OpenStudy (moongazer):

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

4 years ago
OpenStudy (moongazer):

where did you get that fromula?

4 years ago
OpenStudy (moongazer):

@Yahoo!

4 years ago
OpenStudy (anonymous):

u know Distance Formula

4 years ago
OpenStudy (moongazer):

yup :)

4 years ago
OpenStudy (anonymous):

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

4 years ago
OpenStudy (moongazer):

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

4 years ago
OpenStudy (moongazer):

@Yahoo! are you still there?

4 years ago
OpenStudy (moongazer):

help please :)

4 years ago
OpenStudy (anonymous):

I got \[(-1-\frac{ \sqrt{30} }{ 4 }, -1-12\frac{ \sqrt{30} }{ 4 })\] and \[(-1+\frac{ \sqrt{30} }{ 4 }, -1+12\frac{ \sqrt{30} }{ 4 })\]

4 years ago
OpenStudy (moongazer):

@philo1234 how did you do it?

4 years ago
OpenStudy (anonymous):

Those 12 should be 3

4 years ago
OpenStudy (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?

4 years ago
OpenStudy (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: \[12 =10x^{2}+20x+10\]

4 years ago
OpenStudy (anonymous):

Do you understand how I got this?

4 years ago
OpenStudy (anonymous):

@moongazer

4 years ago
OpenStudy (anonymous):

are u there @moongazer

4 years ago
OpenStudy (moongazer):

I'm back sory for the late reply :)

4 years ago