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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.
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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moongazer
 3 years ago
Best ResponseYou've already chosen the best response.0where did you get that fromula?

Yahoo!
 3 years ago
Best ResponseYou've already chosen the best response.0Let that Point Be B (x , y ) d = 2sqrt3 A (1,1) nw use that Formula

moongazer
 3 years ago
Best ResponseYou've already chosen the best response.0I got x^2 + 2x  10 + y^2  2y = 0 @Yahoo!

moongazer
 3 years ago
Best ResponseYou've already chosen the best response.0@Yahoo! are you still there?

philo1234
 3 years ago
Best ResponseYou've already chosen the best response.1I got \[(1\frac{ \sqrt{30} }{ 4 }, 112\frac{ \sqrt{30} }{ 4 })\] and \[(1+\frac{ \sqrt{30} }{ 4 }, 1+12\frac{ \sqrt{30} }{ 4 })\]

moongazer
 3 years ago
Best ResponseYou've already chosen the best response.0@philo1234 how did you do it?

philo1234
 3 years ago
Best ResponseYou've already chosen the best response.1I made 2 equations: 1. Using the slope equation: \[\frac{ y1 }{ x+1 } = 3\] 2. Then used the distance formula to make the second equation: \[2\sqrt{3} = \sqrt{(y1)^2 + (x+1)^2}\] Do you follow so far?

philo1234
 3 years ago
Best ResponseYou've already chosen the best response.13. The I solve for y in the first equation and substitute it in equation 2: \[y = 3x+4\] Substitute in equations 2: \[2\sqrt{3} = \sqrt{(3x+41)^2 +(x+1)^2}\] 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
 3 years ago
Best ResponseYou've already chosen the best response.1Do you understand how I got this?

philo1234
 3 years ago
Best ResponseYou've already chosen the best response.1are u there @moongazer

moongazer
 3 years ago
Best ResponseYou've already chosen the best response.0I'm back sory for the late reply :)
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