What part of the coordinate plane is equidistant from the points A(-3,2) and B(3,2) explain?

Question

I'm completely confused on this question & my family doesn't understand it either.

Answer 1

Let point A = (3,2) and point B = (-3,2).  Now imagine a point P with coordinates (x,y) in the coordinate plane.  The distance from point A to P is:

 

d_A² = (y-2)² + (x-3)²

 

The distance from from point B to point P is:

 

d_B² = (y-2)² + (x-(-3))² = (y-2)² + (x+3)²

 

Point P is equidistant from points A and B if d_A = d_B (or d_A² = d_B²).  So let's set them equal and solve for x and y:

 

(y-2)² + (x-3)² = (y-2)² + (x+3)²

(x-3)² = (x+3)²

x2 - 6x + 9 = x2 + 6x + 9

0 = 12x

 

x = 0

 

This is the equation of a vertical line through the point x=0.  In other words, it is the y-axis.  So the part of the plane equidistant from points A and B are all points along the y-axis.

 

If you plot points A and B on graph, you can see that point B is a reflection of point A across the y-axis.  The y-axis is thus the axis of symmetry and all points along it are equidistant from points A and B.  If you draw a line from point A and point B to any point P on the y-axis, you can see that the distances are equal.