Which set of points are collinear?
a.)A(-1,5); B(2,6); C(4,7)
b.)A(0,0); B(1/2,3); C(1,5)
c.)A(-4,-2); B(0,-4); C(2,-5)
when i graph them they all look straight to me but the one I pick is
A(-1,5); B(2,6); C(4,7)
To be collinear, line segments have to be parallel, and a point must be common
for your choice
slopeAB=1/3, slope BC=1/2
so they are not parallel
b) slopeAB = 3/(1/2) = 6 , slope BC = 2/(1/2) = 4
not parallel
c) slopeAB = -2/4 = -1/2 , slope BC = -1/2
so the correct choice is c)
To determine if a set of points are collinear, you can use the slope formula. If the slopes between any two points are equal, then the points are collinear.
Let's calculate the slopes for each set of points:
a.) A(-1,5); B(2,6); C(4,7)
The slope between A and B:
m₁ = (y₂ - y₁) / (x₂ - x₁)
m₁ = (6 - 5) / (2 - (-1))
m₁ = 1 / 3
The slope between B and C:
m₂ = (y₂ - y₁) / (x₂ - x₁)
m₂ = (7 - 6) / (4 - 2)
m₂ = 1 / 2
Since the slopes between A and B (1/3) and B and C (1/2) are not equal, points A, B, and C are not collinear.
b.) A(0,0); B(1/2,3); C(1,5)
The slope between A and B:
m₁ = (y₂ - y₁) / (x₂ - x₁)
m₁ = (3 - 0) / (1/2 - 0)
m₁ = 3 / (1/2)
m₁ = 6
The slope between B and C:
m₂ = (y₂ - y₁) / (x₂ - x₁)
m₂ = (5 - 3) / (1 - 1/2)
m₂ = 2 / (1/2)
m₂ = 4
Since the slopes between A and B (6) and B and C (4) are not equal, points A, B, and C are not collinear.
c.) A(-4,-2); B(0,-4); C(2,-5)
The slope between A and B:
m₁ = (y₂ - y₁) / (x₂ - x₁)
m₁ = (-4 - (-2)) / (0 - (-4))
m₁ = (-4 + 2) / (0 + 4)
m₁ = -2 / 4
m₁ = -1/2
The slope between B and C:
m₂ = (y₂ - y₁) / (x₂ - x₁)
m₂ = (-5 - (-4)) / (2 - 0)
m₂ = (-5 + 4) / (2 - 0)
m₂ = -1 / 2
Since the slopes between A and B (-1/2) and B and C (-1/2) are equal, points A, B, and C are collinear.
Therefore, the set of points that are collinear is:
c.) A(-4,-2); B(0,-4); C(2,-5)