anonymous
  • anonymous
In 4a) http://ocw.mit.edu/courses/mathematics/18-01sc-single-variable-calculus-fall-2010/exam-3/materials-for-exam-3/MIT18_01SCF10_exam3sol.pdf How do we find the formula for hypotenuse of the triangle? (h-hx/r) (Just for x) (Sorry, somehow this question got closed before)
Mathematics
schrodinger
  • schrodinger
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anonymous
  • anonymous
you ask how they got y = h- hx/r ?
anonymous
  • anonymous
yes
anonymous
  • anonymous
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anonymous
  • anonymous
we have two points : ( 0,h) and (r,0)
anonymous
  • anonymous
hence the slope : (h-0) /(0-r) = -h/r and using y = y1 + m(x-x1) we get : y = h -hx/r
anonymous
  • anonymous
sorry, why are we using h for y1?
anonymous
  • anonymous
i can see how we get the slope
anonymous
  • anonymous
we can take one of the points : (0,h) or (r,0) to be (x1,y1) in both cases we will get the same answer
anonymous
  • anonymous
lets take the (r,0) so y1 = 0 : y = 0 + -h/r(x-r) y = h -hx/r
anonymous
  • anonymous
indeed O_O
anonymous
  • anonymous
we find slope then we can choose any given point on the line to construct our line equation :)
anonymous
  • anonymous
yes
anonymous
  • anonymous
great, got it, thanks :)
anonymous
  • anonymous
yw

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