moongazer
  • moongazer
The slope of a line through A(-1,1) is 3. Locate the point on this line that is 2sqrt3 from A.
Mathematics
jamiebookeater
  • jamiebookeater
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moongazer
  • moongazer
where did you get that fromula?
moongazer
  • moongazer
anonymous
  • anonymous
u know Distance Formula

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moongazer
  • moongazer
yup :)
anonymous
  • anonymous
Let that Point Be B (x , y ) d = 2sqrt3 A (-1,1) nw use that Formula
moongazer
  • moongazer
I got x^2 + 2x - 10 + y^2 - 2y = 0 @Yahoo!
moongazer
  • moongazer
@Yahoo! are you still there?
moongazer
  • moongazer
help please :)
anonymous
  • anonymous
I got \[(-1-\frac{ \sqrt{30} }{ 4 }, -1-12\frac{ \sqrt{30} }{ 4 })\] and \[(-1+\frac{ \sqrt{30} }{ 4 }, -1+12\frac{ \sqrt{30} }{ 4 })\]
moongazer
  • moongazer
@philo1234 how did you do it?
anonymous
  • anonymous
Those 12 should be 3
anonymous
  • anonymous
I made 2 equations: 1. Using the slope equation: \[\frac{ y-1 }{ x+1 } = 3\] 2. Then used the distance formula to make the second equation: \[2\sqrt{3} = \sqrt{(y-1)^2 + (x+1)^2}\] Do you follow so far?
anonymous
  • anonymous
3. The I solve for y in the first equation and substitute it in equation 2: \[y = 3x+4\] Substitute in equations 2: \[2\sqrt{3} = \sqrt{(3x+4-1)^2 +(x+1)^2}\] 4. Square both sides to get rid of the parentheses: \[(2\sqrt{3})^2 = (3x+3)^2 +(x+1)^2\] 5. Multiply out everything, put all the values on one side then solve for x using the quadratic formula: \[12 =10x^{2}+20x+10\]
anonymous
  • anonymous
Do you understand how I got this?
anonymous
  • anonymous
anonymous
  • anonymous
are u there @moongazer
moongazer
  • moongazer
I'm back sory for the late reply :)

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