Answer
$\left( 0,\,-3+2\sqrt{5} \right)$ and $\left( 0,\,-3-2\sqrt{5} \right)$
Work Step by Step
The points on the $y-$axis are represented by the point$\left( 0,y \right)$.
By using the distance formula with \[{{P}_{1}}=\left( {{x}_{1}},{{y}_{1}} \right)=\left( 4,-3 \right)\] , ${{P}_{2}}=\left( {{x}_{2}},{{y}_{2}} \right)=\left( 0,y \right)$ and \[d\left( {{P}_{1}},\,{{P}_{2}} \right)=6\],
$6=\sqrt{{{\left( 0-4 \right)}^{2}}+{{\left( y-\left( -3 \right) \right)}^{2}}}$
By simplifying,
$6=\sqrt{{{\left( -4 \right)}^{2}}+{{\left( y+3 \right)}^{2}}}$
Solving for $y$,
$6=\sqrt{16+{{\left( y+3 \right)}^{2}}}$
Squaring on both sides,
${{6}^{2}}=16+{{\left( y+3 \right)}^{2}}$
$36=16+{{\left( y+3 \right)}^{2}}$
Subtract $16$on both sides,
$36-16=16-16+{{\left( y+3 \right)}^{2}}$
$\Rightarrow 20={{\left( y+3 \right)}^{2}}$
$\Rightarrow {{\left( y+3 \right)}^{2}}=20$
By square root method,
$\Rightarrow \left( y+3 \right)=\pm \sqrt{20}$
$\Rightarrow \left( y+3 \right)=\pm 2\sqrt{5}$
Subtract $3$ on both sides,
$\Rightarrow y=-3\pm 2\sqrt{5}$
$\Rightarrow y=-3+2\sqrt{5}$ and $y=-3-2\sqrt{5}$
Therefore, the points are $\left( 0,\,-3+2\sqrt{5} \right)$ and $\left( 0,\,-3-2\sqrt{5} \right)$.
