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Classification of lines: 1. Parallel and Distinct

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Presentation on theme: "Classification of lines: 1. Parallel and Distinct"— Presentation transcript:

1 Lines can be classified four different ways according to their relationship to each other.
Classification of lines: 1. Parallel and Distinct -slopes are equal and y-intercepts are different y = m1x + b1 y = m2x + b2 m1 = m2 b1 ≠ b2

2 2. Coincident -slopes are equal and y-intercepts are equal m1 = m2 y = m1x + b1 b1 = b2 y = m2x + b2 -the equations are identical -one line sits on the other -the ordered pairs for both lines are the same

3 3. Perpendicular -product of slopes = -1 m1 • m2 = -1 y = m1x + b1 y = m2x + b2 -slopes are negative reciprocals of each other -because slopes are different the lines are intersecting All perpendicular lines are also concurrent. These two lines are concurrent at (5,0).

4 4. Intersecting -slopes are different y = m1x + b1 m1 ≠ m2 y = m2x + b b1 ≠ b2 They are concurrent at the point (-4,-3).

5 4. Intersecting cont’d y = 2x + 5 -3 = 2(-4) + 5 -3 = -8 + 5 -3 = -3
When 2 lines are intersecting, the point of intersection, when substituted for x and y in both equations will make them true statements. (-4,-3) y = 2x + 5 -3 = 2(-4) + 5 -3 = -3 = -3 -3 -4

6 = y = 2x + 5 y = 2(-4) + 5 y = -8 + 5 2(2x + 5) = 1(-5x – 26)
Not only can we prove that a point belongs to both lines but if we have both equations, we can determine the point of intersection. y = 2x + 5 = y = 2(-4) + 5 y = y = -3 2(2x + 5) = 1(-5x – 26) 4x + 10 = -5x – 26 4x + 5x = -26 – 10 9x = -36 x = -4


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