The coordinates of a point on x-axis which lies on the perpendicular bisector of the line segment joining the points (7, 6) and (−3, 4) are

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TO FIND: The coordinates of a point on x axis which lies on perpendicular bisector of line segment joining points (7, 6) and (−3, 4). Let P(x, y) be any point on the perpendicular bisector of AB. Then, PA=PB `sqrt((x -7)^2 + (y -6)^2) = sqrt((x-(-3))^2+(y-4)^2) (x-7)^2+ (y - 6)^2 = (x +3)62 + (y-4)^2 x^2 - 14x + 49 +y^2 - 12y +36 = x^2 +6x +9 +y^2 -8y + 16` -14x - 6x - 12y - 8y + 49 +36 -9 - 16 = 0 - 20x + 20y + 60 = 0 x - y - 3 = 0 x - y = 3 On x-axis y is 0, so substituting y=0 we get x= 3 Hence the coordinates of point is (3,0) .

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