Q24 of 50 Page 1

Find a point P on the y-axis which is equidistant from the points A (4, 8) and B (-6, 6). Also find the distance AP.

Given A (4, 8) and B (-6, 6)


We know that a point on the y-axis is of the form (0, y). So, let the point P (0, y) be equidistant from A and B.


⇒ AP = BP


We know that the distance between the points P (x1, y1) and Q (x2, y2) is


So, AP2 = BP2


⇒ (0 – 4)2 + (y – 8)2 = (0 – (-6))2 + (y – 6)2


⇒ 16 + y2 + 64 – 16y = 36 + y2 + 36 – 12y


⇒ 8 = 4y


⇒ y = 2


∴ The required point P is (0, 2).


Now,



= √52


= 2√13


∴ Distance AP = 2√13


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