anonymous
  • anonymous
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!
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
I need a smart person to answer this, please, I REALLY need to understand this
amistre64
  • amistre64
a smart person eh..... hmmm
anonymous
  • anonymous
yeah

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More answers

amistre64
  • 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
  • amistre64
y=x has a slope of 1, which is a tan(45)
anonymous
  • anonymous
ok, I know this, but where do I go from here?
amistre64
  • amistre64
y = 2x... has a slope of 2....
anonymous
  • anonymous
so it is 2 :)
amistre64
  • amistre64
i aint determined that yet :)
amistre64
  • amistre64
it is the angle of slope 2 - slope1 but you use the "angles" and not the slopes
amistre64
  • amistre64
what angle has a tan^-1(2)?
anonymous
  • anonymous
oh, so, basically, the slope of L1 in relation to slope L2 will be 1, right?
anonymous
  • anonymous
I am really bad at this sorry... forgot all the high school stuff
amistre64
  • 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
  • amistre64
.... we dont need the angles perse, just their ration....
anonymous
  • anonymous
isn't there a simple way?
amistre64
  • amistre64
cos(a-b) = cos(a)cos(b)-sin(a)sin(b)
amistre64
  • amistre64
cos(a) = 1/sqrt(2) cos(b) = 1/sqrt(5) sin(a) = 1/sqrt(2) sin(a) = 2/sqrt(5)
amistre64
  • amistre64
cos(a-b)=1/sqrt(2) * 1/sqrt(5) - 1/sqrt(2) * 2/sqrt(5)
amistre64
  • amistre64
1/10 - 2/10 = -1/10
amistre64
  • amistre64
might be better to work it in tans :) that way we get a new "slope"....tan = slope by the way
amistre64
  • amistre64
tan(a-b) = tan(a)-tan(b) ----------- if I recall correctly 1+tan(a)tan(b)
amistre64
  • amistre64
tan(a) = 1/1 tan(b)=2/1
amistre64
  • amistre64
hold up, make a the bigger one... tan(a) = 2; tan(b)=1 2-1 ---- = 1/3 right? 1+2
anonymous
  • anonymous
yeah
amistre64
  • amistre64
the new slope is 1/3 :) what was the second part?
anonymous
  • anonymous
equation of the line L1 in relation to eq of the line of L2
amistre64
  • amistre64
y=(1/3)x + b just gotta determine where the lines intersect and use that as a parameter for the new system....
amistre64
  • amistre64
they originally intersect at x=-1; y=-1
amistre64
  • amistre64
the new point of intersection is the length of -sqrt(2)....
amistre64
  • amistre64
so the new coord to fill in is (-sqrt(2),0)
anonymous
  • anonymous
why sqrt(2)?
amistre64
  • amistre64
0 = (1/3)(-sqrt(2)) + b b= sqrt(2)/3 y = (1/3)x - (sqrt(2)/3) should be it
amistre64
  • 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
  • anonymous
oh, ok, got it
amistre64
  • amistre64
1-1-sqrt(2) is the triangle of a 45 degree..
anonymous
  • anonymous
how did you get so good at maths?
amistre64
  • amistre64
by messing up alot lol
amistre64
  • amistre64
my son is nagging me for the laptop....ciao :)

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