Q24 of 49 Page 1

Find the point of y – axis which is equidistant from the points (-5, -2) and (3, 2).

Let the point on the y-axis be P(0,y)

Which is equidistant from A(-5,-2) and B(3,2)


Given that AP=BP


Distance between two points (x1,y1) and (x2,y2) is given by
 √(x1-x2)2 + (y1-y2)2


Length of AP=√ (0 + 5)2 + (y + 2)2 =√ 25 + (y + 2)2


Length of BP=√ (0-3)2 + (y-2)2 =√ 9 + (y-2)2


AP + BP


√ 25 + (y + 2)2 =√ 9 + (y-2)2


Squaring both sides,


25 + (y + 2)2 =9 + (y-2)2


25 + y2 + 4 + 4y=9 + y2 + 4-4y


8y=-16


y=-2


∴ The required point on y-axis is (0,-2).


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