Answers

2014-04-23T13:09:33+05:30
Point on y-axis x=0
point=(0,y)
equidistant from (5,-2) and(-3,2)
distance from first point=(25+y^2+y-4y)^1/2
distance from second point=(9+y^2+4-4y)^1/2
both equals
squaring both
29+y^2+4y=13+y^2-4y
8y=-16
y=-2
point=(0,-2)
2 4 2
2014-04-23T14:23:08+05:30
Given,points (5,-2) & (-3,2)
point on y-axis means x-coordinate is zero i.e.,point is (0,y)
given (0,y) is equidistant from the two points
then,
[{(0-5)^2}+{y+2}^2]^1/2=[{(0+3)^2}+(y-2)^2]^1/2
squaring on both sides,
25+{(y+2)^2}=9+{(y-2)^2}
16={(y-2)^2}-{(y+2)^2}
16=-8y
y=-16/8
y=-2
therefore,the point on y-axis equidistant from (5,-2) & (-3,2) is (0,-2).


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