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

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) (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

Plug in , ,

or

or

or

or

So the points in exact radical form are: $\left(-3+2\sqrt{5},0 \right ) \text{ and } \left(-3-2\sqrt{5},0 \right )$

The points in approximate decimal form are: $\left(-3+2\sqrt{5},0 \right ) \text{ and } \left(-3-2\sqrt{5},0 \right )$
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