Question

find a point on the y-axis which is equal distant from the point a (6,5) and b (-4,3)

Answer 1

Step-by-step explanation:

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Answer 2

$\begin{gathered}PA=PB\\\\\tt \sqrt{\bigg(0\ -\ 6\bigg)^2\ +\ \bigg(y\ -\ 5\bigg)^2}\ =\ \sqrt{\bigg(0\ +\ 4\bigg)^2\ +\ \bigg(y\ -\ 3\bigg)^2}\\\\\\\\\sqrt{\bigg(-6\bigg)^2\ +\ y^2\ -\ 10y\ +\ 25}\ =\ \sqrt{\bigg(4\bigg)^2\ +\ y^2\ -\ 6y\ +\ 9}\\\\\\\\\sqrt{\bigg[36\ +\ y^2\ -\ 10y\ +\ 25\bigg]}\ =\ \sqrt{\bigg[16\ +\ y^2\ -\ 6y\ +\ 9\bigg]}\end{gathered}$

On squaring,

$\begin{gathered}\tt 36\ +\ y^2\ -\ 10y\ +\ 25\ =\ 16\ +\ y^2\ -\ 6y\ +\ 9\\\\\\61\ -\ 10y\ =\ 25\ -\ 6y\\\\\\61\ -\ 25\ =\ -6y\ +\ 10y\\\\\\36\ =\ 4y\\\\\\y\ =\ 9\end{gathered}$

So, the point P(0,9) is equidistant from the points A(6,5) and B(-4,3).