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ThingstoKnowChapter2:Geometry
LineEquations(Linear)LinearEquationsofy=mx+cformWhenthelinearequationisintheformofy=mx+cthenthegradientandy–interceptofthegraphcaneasilybefound.Fory=mx+c
• m=gradientàValueofmwillbethegradientoftheline
• c=y–interceptàValueofCwillbethey–intercept,whichisthepointonthey–axiswherethelinecrosses,thusthepointwillhavecoordinatesof(0,C)
“ALWAYSre-arrangelinearequationstohavey=mx+cformatforconvenience”Example:2𝑥 + 5𝑦 = 6
à5𝑦 = −2𝑥 + 6à𝑦 = − !
!𝑥 + !
!
Gradient:− !
!,y–intercept:!
!
!!+ !
!= 1
à!
!𝑥 + 𝑦 = 2
à𝑦 = − !
!𝑥 + 2
Gradient:− !
!,y–intercept:2
2𝑦 + 3 = 3𝑥
à2𝑦 = 3𝑥 − 3à𝑦 = !
!𝑥 − !
!
Gradient:!
!,y–intercept:− !
!
• y–interceptisfoundbyinputtingx=0intotheequation• x–interceptisfoundbyinputtingy=0intotheequation• x–intercepthasay–coordinateof0• y–intercepthasanx–coordinateof0
MidpointGettingamidpointbetweentwopointsissimple.Themidpointofpoint𝐴(𝑥! ,𝑦!)and𝐵(𝑥! ,𝑦!)issimply(
!!!!!!
, !!!!!!)
(!!!!!
! , !!!!!
!)à(x,y)midpoint
Example:Mid-pointof(2,4)and(-6,0)à !!!
! , !!!
!= (−2, 2)
Mid-pointof(-2,4)and(8,3)à !!!!
! , !!!
!= (3, !
!)
Mid-pointof(-3,0)and(3,1)à !!!!
! , !!!
!= (0, !
!)
GradientofLineGradientsofthelinesareindicatedwiththeletter‘m’.Gradientofthelinerepresenttheslopeofthegraph.Thereare2simplewaystofindthegradientsofthegraphs.Findthegradientofaline:1.Whenfindingthegradientofthelinejoiningthetwopoints,point𝐴(𝑥! ,𝑦!)and𝐵(𝑥! ,𝑦!),thenthegradientofthislinecaneasilybefoundwiththeformula:
!!!!!!!!!!
or!!!!!!!!!!
• When𝑦!comesfirstinthenumeratorthen𝑥!hastocomefirstinthedenominator
• BesuretochecktheformulacarefullyExample:(2 , 4)and(3 , 8)
à!!!!!!
= 4à𝑚 = 4
(−1 ,7)and(3 ,−5)
à !!!!!!(!!)
= !!"!= −3
à𝑚 = −3
(𝑎 , 𝑏)and(3𝑎 + 1 ,2𝑏)
à !!!!!!!!!!
= !!!!!
à𝑚 = !
!!!!
2.Findingthevalueof‘m’straightfromtheequationbyre-arrangingtheequationintheformofy=mx+c,thenm=gradient(thisisexplainedinthe‘lineequations’section)
ParallelLines:Whenthelinesareparalleltoeachother,thentheyhavethesamegradient
PerpendicularLines:Whenthelinesareperpendiculartooneanother,thefollowingformulahastobesatisfied:
𝑚! × 𝑚! = −1
𝑚!isthegradientofonelineand𝑚!isthegradientofanotherline,iftheselinesareperpendiculartoeachotherthen𝑚! × 𝑚!shouldbeequalto−1Thisformulaisusedwheneveragradientofaperpendicularlinehastobefound.Example:Findthegradientofthelineperpendicularto2𝑥 + 3𝑦 = 1
à2𝑥 + 3𝑦 = 1à3𝑦 = −2𝑥 + 1à𝑦 = − !
!𝑥 + !
!,thegradientofthislineis− !
!
à− !
! × 𝑚! = −1
à 𝑚! =
!!
àThegradientoftheperpendicularis!
!
LineEquation
Equationofalinecaneasilybefoundaslongasyouknowthegradientofthelineandanyonepointthatisontheline.Usuallythequestionswillaskyoutofindtheequationofalinethatispassingthroughtwopoints,pointAandpointB.Getthegradientofthelineusing!!!!!
!!!!!thenuseeitherpointAorpointBtoget
theequationoftheline.Somequestionsdoprovidethevalueofthegradientofthelinewhiletherearealsoquestionsthatprovidethegradientofthelinethatisperpendiculartothelinethatyouarelookingfor.Forsuchacase,use𝑚! × 𝑚! = −1togetgradientoflinethatyouarelooking.
• Requirements:Apointonline(𝑥!,𝑦!)andthegradientofline‘m’
• Formula:𝑦 − 𝑦! = 𝑚(𝑥 − 𝑥!)thenre-arrangetheformulaintoy=mx+cform
Example:Findthelinewithagradientof3andpoint(2,5)
à𝑦 − 5 = 3(𝑥 − 2)à𝑦 − 5 = 3𝑥 − 6à𝑦 = 3𝑥 − 1
Findthelinewithagradientof−2andpoint(2,−4)
à𝑦 − −4 = −2(𝑥 − 2)à𝑦 + 4 = −2𝑥 + 4à𝑦 = −2𝑥
• Alotofquestionsareprovidedonthis.Usetheanswersheetformore
guidanceifyouneeditDistanceBetweenPointsFindingthedistancebetweentwopoints,pointA(𝑥!,𝑦!)andpointB(𝑥!,𝑦!)orfindingthelengthofthesideofashapewithvertexesAandBcansimplybedonewithaformula.WithpointA(𝑥!,𝑦!)andB(𝑥!,𝑦!)
𝑥! − 𝑥! ! + 𝑦! − 𝑦! !or 𝑥! − 𝑥! ! + 𝑦! − 𝑦! !
Example:Distancebetween(2, 3)and(−2, 5)
à 2− (−2) ! + 3− 5 ! = 2 5 = 4.47Distancebetween(−3, 9)and(−1,−2)
à −3− (−1) ! + 9− (−2) ! = 5 5 = 11.2
• Thisformulaisusedinfindingthesidesofashapetogettheareaorperimeter.
PerpendicularBisectorPerpendicularbisectorofalinejoiningtwopoints,pointAandpointB,caneasilybefoundwhenyouknowthegradientoftheperpendicularbisectorandthepointitpassesthrough.Theperpendicularbisectorisalsoalinesoitrequiresthegradientofthelineandapointthatisonthelinetouse 𝑦 − 𝑦! = 𝑚(𝑥 − 𝑥!).However,fortheperpendicularbisectorofalinejoiningpointsAandB,youcannotuseeitherpointAorpointBbecausethosepointsdonotlieontheperpendicularbisector.Also,thegradientofthelinejoiningAandBisnotthegradientoftheperpendicularbisector.TwomainrequirementsarethemidpointofAandBandthegradientoftheperpendicularline.
PerpendicularbisectorpassesthroughthemidpointandisperpendiculartolineAB.Forgradient:𝑚! × 𝑚! = −1à𝑚!" × 𝑚! = −1Formidpoint:(!!!!!
!, !!!!!
!)à(𝑥!,𝑦!)
Thenusethelineequation:𝑦 − 𝑦! = 𝑚!(𝑥 − 𝑥!)Example:FindtheperpendicularbisectorofthelinejoiningpointC(2, 6)andpointD(−4,−2)àGetthegradientoflineCD
à!!!!!!!!
= !!!!= !
!
àGetthegradientoftheperpendicularbisector
à!!
× 𝑚! = −1à 𝑚! = − !
!
àGetthemidpointofCandD
à !!!!, !!!!
= (−1, 2)
àNowusethelineequationwithgradient− !
!andmidpoint(−1, 2)
à𝑦 − 2 = − !
!(𝑥 − (−1))
à𝑦 − 2 = − !
!(𝑥 + 1)
à𝑦 − 2 = − !
!𝑥 − !
!
à𝑦 = − !
!𝑥 + !
!
• Followthestepsintheexampletosolveactualquestions:o Getthegradientofthelineo Gettheperpendiculargradiento Getthemidpointo Getthelineequation