Question 751242
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 = (-b+-sqrt(b^2-4ac))/(2a)}}}


{{{y = (-(6)+-sqrt((6)^2-4(1)(-11)))/(2(1))}}} Plug in {{{a = 1}}}, {{{b = 6}}}, {{{c = -11}}}


{{{y = (-6+-sqrt(36-(-44)))/(2)}}}


{{{y = (-6+-sqrt(36+44))/(2)}}}


{{{y = (-6+-sqrt(80))/2}}}


{{{y = (-6+sqrt(80))/2}}} or {{{y = (-6-sqrt(80))/2}}}


{{{y = (-6+4*sqrt(5))/2}}} or {{{y = (-6-4*sqrt(5))/2}}}


{{{y = -3+2*sqrt(5)}}} or {{{y = -3-2*sqrt(5)}}}


{{{y = 1.472136}}} or {{{y = -7.472136}}}


So the points in exact radical form are: <img src="http://latex.codecogs.com/gif.latex?\left(0, -3+2\sqrt{5} \right ) \text{ and } \left(0,-3-2\sqrt{5} \right )" title="\left(-3+2\sqrt{5},0 \right ) \text{ and } \left(-3-2\sqrt{5},0 \right )" />



The points in approximate decimal form are: <img src="http://latex.codecogs.com/gif.latex?(0, 1.472136) \text{ and } (0, -7.472136)" title="\left(-3+2\sqrt{5},0 \right ) \text{ and } \left(-3-2\sqrt{5},0 \right )" />