Question

Find the point on y-axis which is equidistant from the points (2,2) and (-4,4 ​

Answer 1

Answer:

Let B be the point equidistant from A and C

A=(2,2), B=(0,y), C=(-4,4)

Distance between AB=Distance between BC

$\sqrt{ {(x2 - x1)}^{2} + {(y2 - y1)}^{2} } = \sqrt{ {(x2 - x1)}^{2} + {(y2 - y1)}^{2} }$

$\sqrt{ {(0 - 2)}^{2} + {(y - 2)}^{2} } = \sqrt{ {( - 4 - 0)}^{2} + {(4 - y)}^{2} }$

$\sqrt{ {( - 2)}^{2} + {(y - 2)}^{2} } = \sqrt{ {( - 4)}^{2} + {(4 - y)}^{2} }$

$\sqrt{4 + {y}^{2} - 4y + 4 } = \sqrt{16 + 16 - 8y + {y}^{2} }$

$\sqrt{ {y}^{2} - 4y + 8 } = \sqrt{ {y}^{2} - 8y + 32}$

$squaring \: on \: both \: the \: sides$

${ (\sqrt{ {y}^{2} - 4y + 8 } )}^{2} = { (\sqrt{ {y}^{2} - 8y + 32 } )}^{2}$

${y}^{2} - 4y + 8 = {y}^{2} - 8y + 32$

${y}^{2} - {y}^{2} - 4y + 8y + 8 - 32 = 0$

$4y - 24 = 0$

$4y = 24$

$y = \frac{24}{4} = 6$

Therefore, (0,6) is the point on y-axis which is equidistant from the given points.