what part of the coordinate plane is equidistant from the points a(-3 2) and b(3,2)? Explain
draw a line through a and b
m = (2-2)/(3+3) = 0 horizontal
so we want a vertical line halfway between
so
x = (-3+3) /2 = 0
looks like the y axis to me.
so you do dat and dis and boom and babys are born from trees
pretty much ya
Yup
ye
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To find the part of the coordinate plane that is equidistant from points A(-3,2) and B(3,2), we need to find the line that is equidistant from these two points. This line is the perpendicular bisector of the line segment joining points A and B.
To begin, let's find the midpoint of the line segment AB. The midpoint is the average of the x-coordinates and the average of the y-coordinates of points A and B.
Midpoint x-coordinate = (x₁ + x₂) / 2 = (-3 + 3) / 2 = 0 / 2 = 0
Midpoint y-coordinate = (y₁ + y₂) / 2 = (2 + 2) / 2 = 4 / 2 = 2
Therefore, the midpoint of AB is M(0, 2).
Next, we need to find the slope of AB. The slope is given by the formula:
slope = (y₂ - y₁) / (x₂ - x₁)
= (2 - 2) / (3 - (-3))
= 0 / 6
= 0
Since the slope of AB is 0, the slope of the perpendicular bisector will be undefined (as it is perpendicular to a line with 0 slope).
Now, using the midpoint M(0, 2) and an undefined slope, we can write the equation of the perpendicular bisector.
An equation with an undefined slope is a vertical line in the form x = a, where 'a' is a constant. In this case, the equation of the perpendicular bisector is x = 0.
So, the part of the coordinate plane that is equidistant from points A(-3,2) and B(3,2) is the vertical line x = 0, which passes through the point (0, 2).