Find the point on the 𝑥 x -axis that is equidistant from the points (7,3) ( 7 , 3 ) and (1,7) ( 1 , 7 ) .

Question

 need help. tried to use the midpoint and distance formula and nothing is working

Answer 1

The set of points that are equidistant from A = (7,3) and B = (1,7) is the perpendicular bisector of line segment AB.

 

    Slope of AB = (7-3) / (1-7) = -2/3

 

So, slope of perpendicular bisector = -1 / (-2/3) = 3/2

 

Perpendicular bisector passes through the midpoint of AB.  Midpoint of AB = ( (7+1)/2, (3+7)/2 ) = (4, 5)

 

Equation of perpendicular bisector:   y - 5 = (3/2)(x - 4)

 

                                                     y - 5 = (3/2)x - 6

 

                                                          y = (3/2)x - 1

 

On the x-axis, y = 0.  So, 0 = (3/2)x - 1

 

                                      1 = (3/2)x     x = 2/3

 

The point on the x-axis equidistant form (7,3) and (1,7) is (2/3, 0).