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Higher Mathematics CfE EditionThis documentproducedspecially for the H.uk.netwebsite, and we require that any copies or derivative works attribute the work to Higher Still Notes hsn .uk.net Straight Lines Contents Straight LinesThe Distance Between Points HigherMathematicsStraight Lines Page CfE Edition hsn .uk.net Straight Line The Distance Between Points Points on Horizontal or Vertical LinesIt is relatively straightforward to work out the distance between two points which lie on a line parallel to the axis. In the diagram to the left, the points 11,xy and 22,xy lie on a line parallel to theaxis, i 12yy Th攠distanc攠b整w敥n the⁰o楮tss業ply th攠diff敲敮c攠in⁴h攠coordinates, i =−dxx where 21xx In⁴heiag牡m⁴漠瑨ee晴,⁴he⁰潩n瑳 11,xy and 22,xy lie on a line parallel to theaxis, i xx Th攠distanc攠b整w敥n th攠灯intssim灬y th攠diff敲敮c攠in⁴h攠coordinates, i =−dyy where 21yy EXAMPLE Calculate the distance between the points 7,3 慮d 16,3 The distance is 16716723 units−−=+=. x 11,xy 22,xy y O d x 11,xy 22,xy O y d HigherMathematicsStraight Lines Page CfE Edition hsn .uk.net The Distance FormulaThe distance formula gives us a method for workingout the length of the straight line between anytwopoints. It is based on PythagorasTheorem. The distancebetween the points 11,xy and 22,xy is ㈱21dxxyy=−+− 畮i瑳 EXAMPLE is the point 2,4 a湤 B3,1 Calculatethe length of the line ㈱21Th攠l敮gths 32ㄴ㈵934⁵湩tsxxyy−+−=−−+−=+−=+=. Calculate the distance between the points 15124, 慮d 1,1−− ㈱21㈱422ㄶㄶTheista湣es ⁵湩tsxxyy−+−=−−+−+=−−+−+=−+=+=+=. y O x 22,xy 11,xy 21xx yy d Note The yy 鐠and xx 鐠comerom⁴he 浥tho搠above. Note You need to become confident working with fractions and surds so practise! HigherMathematicsStraight Lines Page CfE Edition hsn .uk.net The MidpointFormula The point halfbetween two pointsis called themidpointIt is calculated as follows The midpoint 11,xy and 22,xy is 1212xxyy . It may be helpful to think of the midpoint as the average of two points. EXAMPLE Calculate the midpoint of the points 1,4 a湤 7,8 . 1212The midpoint is ,4,2xxyy=. In the diagram below A9,2 汩esn⁴hei牣畭晥牥nce映瑨ei牣汥 wi瑨en瑲e C17,12 , and the line AB is the diameter of the circle. Find the coordinates of B. Since C is the centre of the circle and AB is the diameter, C is the midpoint of AB. Using the midpoint formula, we have: 17,12,where B is the point ,+−+ B礠comp慲in朠and coordinates, we have: 934+= −+=and 12224 So B is the point 25,26 . A B C Note Simply writing The midpoint is (4, 2) would be acceptable in an exam. HigherMathematicsStraight Lines Page CfE Edition hsn .uk.net Gradients Consider a straight line passing through the points 11,xy and 22,xy : The gradientof the line through 11,xy and 22,xy is change in vertical height for change in horizontal distance 䅬s漬ince O灰os楴et慮䅤j慣ent== we扴ai渺 tan where is the angle between the line andthe positive direction of the axis NoteAs a result of the above definitions: lines with positive gradients slope up, from left to right lines with negative gradients slope down, from left to right lineparallel to the axis hagradient of zerolineparallel to the axis haan undefined gradientWe may also use the fact that: Two distinct lines are said to be parallelwhen they have the same gradient (or when both lines are vertical) y O x 22,xy 11,xy 21xx yy positive direction x Note ”s⁴h攠G牥敫整t敲 鍴he瑡”. I琠i猠of瑥n u獥d⁴o瑡nd for渠a湧le. HigherMathematicsStraight Lines Page CfE Edition hsn .uk.net EXAMPLES Calculate the gradient of the straight lineshownin the diagrambelow tantan320·62 (to 2 d.p.)=° Find the angle that the line joining P2,2 a湤 Q1,7 makes with the positive direction of the axis. The line has gradient === 䅮d漠 ==°tantan3tan371·57 (to 2 d.p.). Find the size of angle s桯睮n⁴桥i慧r慭elo眮 坥敥d⁴o攠car敦ul散aus攠the 楮⁴heⁱuest楯ns notthe i渠 t慮 S漠we⁷潲k畴⁴heng汥 and use this to find tantan578·690=°. 卯 㤰㜸·㘹0ㄱ·㌱ to′⸩=°−°=° m x O y a m x O y 32 x y O HigherMathematicsStraight Lines Page CfE Edition hsn .uk.net Collinearity Points whichlie on the same straight line are said to be collinear To test if three pointsA, B and Care collinear we canWork out ABm Work out BCm (or ACm If the gradients from 1. and 2. are the same then A, B and C are collinear.If the gradients are different then the points are not collinear.This test for collinearity can only be used in two dimensions. EXAMPLE Showthat the points P6,1 Q0,2 and R8,6 are collinear. −−=== === Si湣e 偑QR a湤 楳o浭on⁰o楮t, a湤 a牥潬汩near The points A1,1 B1, a湤 C5,7 are collinear. ind the value of Since the points are collinear ABAC 171115124122−−−−−−−+=×−=− 䅂BC s漠䄬⁂nd⁃牥潴潬汩nea爮 m BCm ABBC s漠䄬⁂nd⁃牥潬linea爮 m BCm HigherMathematicsStraight Lines Page CfE Edition hsn .uk.net Gradients of PerpendicularLines Two lines at rightangles to each other are said to beperpendicular perpendicularlines havegradients and m 瑨en ×=− C潮ve牳e汹,映 ×=− then the lines are perpendicular. The simple rule isif you know the gradient of one of the lines, then the gradient of the otheris calculated by inverting the gradient (i.e. flipping the fraction) and changingthe sign. For example: =−32if then . Note that this rule cannot be used if the line is parallel to the axis. If a line is parallel to the axis (0) Ⱐt桥n t桥pe牰endic畬a爠汩nes⁰a牡汬el 瑯⁴he axis it has an undefined gradient.f a lineis parallel to the axis then the perpendicular line is parallel to the axisit has a gradient of zero EXAMPLES Given that T is the point 1,2 a湤匠is 4,5 ,ind⁴he g牡dien琠潦 汩ne⁰e牰endic畬a爠瑯 ST. −−==− So 57m since ×=−mm . TriangleP has vertices M3,9 O0,0 and P12,4 . Show that the triangle is rightangled. Sketch: −−=− MP3ㄲ−−=−=− ㄲ0 Si湣e 位佐×=− is⁰敲灥ndicular⁴o w桩c栠me慮s 䵏P isi杨t慮杬edt NoteThe converse of Pythagorass Theorem could also be used here: P12,4 O0,0 M3,9 HigherMathematicsStraight Lines Page CfE Edition hsn .uk.net 2221241603990=+==−+= MP12349155=−−+−=+− Si湣e 222OPOMMPddd , triangle MOP is rightangled at O. The Equation of a StraightLine To work out the equation of a straight line, we need to know two things: the gradient of the lineand a point which lies on the line. straight linethrougthe point ,ab withgradient has the equation ybmxa−=− N潴ice⁴ha琠if⁷e慶e 愠point 0,c the axis intercept then the equation becomes ymxc=+ .⁙潵h潵ld a汲eady be慭ili慲⁷it栠t桩sI琠is潯d⁰牡c瑩ce⁴漠rea牲ange⁴heq畡瑩潮 o映a瑲aigh琠汩nen瑯⁴he潲m axbyc++= wh敲攠is positive. This is known as the general form of the equation of a straight line.Lines Parallel to AxesIf a line is parallel to the axis (i.e. 0m ),ts煵at楯ns yc I映aines⁰a牡汬e氠瑯⁴he axis (i.e. is undefined), its equation is xk x O y k xk x O y c yc HigherMathematicsStraight Lines Page CfE Edition hsn .uk.net EXAMPLES Find the equation of theline with gradient 13 passing through the point 3,4 43(3)31233153150.ybmxayx−=−−−=−×+=−=−−−= Find the equation of the line passing through A3,2 and B2,1 T漠w潲k畴⁴heq畡ti潮,⁷e畳琠fi牳琠晩nd⁴he牡dien琠潦⁴heine⁁B: ==−− No眠睥慶er慤ient,andan⁵se⁴his⁷i瑨ne映瑨eiven⁰潩n瑳: 23usi湧⁁3,2湤 5357〮ybmxa−=−−=−−=−=+−+= Find the equation of the line passing through 35,4 a湤 35,5 Th攠gradi敮ts⁵nd敦in敤inc攠th攠coordinates are equal.So the equation of the line is =− Note It is usually easier to multiply out the fraction before expanding the brackets. HigherMathematicsStraight Lines Page CfE Edition hsn .uk.net Extracting the radientYou should already be familiar with the following fact. The line with equation ymxc=+ 桡sr慤ient It is important to remember that you must rearrangethe equation into this form beforeextracting the gradient. EXAMPLES Find the gradient of the line with equation 3240++= Weave⁴漠牥a牲ange⁴heq畡瑩潮: 32㐰234yx++==−−=−− 卯⁴桥r慤ients The line through points A3,3 慮d⁂慳qu慴ion 5ㄸ0−−= Find⁴heq畡瑩潮映theine⁴h牯畧h⁁⁷hichs⁰e牰endic畬a爠瑯 䅂. Fi牳琬ind⁴he牡dient映䅂: 5ㄸ05ㄸ.−−==− 卯 m a湤 =− The牥f潲e⁴he煵at楯ns: 515351535120.+=−−+=−−+=−+++= HigherMathematicsStraight Lines Page CfE Edition hsn .uk.net Medians medianof a triangle is a line throughvertandthe midpointof the opposite side is median ABC The瑡nda牤⁰牯cess潲inding⁴heq畡瑩on映aedians s桯睮be汯w. EXAMPLE riangle ABChas vertices A4,9 B10,2 and C4,4 Find⁴heq畡瑩潮映theedian牯m 却慲t⁷it栠愠 s步tch: Step 1 C alculate the midpoint of the levantline Using B10,2 and C4,4 1041427,1.=− Step 2 C alculate the gradient of the line between the midpointand the opposite vertex Using A4,9 a湤 M7,1 19−−−== Step 3 Find the equation using this gradient and either of the two points used inStep 2 ing A4,9 a湤 m 94(3)327832385983590.ybmxa−=−+=−×+=−=−−−= HigherMathematicsStraight Lines Page CfE Edition hsn .uk.net Altitudes An altitudeof a triangle is a line through a vertex, perpendicularto the opposite side. BD is altitude ABC The瑡nda牤⁰牯cess潲 晩nding⁴heq畡瑩on映an汴i瑵de 楳 s桯睮be汯w. EXAMPLE Triangle ABC has vertices A3,5 , B4,3 and C7,2 Find⁴heq畡瑩潮映the汴i瑵de牯m 䄮 却慲t⁷it栠愠sketc栺 Step 1 C alculate the gradient of the sidewhich is perpendicular to the altitude Using B4,3 and C7,2 == Step 2 C alculate the gradient of the altitudeusing ×=− 啳in朠 BC䅄×=− =− Step 3 Find the equation using this gradient and the point that the altitude passes through Using A3,5 a湤 =− 5113112811280.ybmxa−=−+=−−=−++−= HigherMathematicsStraight Lines Page CfE Edition hsn .uk.net Perpendicular Bisectors perpendicular bisectoris a line which cuts through the midpoint line segment at rightangles. The standard process for finding the equation of aperpendicular bisectoris shownbelow. EXAMPLE A is the point 2,1 and B is the point 4,7 . Find the equation of the perpendicular bisector of Start with a sketch: Step 1 C alculate the midpoint of the line segment beingbisected. Using A2,1 a湤 B4,7 : 2417Midpoint1,4.−++ Step 2 C alculate the gradient of the line used in tep then find the gradient of itsperpendicular bisectorusing ×=− 啳in朠 A2,1 a湤 B4,7 : 71−− =− 獩n捥 ×=− Step 3 Find the equation of the perpendicular bisector using the point from Step 1and the gradient from Step 2. Using 1,4 and =− 4150.ybmxa−=−−=−−=−++=−++−= Inot栠c慳esⰠC is⁴h攠 pe牰endic畬a爠bisec瑯r映䅂. HigherMathematicsStraight Lines Page CfE Edition hsn .uk.net Intersectionof Lines Many problems involve lines which intersect (cross each otherOnce we have equations for the lines, the problem is to find values for and which satisfy both equations, i.e. solve simultaneous equations.here are three different techniquesand,depending on the form of the equations, one may be more efficient than the othersWe will demonstrate thetechniques by finding the point of intersection of the lines with equations 315=+ a湤 =− liminationThis should be a familiar method, and can be used in all cases. 315 =+=− :†2ㄸ P畴 y in瑯 =+ S漠瑨einesn瑥牳ec琠a琠瑨e⁰潩n琠 12,9 quatingThis method can be used when both equations have a commoncoefficientIn this case, both equations have acoefficient of Make the subject of both equations: 315=− =+ Eq畡瑥 3ㄵ32ㄸ−=+ 卵b s瑩瑵瑥 y in瑯: =−=+ S漠瑨einesn瑥牳ec琠a琠瑨e⁰潩n琠 12,9 . HigherMathematicsStraight Lines Page CfE Edition hsn .uk.net ubstitutiThis method can be used when one equation has coefficient of i.e.just an with no multiplier) Substitute =− in瑯: 315 33153915224 12.yxxxxx x =+−=+ −=+ Substitute 12x in瑯: ㄲ3=−=− S漠瑨einesn瑥牳ec琠a琠瑨e⁰潩n琠 12,9 . EXAMPLE Find the point of intersection of the lines 2110−+= a湤 2㜰+−= El業楮ate 2110 270 −+=+−= 2:†5ㄵ0×++==− P畴 =− in瑯 6ㄱ0−−+= S漠瑨e⁰潩n琠潦n瑥牳ec瑩潮s 3,5 Triangle PQR has vertices P8,3 , Q1,6 a湤 R2,3 (a)⁆ind⁴heq畡瑩潮 潦汴i瑵de⁑S. ⡢⤠Fi湤⁴hequatio渠ofedia渠剔. (c)⁈enceind⁴he潯牤ina瑥s映M HigherMathematicsStraight Lines Page CfE Edition hsn .uk.net (a)Find the gradient of PR −−=== S漠瑨e牡dien琠潦⁑Ss =− ,inc攠th攠 偒QS×=− Fi湤 t桥eq畡瑩潮映usi湧 Q1,6 a湤 =− 6116150.−=−+−=−−+−= Findthe coordinates of T, the midpoint of P 8136T,,. Find⁴he 杲慤ient吠usin朠 R2,3 a湤 7922T, : −−==== Find⁴he eq畡瑩潮映R吠usin朠 R2,3 a湤 m 35235105130.+=−+=−−−= ⡣)Now solve the equations simultaneouslyto find MEliminate 50 5130 +−=−−= :†6ㄸ0+−= P畴 x in瑯 3㔰+−= S漠瑨e⁰潩n琠潦n瑥牳ec瑩潮s M3,2 . Note This is the standardmethod for finding the equation of an altitude. Note Any of the three techniques could have been used here. Note This is the standard method for finding the equation of a median. HigherMathematicsStraight Lines Page CfE Edition hsn .uk.net Concurrency Any number of lines are said to be concurrentif there is a point through which they all pass. So in the previous section, by finding a point of intersection of two lines, we showed that the two lines were concurrent.For three lines to be concurrent, they must all pass through a single point.The three lines are concurrentThe three lines arenot concurrentA surprising fact is thatthe following lines in a triangle are concurrent. The three mediansof a triangleare concurrent. The three altitudesof a triangleare concurrent.The three perpendicular bisectorsin a triangle are concurrent.The threeangle bisectorsof a triangle are concurrent.