Question

Consider two straight lines L1 (y = 2x +1) and L2 (y=x).
a) What is the slope of line L1 in respect to line L2?
b) What is the equation of line L1 in respect to line L2?
NOTE. Think that the original xy co-ordinate system has been
rotated in respect to the origin so that line L2 defines the new
x-axis x´ and y´ is the new y-axis!

Answer 1

I need a smart person to answer this, please, I REALLY need to understand this

Answer 2

a smart person eh..... hmmm

Answer 3

yeah

Answer 4

the slope of L1 with respect to L2 suggests to me that it is a relative answer. If L2 were flat, what would the slope of L1 be with it..

Answer 5

y=x has a slope of 1, which is a tan(45)

Answer 6

ok, I know this, but where do I go from here?

Answer 7

y = 2x... has a slope of 2....

Answer 8

so it is 2 :)

Answer 9

i aint determined that yet :)

Answer 10

it is the angle of slope 2 - slope1 but you use the "angles" and not the slopes

Answer 11

what angle has a tan^-1(2)?

Answer 12

oh, so, basically, the slope of L1 in relation to slope L2 will be 1, right?

Answer 13

I am really bad at this sorry... forgot all the high school stuff

Answer 14

tan(63.43) = 2

not really, since we have rotated the axis by a certain degree, we need to determine the degrees involved, then you trig stuff to re dress the problem

Answer 15

.... we dont need the angles perse, just their ration....

Answer 16

isn't there a simple way?

Answer 17

cos(a-b) = cos(a)cos(b)-sin(a)sin(b)

Answer 18

cos(a) = 1/sqrt(2) cos(b) = 1/sqrt(5)

sin(a) = 1/sqrt(2) sin(a) = 2/sqrt(5)

Answer 19

cos(a-b)=1/sqrt(2) * 1/sqrt(5) - 1/sqrt(2) * 2/sqrt(5)

Answer 20

1/10 - 2/10 = -1/10

Answer 21

might be better to work it in tans :) that way we get a new "slope"....tan = slope by the way

Answer 22

tan(a-b) = tan(a)-tan(b)
----------- if I recall correctly
1+tan(a)tan(b)

Answer 23

tan(a) = 1/1 tan(b)=2/1

Answer 24

hold up, make a the bigger one...

tan(a) = 2; tan(b)=1

2-1
---- = 1/3 right?
1+2

Answer 25

yeah

Answer 26

the new slope is 1/3 :)

what was the second part?

Answer 27

equation of the line L1 in relation to eq of the line of L2

Answer 28

y=(1/3)x + b

just gotta determine where the lines intersect and use that as a parameter for the new system....

Answer 29

they originally intersect at x=-1; y=-1

Answer 30

the new point of intersection is the length of -sqrt(2)....

Answer 31

so the new coord to fill in is (-sqrt(2),0)

Answer 32

why sqrt(2)?

Answer 33

0 = (1/3)(-sqrt(2)) + b

b= sqrt(2)/3

y = (1/3)x - (sqrt(2)/3) should be it

Answer 34

why sqrt(2)?

the original lines cross at (-1,-1) when we rotate the axis to accomodate the new axises.... that point is still there, but we use the distance of it from the origin as the x intercept

Answer 35

oh, ok, got it

Answer 36

1-1-sqrt(2) is the triangle of a 45 degree..

Answer 37

how did you get so good at maths?

Answer 38

by messing up alot lol

Answer 39

my son is nagging me for the laptop....ciao :)