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

1. moongazer Group Title

where did you get that fromula?

2. moongazer Group Title

@Yahoo!

3. Yahoo! Group Title

u know Distance Formula

4. moongazer Group Title

yup :)

5. Yahoo! Group Title

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

6. moongazer Group Title

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

7. moongazer Group Title

@Yahoo! are you still there?

8. moongazer Group Title

9. philo1234 Group Title

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 Group Title

@philo1234 how did you do it?

11. philo1234 Group Title

Those 12 should be 3

12. philo1234 Group Title

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 Group Title

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 Group Title

Do you understand how I got this?

15. philo1234 Group Title

@moongazer

16. philo1234 Group Title

are u there @moongazer

17. moongazer Group Title

I'm back sory for the late reply :)