what part of the coordinate plane is equidistant from the points a(-3 2) and b(3,2)? Explain
all points on a line perpendicular to the line between those two points and through a point halfway between the two points
m = (Y2-Y1)/(X2-X1) slope between
m' = -1/m slope of perpendicular, the line we want
y = m' x + b
to get b (point halfway between)
average x = (-3+3)/2 = 0
average y = (2+2)/2 = 2
so through (0 , 2)
2 = m' * 0 + b
b = 2
y = m' x + 2
By the way it is the y axis.
The given points represent a hor. line.
A(-3, 2), M(x, y), B(3, 2). M is the mid-point of the given line.
3 - (-3) = 2(x - (-3)).
6 = 2x + 6,
X = 0..
Y = 2 = k(constant).
M(0, 2).
Idk
To determine the part of the coordinate plane that is equidistant from points A(-3, 2) and B(3, 2), we need to find the set of points that are the same distance away from both A and B.
Since the y-coordinates of points A and B are the same, we know that the line connecting them is a horizontal line. Therefore, any point on this line will have the same y-coordinate as A and B.
To find the x-coordinate of the equidistant point, we can analyze the x-coordinates of A and B. As A is to the left of B, any point with an x-coordinate greater than -3 and less than 3 will be equidistant to A and B.
Combining both the y-coordinate and x-coordinate, we can conclude that the part of the coordinate plane equidistant from points A(-3, 2) and B(3, 2) is the horizontal line y = 2 between x = -3 and x = 3.