Asked by my_crafting
what part of the coordinate plane is equidistant from the points a(-3 2) and b(3,2)? Explain
Answers
Answered by
Damon
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
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
Answered by
Damon
By the way it is the y axis.
Answered by
Goat mom
THanks
Answered by
Ms. Sue
Goat mom/my_crafting -- Please use the same name for your posts.
Answered by
Henry2
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).
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).
Answered by
Rosemary
Idk
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