Graphing - Parallel and Perpendicular Lines

1 - Parallel and Perpendicular LinesObjective: Identify the equation of a line given a Parallel or perpendic -ular is an interesting connection between the slope of Lines that are Parallel andthe slope of Lines that are Perpendicular (meet at a right angle). This is shown inthe following above graph has two parallellines. The slope of the top line isdown 2, run 3, or 23. The slope ofthe bottom line is down 2, run 3 aswell, or above graph has two perpendic -ular Lines . The slope of the flatter lineisup2, slope of the steeper line is down3,run2or View Note:Greek Mathematician Euclid lived around 300 BC and pub-lished a book titled,TheElements.

2 In it is the famous Parallel postulate whichmathematicians have tried for years to drop from the list of postulates. Theattempts have failed, yet all the work done has developed newtypes of geome-tries!As the above graphs illustrate, Parallel Lines have the sameslope and perpendic -ular Lines have opposite (one positive, one negative) reciprocal (flipped fraction)slopes. We can use these properties to make conclusions about Parallel and per-pendicular the slope of a line Parallel to5y 2x= 2x= 7To find the slope we will put equation in slope intercept form+ 2x+ 2xAdd2xto both sides15y= 2x+ 7 Putxterm first555 Divide each term by5y=25x+75 The slope is the coefficient ofxm=25 Slope of first Lines have the same slopem=25 Our SolutionExample the slope of a line Perpendicular to3x 4y= 23x 4y= 2To find slope we will put equation in slope intercept form 3x 3xSubtract3xfrom both sides 4y= 3x+ 2 Putxterm first 4 4 4 Divide

3 Each term by 4y=34x 12 The slope is the coefficient ofxm= 43 Our SolutionOnce we have a slope, it is possible to find the complete equation of the secondline if we know one point on the second the equation of a line through(4, 5)and Parallel to2x 3y= 3y= 6We first need slope of Parallel line 2x 2xSubtract2xfrom each side 3y= 2x+ 6 Putxterm first 3 3 3 Divide each term by 3y=23x 2 Identify the slope,the coefficient ofx2m=23 Parallel Lines have the same slopem=23We will use this slope and our point(4, 5)y y1=m(x x1)Plug this information into point slope formulay ( 5) =23(x 4)Simplify signsy+ 5 =23(x 4)Our SolutionExample the equation of the line through(6, 9) Perpendicular toy= 35x+ 4inslope-intercept 35x+ 4 Identify the slope,coefficient ofxm= 35 Perpendicular Lines have opposite reciprocal slopesm=53We will use this slope and our point(6, 9)y y1=m(x x1)Plug this information into point slope formulay ( 9) =53(x 6)Simplify signsy+ 9 =53(x 6)

4 Distribute slopey+ 9 =53x 10 Solve fory 9 9 Subtract9from both sidesy=53x 19 Our SolutionZero slopes and no slopes may seem like opposites (one is a horizontal line , one isa vertical line ). Because a horizontal line is Perpendicular to a vertical line we cansay that no slope and zero slope are actually Perpendicular slopes!Example the equation of the line through (3, 4) Perpendicular tox= 2x= 2 This equation has no slope, avertical lineno slope Perpendicular line then would haveazero slopem= 0 Use this and our point(3,4)y y1=m(x x1)Plug this information into point slope formulay 4 = 0(x 3)

5 Distribute slopey 4 = 0 Solve fory+ 4 + 4 Add4to each sidey= 4 Our SolutionBeing aware that to be Perpendicular to a vertical line meanswe have a hori-zontal line through ayvalue of 4, thus we could have jumped from this pointright to the solution,y= and Intermediate Algebra by Tyler Wallace is licensed under a Creative CommonsAttribution Unported License. ( ) Practice - Parallel and Perpendicular LinesFind the slope of a line Parallel to each given )y= 2x+ 43)y= 4x 55)x y= 47)7x+y= 22)y= 23x+ 54)y= 103x 56)6x 5y=208)3x+ 4y= 8 Find the slope of a line Perpendicular to each given )x= 311)y= 13x13)x 3y= 615)x+ 2y= 810)y= 12x 112)y=45x14)3x y= 316)8x 3y= 9 Write the point-slope form of the equation of the line ) through: (2,5), Parallel tox= 018) through: (5, 2), Parallel toy=75x+ 419) through: (3,4), Parallel toy=92x 520) through:(1, 1), Parallel toy= 34x+ 321) through: (2,3), Parallel toy=75x+ 422) through.

6 ( 1,3), Parallel toy= 3x 123) through: (4,2), Parallel tox= 024) through: (1,4), Parallel toy=75x+ 225) through: (1, 5), Perpendicular to x+y= 126) through: (1, 2), Perpendicular to x+ 2y= 227) through: (5,2), Perpendicular to5x+y= 3528) through: (1, 3), Perpendicular to x+y= 129) through: (4,2), Perpendicular to 4x+y= 030) through: ( 3, 5), Perpendicular to3x+ 7y= 031) through: (2, 2) Perpendicular to3y x= 032) through:( 2,5). Perpendicular toy 2x= 0 Write the slope-intercept form of the equation of the line ) through: (4, 3), Parallel toy= 2x34) through: ( 5,2), Parallel toy=35x35) through: ( 3,1), Parallel toy= 43x 136) through: ( 4,0), Parallel toy= 54x+ 437) through: ( 4, 1), Parallel toy= 12x+ 138) through: (2,3), Parallel toy=52x 139) through: ( 2, 1), Parallel toy= 12x 240) through: ( 5, 4), Parallel toy=35x 241) through: (4,3), Perpendicular tox+y= 142) through: ( 3, 5), Perpendicular tox+ 2y= 443) through: (5,2), Perpendicular tox= 044) through.

7 (5, 1), Perpendicular to 5x+ 2y=1045) through: ( 2,5), Perpendicular to x+y= 246) through: (2, 3), Perpendicular to 2x+ 5y= 1047) through: (4, 3), Perpendicular to x+ 2y= 648) through: ( 4,1), Perpendicular to4x+ 3y= 9 Beginning and Intermediate Algebra by Tyler Wallace is licensed under a Creative CommonsAttribution Unported License. ( ) - Parallel and Perpendicular Lines1) 22) 233) 44) 1035) 16)657) 78) 349) 010) 211) 312) 5413) 314) 1315) 216) 3817)x= 218)y 2 =75(x 5)19)y 4 =92(x 3)20)y+ 1 = 34(x 1)21)y 3 =75(x 2)22)y 3 = 3(x+ 1)23)x= 424)y 4 =75(x 1)25)y+ 5 = (x 1)26)y+ 2 = 2(x 1)27)y 2 =15(x 5)28)y 3 = (x 1)29)y 2 = 14(x 4)30)y+ 5 =73(x+ 3)31)y+ 2 = 3(x 2)32)y 5 = 12(x+ 2)33)y= 2x+ 534)y=35x+ 535)y= 43x 336)y= 54x 537)y= 12x 338)y=52x 239)y= 12x 240)y=35x 141)y=x 142)y= 2x+ 143)y= 244)y= 25x+ 145)y= x+ 346)y= 52x+ 247)y= 2x+ 548)

8 Y=34x+ 4 Beginning and Intermediate Algebra by Tyler Wallace is licensed under a Creative CommonsAttribution Unported License. ( )7