SOLUTION: can someone show me how to "Find all points on the y-axis that are 6 units from the point (4, -3)"? I put this up earlier and got some help, but would like to see an example of how

Algebra ->  Linear-equations -> SOLUTION: can someone show me how to "Find all points on the y-axis that are 6 units from the point (4, -3)"? I put this up earlier and got some help, but would like to see an example of how      Log On


   

Question 751242: can someone show me how to "Find all points on the y-axis that are 6 units from the point (4, -3)"? I put this up earlier and got some help, but would like to see an example of how exactly to do it. thank you!
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
Any point on the y axis is of the form (0,y)

So we have 2 points: (0,y) and (4, -3)

and the distance between the two must be 6 units, so...

d = sqrt( (x1-x2)^2 + (y1-y2)^2 )

6 = sqrt( (0-4)^2 + (y-(-3))^2 )

6 = sqrt( (-4)^2 + (y+3)^2 )

6 = sqrt( 16 + y^2 + 6y + 9 )

6 = sqrt( y^2 + 6y + 25 )

6^2 = y^2 + 6y + 25

36 = y^2 + 6y + 25

0 = y^2 + 6y + 25 - 36

0 = y^2 + 6y - 11

y^2 + 6y - 11 = 0


Use the quadratic formula to solve for y

y+=+%28-b%2B-sqrt%28b%5E2-4ac%29%29%2F%282a%29

y+=+%28-%286%29%2B-sqrt%28%286%29%5E2-4%281%29%28-11%29%29%29%2F%282%281%29%29 Plug in a+=+1, b+=+6, c+=+-11

y+=+%28-6%2B-sqrt%2836-%28-44%29%29%29%2F%282%29

y+=+%28-6%2B-sqrt%2836%2B44%29%29%2F%282%29

y+=+%28-6%2B-sqrt%2880%29%29%2F2

y+=+%28-6%2Bsqrt%2880%29%29%2F2 or y+=+%28-6-sqrt%2880%29%29%2F2

y+=+%28-6%2B4%2Asqrt%285%29%29%2F2 or y+=+%28-6-4%2Asqrt%285%29%29%2F2

y+=+-3%2B2%2Asqrt%285%29 or y+=+-3-2%2Asqrt%285%29

y+=+1.472136 or y+=+-7.472136

So the points in exact radical form are:


The points in approximate decimal form are: