Question

find the point on y axis equidistant from (2,-3)and(4,6)​

Answer 1

Answer:

$point \: is \: \: \: (0 \: \: \: \: \: \: \: \: \: \frac{39}{3} )$

Step-by-step explanation:

$(0 \: \: \: \: \: y) \\ \\ \\ \\ \\ \\ \\ \sqrt{ {(4 - 2)}^{2} + {(y + 3)}^{2} } = \sqrt{ {(0 - 4)}^{2} + {(y - 6)}^{2} } \\ 4 + {(y + 3)}^{2} = 16 + {(y - 6) }^{2} \\ \\ {(y + 3)}^{2} = {(y - 6)}^{2} + 12 \\ \\ \\ \\ {y}^{2} + 9 + 6y = {y}^{2} + 36 - 12y + 12 \\ \\ 9 + 6y = 36 - 12y + 12 \\ \\ 6y + 12y = 48 - 9 \\ \\ 18y = - 39 \\ \\ y = \frac{-39}{18} = \frac{-13}{6} \\ \\ \\ \\ \\ point \: is \: \: \: (0 \: \: \: \: \: \: \: \: \: \frac{-13}{6} )$