1.4 Lines1.4 LinesEssential Question: How can you use the equations of

two non-vertical lines to tell whether the lines are parallel or perpendicular?

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An arithmetic sequence is nothing more than a linear equation…◦{3, 8, 13, 18, … }

→ un = 3 + (n - 1)(5)→ un = 3 + 5n – 5→ un = 5n – 2

◦If we replace n with x and un with y, we have the linear equation: y = 5x – 2

◦The sequence above corresponds to the points { (1, 3), (2, 8), (3, 13), (4, 18), … }

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Slope◦Change in y ÷ change in x◦Δ y ÷ Δx (delta y ÷ delta x)◦

Example 1◦Find the slope of the line that passes through

(0,-1) and (4,1)

1 ( 1) 2 1

4 0 4 2

2 1

2 1

y y

x x

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Finding Slope From a Graph◦Find two points on the coordinate plane, and

use the slope formulaProperties of Slope

◦If m > 0, the line rises from left to right. The larger m is, the more steeply the line rises.

◦If m = 0, the line is horizontal◦If m < 0, the line falls from left to right. The

larger m is, the more steeply the line falls.

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Slope-Intercept Form◦y = mx + b

“m” is the slope “b” is the y-intercept (the point where the graph

crosses the y-axis)Example 3: Graphs of Arithmetic

Sequences◦1st three terms of an arithmetic sequence are

-2, 3, and 8. Use an explicit function and compare it to slope-intercept form

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Example 3 (Continued)◦Sequence: -2, 3, 8, …◦Explicit form: un = u1 + (n-1)(d)

◦u1 = , d =

◦What connection(s) do you see between the slope-intercept form and the original sequence?

-2 5un = -2 + (n-1)(5)un = -2 + 5n – 5un = 5n – 7

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Connection between Arithmetic Sequences and Lines◦Explicit Form: un = u1 + (n-1)(d)◦Slope Intercept Form: y = mx + b

◦The slope corresponds to the common difference (m = d)

◦The y-intercept represents the value of u0, or the term before the sequence started, which is u1 less one common difference (u1 – d)

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Graphing a Line◦Solve an equation for y (i.e. get y by itself)◦Plot the y-intercept◦Use the slope (rise over run) to make a 2nd plot◦Draw a line which connects the two dots (a line,

not a segment)

Graphing calculator◦GRAPH → F1, input in the equation, solved for y◦2nd, F5

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Assignment◦Page 40

Problems 3 – 14 (all problems) Show your work

1.4 Lines1.4 Lines(Day 2)(Day 2)

Essential Question: How can you use the equations of two non-vertical lines to tell whether

the lines are parallel or perpendicular?

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Point-Slope Form◦Use point slope form when you’re given a point

and a slope (SURPRISE!) or two points to determine an equation (and use those points to determine the slope)

◦y – y1 = m(x – x1)

◦Any point given can be used for (x1, y1)

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Point-Slope Form (Example 6)◦Find the equation of the line that passes

through the point (1, -6) with a slope 2. y – y1 = m(x – x1)

y – (-6) = 2(x – 1)

y + 6 = 2x – 2

– 6 – 6

y = 2x – 8

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Vertical and Horizontal Lines◦Equation of a Horizontal Line

Every point along a horizontal line will have the same y value.

Written as y = b (where b is the y-intercept) The slope of a horizontal line = 0

◦Equation of a Vertical Line Every point along a vertical line will have the

same x value. Written as x = c (where c is some constant) The slope of a vertical line is undefined

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Parallel & Perpendicular Lines◦Parallel lines have the same slope◦Perpendicular lines have inverse reciprocal

slopes That means the product of the slopes of two

perpendicular lines is -1 Take the slope of one line, flip as a fraction and

flip sign.

Parallel & Perpendicular Lines (Example 9)◦Given line M whose equation is 3x - 2y + 6=0, find

the equation of the parallel and perpendicular lines which go through the point (2, -1)

a)Parallel Line◦ Get y by itself to find

the slope of line M◦ The slope of M is 3/2

◦ Use point-slope formy – y1 = m(x – x1)

y – (-1) = 3/2(x – 2)

y + 1 = 3/2x – 3

– 1 – 1

y = 3/2x – 4

3x – 2y + 6 = 0

-2y = -3x – 6

y = 3/2x + 3

–3x –6 –3x – 6

-2 -2 -2

Parallel & Perpendicular Lines (Example 9)◦Given line M whose equation is 3x - 2y + 6=0, find

the equation of the parallel and perpendicular lines which go through the point (2, -1)

a)Perpendicular Line◦ Slope of M = 3/2, so the perpendicular slope is -2/3

◦ Use point-slope form

y – y1 = m(x – x1)

y – (-1) = -2/3(x – 2)

y + 1 = -2/3x + 4/3

– 1 – 1

y = -2/3x + 1/3

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Standard Form of a Line◦Written as Ax + By = C, where A, B, and C are

integers◦We can convert our perpendicular line to

standard form by removing any fractions2 1

3 32

(3) (31

3 33 2 1

)

3 1

)

2

(3

y x

y x

y x

x y

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Forms of Linear Equations◦Standard Form: Ax + By = C

All integers, x and y terms on same side◦Slope-Intercept Form: y = mx + b

Best for quickly identifying slope and graphing◦Point-Slope Form: y – y1 = m(x – x1)

Best for writing equations◦Horizontal Lines: y = b

Slope = 0◦Vertical Lines: x = c

Slope is undefined

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Assignment◦Page 40 – 41

Problems 17 – 49 (odd problems) Show your work