Question

Find the coordinate of the point on Y-axis which is equidistant from point P(6,5)and J(-4,3)

Answer 1

$\mathbf{Answer}$

$\boxed{\mathbf{y = - \frac{9}{4}}}$

$\underline{\underline{\huge\mathfrak{Solution}}}$

Formula used :

$\boxed{\mathbf{Distance = \sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2}}}}$

Let the point be A and the coordinates be x and y.

Now, since we have to find the coordinate of the point A on Y axis, we can conclude that x = 0.

Given that,
Distance of A(0, y) and P(6, 5) = Distance of A(0, y) and J(-4, 3)

P(6, 5) = $P(x_{1} , y_{1})$
A(0, y) = $A(x_{2} , y_{2})$
J(-4, 3) = $J(x_{3} , y_{3})$

Now,

$\bf Distance \: of \: AP = Distance \: of \: AJ \\ \\ \sqrt{ {( x_{1} - x_{2}) }^{2} + {( y_{1} - y_{2} )}^{2} } = \sqrt{ {( x_{3} - x_{2}) }^{2} + {( y_{3} - y_{2} )}^{2} } \\ \\ \sqrt{ {(6 - 0)}^{2} + {(5 - y)}^{2} } = \sqrt{ {( - 4 - 0)}^{2} + {(3 - y)}^{2} } \\ \\ \sqrt{ 36 + 25 + {y}^{2} - 10y } = \sqrt{16 + 9 + {y}^{2} - 6y } \\ \\ 36 + 25 + {y}^{2} - 10y = 25 + {y}^{2} - 6y \\ \\ 36 + 10y = - 6y \\ \\ 10y + 6y = - 36 \\ \\ 16y = 36 \\ \\ y = - \frac{36}{16} \\ \\ \bf y = - \frac{9}{4}$

Answer 2

Answer:

Step-by-step explanation:

Hey I have taken A and B instead of P and J