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Geometry. Unit IIB 3.7: Writing Equations of Lines. Coordinates of a point on a graph. =-1. ). Y. - PowerPoint PPT PresentationTRANSCRIPT
Geometry
Unit IIB3.7: Writing Equations of Lines
While the slope-intercept form for the equation of a line is very useful in graphing, it is not nearly as useful for writing the equation of a line (unless you happen to be given the slope and the y-intercept). The point-slope form for the equation of a line is _____________________________________
In this equation m is, again, the slope and x1 and y1 are _________________________________________
__________________________________
π¦β π¦1=π (π₯βπ₯1)Coordinates of a point on a graph
Example: Write an equation for the line in point-slope form. 1. slope of -1 and contains the point (-3, 9) 2. passes through the points (1, 4) and (-3, 8)
3. slope of 3 and a y-intercept of 15 4. passes through the point (5, 1) and is perpendicular to the line π¦β15=3(π₯β0)
π¦β π¦1=π (π₯βπ₯1)
π¦β π¦1=π (π₯βπ₯1)
π¦β π¦1=π (π₯βπ₯1))
π¦β9=βπ₯β3π¦=βπ₯+6
=-1
π¦β4=β1(π₯β1)π¦β4=βπ₯+1π¦=βπ₯+5
π¦β1=β3/5(π₯β5)π¦β15=3π₯π¦=3 π₯+15
π¦β1=β35π₯+3
If you are specifically asked to write an equation in slope-intercept form, it is generally easiest to write it in point-slope form first and then change it to slope-intercept. Example: Write an equation of the line in slope-intercept form. 5. passes through the points (-4, 7) and (-2, 12) 6. passes through the point (12, 15) and is parallel to the line
π=12β7β2+4
=52
π¦=ππ₯+π
7=52
(β4 )+π
7=β10+π
17=π
π¦=52π₯+17
π¦β π¦1=π (π₯βπ₯1)π¦β15=1 /3(π₯β12)
π¦β15=1 /3π₯β4
Y π¦β7=
52(π₯+4 )
π¦β π¦1=π (π₯βπ₯1)
π¦β7=52π₯+10
π¦=52π₯+17