moongazer
The slope of a line through A(-1,1) is 3. Locate the point on this line that is 2sqrt3 from A.
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moongazer
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where did you get that fromula?
moongazer
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@Yahoo!
Yahoo!
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u know Distance Formula
moongazer
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yup :)
Yahoo!
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Let that Point Be B (x , y ) d = 2sqrt3 A (-1,1) nw use that Formula
moongazer
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I got x^2 + 2x - 10 + y^2 - 2y = 0
@Yahoo!
moongazer
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@Yahoo! are you still there?
moongazer
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help please :)
philo1234
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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
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@philo1234 how did you do it?
philo1234
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Those 12 should be 3
philo1234
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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?
philo1234
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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:
\[12 =10x^{2}+20x+10\]
philo1234
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Do you understand how I got this?
philo1234
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@moongazer
philo1234
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are u there @moongazer
moongazer
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I'm back sory for the late reply :)