Question

hey guys! I hope you can help me with this question.: find a point on the y axis which is equidistant from the point A(5,6) and (3,-4).​

Answer 1

$\huge \mathrm{ \pink{A} \purple{n} \blue{s} \color{navy}{w} \orange{e} \color{teal}{r \: \: :- }}$

Let the point on y axis is P(0,y) which is equidistant from the points A(6,5) and B(-4,3). The distance between PA is equal to PB such that,

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

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).

Answer 2

★Given:-

• a point on the y axis equidistant from the point A(5,6) and (3,-4).

★find:-

• point on the y axis

★solution:-

• we know that y axis x value o so cordinet on y axis (0,y)

let P=(0,y). and Q =(5,6) R=(3,-4)

• and given that the point (5,6) and(3,-4) y axis which is equidistant from the

it means

$\implies PQ=PR$

we know that two point PQ between distance formula

$\implies \pink{ PQ = \sqrt{( x_{1} - x_{2}) {}^{2} + (y_{1} - y_{2}) {}^{2} } }$

question according:-

$\implies \: \sqrt{(0 - 5) {}^{2} + (y-6) {}^{2} } = \sqrt{( 0-3) {}^{2} + (y+4) {}^{2} }$

squaring both sides

$\implies \: ( - 5) {}^{2} + 6 {}^{2} + y {}^{2} - 12y = ( - 3 ){}^{2} + {y}^{2} + 4 {}^{2} + 8y) \\ \\ \implies \: 25 + 36 + y {}^{2} =9 + 12y + 8y + y {}^{2} + 16 \\ \\ \implies \cancel{y {}^{2} - y {}^{2} } + 25 +36 = 20y+25 \\ \\ \implies \: 20y = 36 \\ \\ \implies \: y = \cancel{\frac{36}{20} } \\ \\ \implies \: y = 1.3$

(0,1.3) a point on the y axis which is equidistant from the point A(5,6) and (3,-4).

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I hope u help you