Find the coordinates of a point P on Y axis equidistant from two points A(-3,4) and B(3,6) on the same plane

Hi Sree, let the point P be (0, a). As it lies on Y axis its abscissa is to be 0 
Now given P is equidistant from A(-3,4) and B(3,6)
This means PA^2 = PB^2
(-3)^2 + (a-4)^2 = (3)^2 + (a - 6)^2
==> a - 4 = + (a - 6) which is not valid
But a - 4 = -(a-6) 
==> a - 4 + a - 6 = 0
OR 2 a = 10 
So a = 5
(0,5) is equidistant from A and B

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HI SREE SHRAVANYA,
LET THE POINT P IS (X,Y).
A(-3,4) AND B(3,6) ARE EQUIDISTANT FROM P.   (GIVEN)
SO, AP = BP
SQUARING BOTH SIDES
AP2 = BP2
NOW BY APPLYING DISTANCE FORMULA
[X-(-3)]2 + [Y-4]2 = [X-3]2 + [Y-6]2
[X+3)]2 + [Y-4]2 = [X-3]2 + [Y-6]2
X2 + 6X + 9 + Y2 - 8Y + 16 = X2 - 6X + 9 + Y2 - 12Y + 36
6X - 8Y + 25 = -6X - 12Y + 45
12X + 4Y = 20
3X + Y = 5
SO, ALL THE POINTS LYING ON THIS LINE 3X - Y = 5 ARE EQUIDISTANT FROM A & B.
IF X=0, THEN Y=5
IF X=1, THEN Y=2
IF X=2, THEN Y=-1

IF X=(5/3), THEN Y=0
IF X=(4/3), THEN Y=1
AND SO ON....
HOPE, THIS ANSWER HELPS!!!
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