=2x-6 & 2x+3y=18

Answer 1

we have 2 vectors that make up the slope

Answer 2

a=<1, 2>; |a| = sqrt(5)
b=<3,-2>; |b| = sqrt(13)
----------
3-4 = -1 = a.b

cos(t) = a.b / |a||b|

t = cos-1 (-1/sqrt{5*13})

t = cos-1 (-1/sqrt{65})

and the other one should be pi - t

Answer 3

if im lucky :)

we get about 97 and 83

Answer 4

yeah the answer is 97.1 and 82.9. Can you explain how you did that?

Answer 5

.....i did lol

Answer 6

but i used vector concepts that you prolly aint used to

Answer 7

right

Answer 8

we can draw it out and get the same idea i spose

Answer 9

how do you know the coordinates i.e. 1,2 3,-2

Answer 10

slopes are vectors; y/x is a slope so: <x,y> is the vector

y/x = 2/1 = <1,2>

y/x = -2/3 = <3,-2>

Answer 11

but despite that; we we draw the lines and slope the we can get the tangents needed here

tan(a) = 2/1
a = tan-1 (2/1)
= 1.1071

tan(b) = 2/3
b = tan-1 (2/3)
= .588

c = pi - 1.1071 - .588 = 1.4465 radians * 180/pi = 82.8

180
- 82.8
-------
97.2

Answer 12

i don't understand greek but thanks

Answer 13

:)

Answer 14

if your material that your working with has a technique or method; we could try to work that out

Answer 15

but just trying to solve this on the fly i have to resort to my own ingenuity

Answer 16

which i understand... but you're using variable which have no significance to me like pi and c... i haven't learned it in school im just trying to get ahead... I have another question though...

Answer 17

for finding the equation of a line such as (-2,4) and (1,2) could you explain how to find b(y-int)?

Answer 18

the most important think about a line are, well, 2 things; slope, and a point to anchor it to

Answer 19

you have 2 points given so we need to determine a slope

Answer 20

okay so delta y/delta x right?

Answer 21

\[slope=\frac{1-Py}{x-Px}\]
which means we subtract one point from the other and stack y/x

Answer 22

y - Py ....

Answer 23

this would be ideal if we moved both point so that one of them was at the origin (0,0)

(-2,4) (1,2)
+2-4 +2-4
------ ------
(0,0) (3,-2)

now we are able to see the slope as -2/3

do you know why?

Answer 24

no that was my next question
lol

Answer 25

from 0,0 to 3,-2 ; what is our change in y?
what is our change in x?

Answer 26

delta x = 3 and delta y= -2

Answer 27

correct; so deltay/deltax as you see it is then: -2/3

Answer 28

do you understand how I was able to get 0,0 and 3,-2 to work from?

Answer 29

yeah because either of the points will yield the same answer

Answer 30

correct, and we are able to move them to a more convient position to measure things from graphically; formulawise its just subtracting points

Answer 31

with this slope; we need to reform our slope equation but with 1 point and a generic point

Answer 32

okay so we can say y=(-2/3)+b as in y=mx+b therefore b= 8/3?

Answer 33

*y=(-2/3)x+b

Answer 34

perhaps, lets see where it leads :) ill use the point(1,2) and have a generic (x,y)
\[slope=\frac{y-Py}{x-Px}\]
\[\frac{-2}{3}=\frac{y-1}{x-2}\]
\[(x-2)*\frac{-2}{3}=\frac{y-1}{\cancel{x-2}}*\cancel{(x-2)}\]
\[\frac{-2}{3}(x-2)=y-1\]
\[\frac{-2}{3}x-\frac{2}{3}(-2)=y-1\]
\[\frac{-2}{3}x+\frac{4}{3}=y-1\]
\[\frac{-2}{3}x+\frac{4}{3}+1=y\cancel{-1+1}^{ 0}\]
\[\frac{-2}{3}x+\frac{4}{3}+1=y\]
\[\frac{-2}{3}x+\frac{4}{3}+1*\frac{3}{3}=y\]
\[\frac{-2}{3}x+\frac{4}{3}+\frac{3}{3}=y\]
\[\frac{-2}{3}x+\frac{4+3}{3}=y\]
\[\frac{-2}{3}x+\frac{7}{3}=y\]
I would say from this that b = 7/3

Answer 35

there are shortcuts you can take; but this IS the details that you should be familiar with

Answer 36

okay i'll look this over thanks for your help

Answer 37

yw

Answer 38

http://www.purplemath.com/modules/strtlneq.htm this says that b=8/3 rather than 7/3?