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