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!

- anonymous

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- anonymous

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

- amistre64

a smart person eh..... hmmm

- anonymous

yeah

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## More answers

- amistre64

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..

- amistre64

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

- anonymous

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

- amistre64

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

- anonymous

so it is 2 :)

- amistre64

i aint determined that yet :)

- amistre64

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

- amistre64

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

- anonymous

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

- anonymous

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

- amistre64

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

- amistre64

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

- anonymous

isn't there a simple way?

- amistre64

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

- amistre64

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

- amistre64

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

- amistre64

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

- amistre64

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

- amistre64

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

- amistre64

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

- amistre64

hold up, make a the bigger one...
tan(a) = 2; tan(b)=1
2-1
---- = 1/3 right?
1+2

- anonymous

yeah

- amistre64

the new slope is 1/3 :)
what was the second part?

- anonymous

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

- amistre64

y=(1/3)x + b
just gotta determine where the lines intersect and use that as a parameter for the new system....

- amistre64

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

- amistre64

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

- amistre64

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

- anonymous

why sqrt(2)?

- amistre64

0 = (1/3)(-sqrt(2)) + b
b= sqrt(2)/3
y = (1/3)x - (sqrt(2)/3) should be it

- amistre64

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

- anonymous

oh, ok, got it

- amistre64

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

- anonymous

how did you get so good at maths?

- amistre64

by messing up alot lol

- amistre64

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

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