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moongazer

  • 3 years ago

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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  1. moongazer
    • 3 years ago
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    where did you get that fromula?

  2. moongazer
    • 3 years ago
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    @Yahoo!

  3. Yahoo!
    • 3 years ago
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    u know Distance Formula

  4. moongazer
    • 3 years ago
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    yup :)

  5. Yahoo!
    • 3 years ago
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    Let that Point Be B (x , y ) d = 2sqrt3 A (-1,1) nw use that Formula

  6. moongazer
    • 3 years ago
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    I got x^2 + 2x - 10 + y^2 - 2y = 0 @Yahoo!

  7. moongazer
    • 3 years ago
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    @Yahoo! are you still there?

  8. moongazer
    • 3 years ago
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    help please :)

  9. philo1234
    • 3 years ago
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    I got \[(-1-\frac{ \sqrt{30} }{ 4 }, -1-12\frac{ \sqrt{30} }{ 4 })\] and \[(-1+\frac{ \sqrt{30} }{ 4 }, -1+12\frac{ \sqrt{30} }{ 4 })\]

  10. moongazer
    • 3 years ago
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    @philo1234 how did you do it?

  11. philo1234
    • 3 years ago
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    Those 12 should be 3

  12. philo1234
    • 3 years ago
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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?

  13. philo1234
    • 3 years ago
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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: \[12 =10x^{2}+20x+10\]

  14. philo1234
    • 3 years ago
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    Do you understand how I got this?

  15. philo1234
    • 3 years ago
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    @moongazer

  16. philo1234
    • 3 years ago
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    are u there @moongazer

  17. moongazer
    • 3 years ago
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    I'm back sory for the late reply :)

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