Question

find the point on the y axis which is equidistance from the points (5,-2) and (-3,2).

Answer 1

Let the point on the y axia be (0 , y)

Given distance between ( 5, – 2) & ( 0 , y) is equal to ( – 3 , 2 ) and ( 0 , y)

Hence

$\sqrt{(0 - 5) {}^{2} + (y + 2) {}^{2} } = \sqrt{(0 + 3) {}^{2} + (y - 2) {}^{2} } \\ 25 + (y + 2) {}^{2} = 9 + (y - 2) {}^{2} \\ 25 + y {}^{2} + 4 + 4y = 9 + y {}^{2} + 4 - 4y \\ 25 + 4y = 9 - 4y \\ 25 - 9 = - 4y - 4y \\ 16 = - 8y \\ y = - 16 \div 8 \\ y = - 2$

Hence the point on y axis is ( 0, – 2)

Answer 2

Answer:

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