summer assignment review writing and graphing linear equations: slope intercept form and point slope...

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Summer Assignment Review Writing and Graphing Linear Equations: Slope Intercept form and Point Slope Form

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Summer Assignment Review

Writing and Graphing Linear Equations:Slope Intercept form and Point Slope Form

Calculating SlopeGiven two points:

The points may have been given as coordinate pairs, or identified from a graph.

π‘š=𝑦2βˆ’ 𝑦1π‘₯2βˆ’ π‘₯1

Ex: Find the slope of the line containing the indicated points:

1) 2) (5 ,2 ) ,(βˆ’3 ,6) (0 ,βˆ’4 ) ,(2 ,8)π‘š=βˆ’12 π‘š=6

Ex: Find the slope of the line:

** you can also identify the sign of the slope visually, and use

Slope Intercept FormWrite an equation in slope-intercept form given the slope and y-intercept.

1) 2) 3)

𝑦=3 π‘₯βˆ’2 𝑦=βˆ’23π‘₯+4 𝑦=

14π‘₯+34

Identify the slope and y intercept for each line. (if the equation is not in slope intercept form, you must solve for y)

4) 5) 6)

𝑦=βˆ’2π‘₯+4 𝑦=53π‘₯βˆ’6 𝑦=βˆ’

18π‘₯+52

π‘š=βˆ’2 ,𝑏=4 π‘š=53,𝑏=βˆ’6 π‘š=βˆ’

18,𝑏=

52

Write an Equation Given A GraphIdentify the y-intercept from the graph ( ** if possible), and calculate the slope using any two points on the line, or by counting the rise and the run.

** If the y-intercept is not an integer value, you must start with point-slope form.

𝑏=βˆ’5

𝑏=1

π‘š=43

π‘š=βˆ’12

𝑦=43π‘₯βˆ’5

𝑦=βˆ’12π‘₯+1

Graphing EquationsGraphing is usually the easiest when the equation is in slope intercept form. If it is not in slope intercept form, solve for y. (If the equation was in standard form, how would you graph it?)

Graph each linear equation:

1) 2) 3) π‘š=

12𝑏=0 π‘š=βˆ’3 5 𝑦=

32π‘₯βˆ’2π‘š=

32 𝑏=βˆ’2

**Plot the y-intercept, then use the slope to find another point. **

Point-Slope FormWhen you have a point and a slope this is the form you start with. You will usually have to then solve for y and get the equation into slope-intercept form.

Write an equation in slope intercept form that has the given slope and contains the given point:

1) 2) 3)

π‘¦βˆ’ 𝑦1=π‘š (π‘₯βˆ’π‘₯1)

π‘¦βˆ’ 𝑦1=π‘š (π‘₯βˆ’π‘₯1)π‘¦βˆ’βˆ’3=2(π‘₯βˆ’4 )𝑦+3=2π‘₯βˆ’8

𝑦=2 π‘₯βˆ’11

π‘¦βˆ’ 𝑦1=π‘š (π‘₯βˆ’π‘₯1)π‘¦βˆ’4=βˆ’2(π‘₯βˆ’βˆ’1)π‘¦βˆ’4=βˆ’2(π‘₯+1)π‘¦βˆ’4=βˆ’2π‘₯βˆ’2

𝑦=βˆ’2π‘₯+2

π‘¦βˆ’ 𝑦1=π‘š (π‘₯βˆ’π‘₯1)

π‘¦βˆ’2=12(π‘₯βˆ’6)

π‘¦βˆ’2=12π‘₯βˆ’3

𝑦=12π‘₯βˆ’1

Parallel and Perpendicular Lines

**Perpendicular lines have opposite reciprocal slopes. **

**Parallel lines have the same slopes and different y-intercepts. **

Parallel and Perpendicular LinesWrite an equation in slope-intercept form for the line that contains the given point and is parallel to the given line.

1) 2) 3)

Write an equation in slope-intercept form for the line that contains the given point and is perpendicular to the given line.

4) 5) 6)

** You can either start with point slope form, or find the y-intercept and use slope-intercept form

𝑦=2 π‘₯+5 𝑦=βˆ’35π‘₯+2 𝑦=2 π‘₯βˆ’8

𝑦=βˆ’12π‘₯+5 𝑦=

13π‘₯+5 𝑦=

72π‘₯+3