What is slope? --=-!-tt r V Slopes of Lines Notes fl1tt--1hu-::t"llg1ll-cr'2.l:l:li.cl1/':.1tmll-lla.y..'-lj""-----.!. '('( ()..,()..,'VL slope of a Line n a coordinate plane, the slope of a line is the ratio of the change along the -axis to the change along the x-axis between an two points on the line. The slope m of a line containing two points with coordinates (x 1,,) and (Xl' Y2) is given b the formula Find the slope of each line. i ( ( A ( ), ( F (0/! x i J / j - :) - Lf - --1 L 0 x "111 - --- r--f-.! 0 x L fv 1--' --'Tp J.:-- ; ll...-- 1 -- o - 0 i Y, i ( i i i, 0 ( 1, 0- -- x -1-1-1--! 1 1 f ( l! R, -1, i--.i
YourTum Find the slope of each line. S l q, 1 P ( 10 -f.-a x f-t'm <3 x ( ( Q.,.Concept Summa.- Classifing Slopes Positive Slope Negative Slope Zero Slope Undefined Slope Yl r 0 x Find the slope of the line containing the given points. (6,-2) and (-3, -5) -5-(-Z2 _ :i-.._ nl,"-=3- l2- - - q --.3 _f _ (4,2) and (4,-3) - 3--:2 -G m -tf 7) YourTum Find the slope of the line containing the given points. (-3,3)and(4,3). 3-3 O_ N, ::..- - 0 Li-L-3j :t- (8,-3) and (-6, -2) Wt:: Z-{-3) ::.--1 - -Ll 3
Partitioning a Segment Notes Can ou find the midpoint of the line segment? () 3) Now with the same line partition the segment in a ratio of 1:1. 1:'50 "'Ve 'S!C o.s r m PO:' - L+ - -z.. Sometimes we need to break down segments into more than just two even pieces. 1 2 1. Find the point, P, that lies along the direct line segment from A = -7 to B = 8 and partitions the segment into the ratio of2:3. o( 234587 2. Find the point, P, that lies alon the direct line segment from A = 9 to B = -4 and partitions'3 the segment into the ratio of2:. (use the vertical number line) D,'s"'(.. k. "Z..t. - b z. l3 _ JJe.. -.!l = U.l. b... - b - 3 3 '3 Lj Y1. lam: {'(,OM q S -% --h"- L..A.t1.[1L{0/3 ] Your Turn.Jp Ned 0 '0-(0.<:: q -..(1/3: l-11./3 3. Find the point, P, that lies alon the direct line segment from A = -6 to B = 8 and partitions the segment into the ratio of 3:.l J 3t2 5 g Ys {.,('", t-s g c<wv-1:. -ht2ft; :2 g --""" -...---...-. S -7-9 4
4. Given the points A{-3,-4) and 6(5, OL find the coordinates of the point P on directed line segment AB that partitions AB in the ratio 2:3. G3:; 7(5 x..-"'l1.e. 8.'-; - bl ' A -/ C. 5-15 ; 3 1.5-3 -) Sol: 11-3 + S 'ls =- 'ls 'U - v' Lt e-)b -l{ 0 ol:q l1 2(5= J/ S = 13/s - Ll t- 3/s = -l1./...1r-_--".<,. 5. Given the points A(8, -5) and 6(4, 7), find the coordinates of the point P on directed line segment AB that partitions AB in the ratio 1:3. Yt "?>;-.'/«( X -V1t e AC. -7L d:l -"o..l-te -7B - 5-7 7 d 12 nol/q =- 3 - t 3.:.-2 Your Turn 6. Find the coordinates of point P that lies on the line segmentmq, M(-9, -5), Q(3, 5), and -- partitions the segment at a ratio of 2 to 5. '"2- "'L x. - va. e. 2+5 - -7 N -e - C( -) 'S 0(:11.,2/7:: 3:4 - q -t 3.L -: -{;;lj -' - V<!, &.t.e. c-? a -5-7 S cj.: 1'0 1D. Zf7 -.: :.l1 7-5+ L :.h :;. - 1/7 5
E.qua t" ons 0fL" mes No tes Slope Point Slope Form Slope ntercept Form (h, k) Form (Yl:: 1j2,-j i -j t::: m['-.- X) :; m)l..+-b -:-ell- h) +k )(;.-x/ Depending upon what ou are given ou can use different equations for find the equation of a line. - Write ation in slope-interept form of the line having the give slope and -intercept. slope: - intercept B slope: 11' (0,-3) m t) W. b, 'j";- 5/ 1 'f.. -:3 Ll-5X-;;L Your Turn "J Write an equation in slope-intercept form of the line having the given slope and -intercept, slope: 12, - intercept slope:, b: 8 5 7 :: (2-)( + '-1/5 :; G"/:t- "X +-? Write an equation in slope-intercept m=2,(3,11) form of the line with the given slope and through the given point m=--,(-3,-6). 5 <j -,; a{. -:3) J,- le z: - 4/ 5 {X!r3) 4 Your Turn Write an equation in slope-intercept form of the line with the given slope and through the given point. 5 m = 4,(-4,8) m = -,(-2,-5) 7 Write an equation of the line through each pair of points in slope-intercept form. (-12, -6) and (8,9) (0,5)and (3,3) n1 =-q+(p '::. /5-3 <if -1-12-- ;;;La - Y - q " 3/lf (X - S)
Your Turn Write an equation of the line through each pair of points in slope-intercept (2,4) and (-4, -11) fl1- _ -1(- i - /5 _ 5",-._-- ':4-2- --u Z. 'j_ 5 ( )(_) '1- '-l :: 6/2 '1-"- 5 Y:- G/Z'/. -- Find the equation of the graphed line. // -5 / form. (-3, -2) and (-3,4) 1./.--2 U J (), _ m :: _ ':: UVltt-tT1VLeo.. -3-3 0 X-=- -3 b::,./ 3./ 5./ 5 / 4 b-3 / /' 1 4 / :::. 3/ 'X.- 2 / -5 4-3 -2 1 2 3 4 5-1 -2-2 -3-4 -3 Y 1 1 2 3 4 5-1 -3 4 m::: --2- -z-)(-3 Your Turn Find the equation of the graphed line. --- :-'l.---!-- -'l'l-l:: -5 4 3 2-1 :1 2-4 -5 3 4 5 5-4 -3 1 1 2 3 4 5 1 / -2-3 / -4 5 7
Geometr Unit 3 Note Sheets ntroduction to Parallel and Perpendicular Lines Notes Determine the slopes of the lines on each graph. 1. 2... and mz = 3-3 3 m1= andmz= -. )' 3. 4. and mz = ml = t.:...9_ and mz = _O-=. 5. J' 6. m1 = _..::.... k. and mz = ---':5... - 7. Given that each set oflines above is parallel, what can ou conclude about the slopes of parallel lines? lh.e.. 7){op.JLS o.r e. t 5G.m 8. What happens if a line has the same slope and the same -intercept? 1hL O-.rt- t SG rn ; a..,., ca OflL G+h.e. r: on e.. 8
Determine the slopes of the lines on each graph. Then use a protractor to measure the angle of intersection. r 9. 10. _x 11. ml = and m2 =...3 Angle of ntersection = <1()" 12....,, "., ml = k and m2 = &- Angle of ntersection = 90" 13. m 1 = -/z. and m2 = 2- Angle of ntersection = 9of) _x 14. ml = 3' and m2 = -k(3 Angle of ntersection = '10 _u 1u 'k.- m 1 = and m2 =.2 m 1 = 5 and m2 = -z,.. Angle oflntersection = C}d' Angle oflntersection = 90 fj 15. What do ou notice about each, set of s!opes in the graphs a,bove? 1,/ _,.f.. '-ht.l.,.0...e., Opp 05 d"-(' i 11') 5 0 f e.,c;...lk, 0 ry,f. r, oo»: po'> 1(, Oil L n.lj +'111(" --rhe...1 "-re.. ('Lc"i P;"'OL<?J of {_G..ch 0 fr. 16. What do ou notice about the angles of intersection in the graphs above? '-1 i"'t c.l 90" 17. What kind of lines would ou classif these as? Explain. ((. r: (l'tn d c:.v-,lg.-r i blmlr. (t,1j c;..( 9
Slopes and Parallel Lines Notes What is the connection between slope and parallel lines? Slope is useful for determining whether two lines are parallel. Slope Criterion for Parallel Lines Because the theorem above has a biconditional (if and onl if) ou can use it in either "direction". Recall the different was to write equations of lines. Slope ntercept Form Point Slope Form (h, k) Form :: t'l"x tb -''J i ::. M (X-X,") '0 -z: (k( l(-) -t. k E X A MP le Writing Equatiorls of 'arauel lines Write the equation of each line in slope-intercept form. A The line parallel to = -2x + 3 that passes through (1, -4) The given line is in slope-intercept form and its slope is --=._. The required line has slope '=-_ because parallel lines have the same slope. Y - Y l = m(x Xl) Use point-slope form. Y - (... <..j} = -2. ex - --L-_) Substitute for m, Xl' and Y,. + _<d =."±".L. Simplif each side of the equation. _.7.. 'B = -.".--.-""..-"-" Write the equation in slope-intercept form. The line that passes through (2,3) and is parallel to the line through (1, -2) and (7, 1) The slope of the line through (1, -2) and (7,1) is So, the required line has slope... - YYl=m(x-xl) 3, -""" "- = (x - " ) Y=_._tx + '2- Use point-slope form. Substitute for m x, v and 1" Simplif and write Slope-intercept form. 10
REFLECT' 3a. n Part A, how can ou check that ou wrote the correct equation? ehlk ±'Q_u +hc_, H-u.....lQ:?_.c,,L...ifu. 5qmR-. ;1frt:l:h... (u----f2-u:>.hcq",,+-j&q()t :">u if... in 3b. n Part A, once ou know the slope of the required line, how can ou finish solving the problem using the slope-intercept form of a linear equation? _il_... J.:n.....±.h±_""'!fL1.J'--f2..oLa.L..._Cu1Qf-j hl()_. sq Q LJ:L.. C.::lt.ceJ2+)"""'-""""""-"""""""'"... _ ""... _. Your Turn '","ritean equation of the Hne that ;llolsscs hrough (-3, -5) and tsjarahcl to thehne=lx-. (Yl 3 'j -,.S- -i'' -5< ::: 3('>< l-3) =. 3x +q -s: -;:: 3)<' +q Write anequation urthe line that passes through (-2, 10 and is parallel to the line J' = X + 5. rn :::-1 _ '" -(l( +.,)) (((-+2-) l() - t ( -::. - X -'1-.. 11 ";'1/ ::: -X -. Your Turn Write an equation of the line that passes through point P and is parallel t() the line with the given equation. P(:8. 7), Y = 3 P{l, -2),)'''''' :'- 8.!;j - (Y1 ::: 0 "1 - "f 0('>( -l) 6-i- 0 l 1- <A- p, - 4 '11"2""-' (X -). r' tt>+l x- -,-' -"L-.,. )(- 11
Slopes and Perpendicular Lines Notes What is the connection between slope and perpendicular lines? Slope is useful for determining whether two lines are perpendicular. Slope Criterion for Jlma»tl Hiles PRrO-'LnolJ ("kfll/ L,'.2.J Because the theorem above has a biconditional (if and onl if) ou can use it in either "direction"..e X A WlP L '1:,' Writing Equations of Perpendicular lines Write the equation of the line perpendicular to = 3x- 8 that passes through (3, 1). Write the equation in slope-intercept form. A First find the slope of the required tne, The given line is in slope-intercept form and its slope is..3 Let the required line have slope m. Since the lines ate perpendicular, their slopes is -1. So,.,. m = -1, and therefore, 111= _.=.!3... the product of a Now use point-slope form to find the equation of the required line. - )'1 = m(x - Xl) Usepoint-slope form. Y - J_ =,-=(x -.._.....:::3""---_)..0...L... :::: Substitute for m, Xl' and Y r.,,:_2c.._.+.j Distributive Propert Y= -X t"l Write the equation in stope-nterceot form. la. How do ou find the slope of the given line? -:! +ODk:_ +Jv._.2.p-OSi'tL2 MP +c:")k i'+s 04 :-o-tvof", -......,.... "-.....-.-...=-,-",,,....... -. -- 3b. How can ou use graphing to check our answer?..... - -..,..-.-..--.<-- ----"'. mm._ _... -_. --._-..,...,,,.,... 10 m""' S!A!"--Lt6.. Z..1, @-,L)l:},,Ll._.ll:b,..9/...- f'1---±b-±b.u+-g-r-l----f=fr ","5lc.JJv.,i t".,lts., 12
{P.. J... b..lc b c. Your Turn Line a: == 5x - 3 Line n x + 5)' = 2 Lince: -lo - 2x = {} LA "'" -? J - 'S t +l.x.. 6. Determine which ines, if an; are parallel or perpendicular. tinea: + 6v "" Line h: Y= 3x - 8 -Lx.' -'L".. Li'1-1.x'- "= - - tj... b bile.. 0..L C V--4 7. Decide whet bel' AC and DB are perpendlcular. :./0';) -10 -v lj -:-Js-:x Llne c:-l5 + 4,,5x "'" 6 -.S'1j =- -4.5'>( (.. 3)( -.f B (- 'L) J D (S-,- ) 1-2.. _ -3 -lz. 5"-(:-1i - 1)?t'-r- f'(>j10""(" r: (. 13
f Perpendicular Bisecto! Nftes Bisec:or - 1+' G.. PO t) -+- /.S ct PO t1 c.. ll,! vd Y Ni S """flncl{«to +h.ed "}LA,. P9rp {'V'lcl.lVA ((l.. ' b:s.e-j (fr. nstructions to Construct a Perpendicul('tor / 1. Place our compass point on A d;e compass MORE THAN halfwa to point B, but not beondb. X _ 2, With this length, swing a large arc that wi,a go BH above and below AS. (f ou do not wish to make one large continuous arc, ou ma sim1(yplace one-small arc above AS and one small arc below AS.) 3. Without changing the span on the.c9xhpass, place the compass point on B and swing the arc again. The two arcs ou have created should ntersect. 4. With our straightedge, connect the two points of intersection, 5. This new straight line bisects AB. Label the point where th new line and AB cross as C. f '. 1.. p. e. a YourTum 2, ------ 3. p, / / / - D N " 14 '.
Construct a perpendicular bisector and then n the relationships that we know about the figure, and what values we cannot state specific relationships about. p----------------------------------------q Relationships What cannot be assumed Oo., oe 11 re.c::- - / -Y r: OR 'id. (cvi..j..- tf."n Jt -.L l c"'+ OOOtr', G.,.r.(... lv1,-g /J 15