## moongazer 2 years ago The slope of a line through A(-1,1) is 3. Locate the point on this line that is 2sqrt3 from A.

1. moongazer

where did you get that fromula?

2. moongazer

@Yahoo!

3. Yahoo!

u know Distance Formula

4. moongazer

yup :)

5. Yahoo!

Let that Point Be B (x , y ) d = 2sqrt3 A (-1,1) nw use that Formula

6. moongazer

I got x^2 + 2x - 10 + y^2 - 2y = 0 @Yahoo!

7. moongazer

@Yahoo! are you still there?

8. moongazer

9. philo1234

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

@philo1234 how did you do it?

11. philo1234

Those 12 should be 3

12. philo1234

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. 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

Do you understand how I got this?

15. philo1234

@moongazer

16. philo1234

are u there @moongazer

17. moongazer

I'm back sory for the late reply :)
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