In order to determine the part of the coordinate plane equidistant from the points A(-3,2) and B(3,2), we need to find the line that is equidistant from both points.
The distance between two points in a coordinate plane is given by the distance formula:
d = √((x2 - x1)^2 + (y2 - y1)^2)
Let's calculate the distance between A(-3,2) and B(3,2):
d = √((3 - (-3))^2 + (2 - 2)^2)
= √((6)^2 + (0)^2)
= √(36 + 0)
= √36
= 6
Since the distance between A and B is 6 units, the line equidistant from A and B must be the line perpendicular to the x-axis passing through the midpoint of the line segment AB.
The midpoint of AB can be calculated by averaging the x-coordinates and y-coordinates of A and B:
Midpoint (M) = ((x1 + x2)/2, (y1 + y2)/2)
= ((-3 + 3)/2, (2 + 2)/2)
= (0, 2)
Therefore, the line equidistant from A and B is the line with y-coordinate 2. This line is parallel to the x-axis and lies on the same y-coordinate as points A and B, which means any point on this line will be equidistant from A and B.
What part of the coordinate plane is equidistant from the points A(–3,2) and B(3,2)? Explain your answer.
(2 points)
1 answer