Www. APP 1.1 Straight Line Applications 1.1 Distance Formula The Midpoint Formula Prior Knowledge...

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Transcript of Www. APP 1.1 Straight Line Applications 1.1 Distance Formula The Midpoint Formula Prior Knowledge...

  • Where is Straight Line TheoryUsed in HigherRecurrence RelationsCircleDifferentiationLogs & ExponentialsVector

  • Straight Liney = mx + cm = tan Possible values for gradientFor Perpendicular lines the following is true.m1.m2 = -1Parallel lines have same gradient

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    Straight Line FactsY axis Intercepty = mx + cAnother version of the straight line formula isax + by + c = 0

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    QuestionsFind the gradient and y intercept for equations.(a) 4x + 2y + 10 = 0(b) 3x - 5y + 1 = 0(c) 4 - x 3y = 0m = -2 c = -5m = 3/5 c = 1/5m = - 1/3 c = 4/3Demo

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    =The Equation of the Straight Liney b = m (x - a)

    The equation of any line can be found if we know the gradient and one point on the line.y - bx ay - b(x a)mGradient, mPoint (a, b)y b = m ( x a )Point on the line ( a, b )axybDemo

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    *www.mathsrevision.com*m < 0m > 0m = 0x = ay = cSloping left to right up has +ve gradientSloping left to right down has -ve gradientHorizontal line has zero gradient.Vertical line has undefined gradient.

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    *www.mathsrevision.com*m > 0Lines with the same gradient means lines are Parallel

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    Straight Line Theory

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    www.mathsrevision.comAPP 1.1

    Straight Line Theory

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    Straight Line Theory

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    Straight Line Theory

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    Find the equation of the straight line which is parallel to the line with equation and which passes through the point (2, 1) . Find gradient of given line:Knowledge: Gradient of parallel lines are the same. Therefore for line we want to find has gradientFind equation: Using y b =m(x - a)Typical Exam Questions

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    HHM Book PracticeHHMEx1AHHMEx1B Q4 , Q5HHMEx1EHHMEx1G Q1 , Q2

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    Distance FormulaLength of a straight lineA(x1,y1)B(x2,y2)x2 x1y2 y1CThis is just Pythagoras Theorem

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    Distance FormulaThe length (distance ) of ANY line can be given by the formula :Just Pythagoras Theorem in disguiseDemo

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    22682226Isosceles Triangle !Demo

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    HHM Book PracticeHHMEx12B

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    Mid-Point of a lineA(x1,y1)B(x2,y2)x1x2My1y2The mid-point (Median) between 2 points is given bySimply add both x coordinates together and divide by 2.Then do the same with the y coordinates.Online Demo

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    HHM Book PracticeUse Show me boards

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    www.mathsrevision.comHigherOutcome 1

    *www.mathsrevision.com*m = tan The gradient of a line is ALWAYSequal to the tangent of the angle made with the line and the positive x-axistan =0o < 180oDemo

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    m = tan m = tan 60o = 3

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    m = tan y = -2x m = tan = tan-1 (-2) = 180 63.4 = 116.6o

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    Find gradient of the line:Use table of exact valuesFind the size of the angle a that the linejoining the points A(0, -1) and B(33, 2) makes with the positive direction of the x-axis. Exam Type Questions

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    The line AB makes an angle of 60o with the y-axis, as shown in the diagram. Find the exact value of the gradient of AB. Find angle between AB and x-axis:Use table of exact values(x and y axes are perpendicular.)Typical Exam Questions60o

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    The lines and make angles of a and b with the positive direction of the x-axis, as shown in the diagram.a)Find the values of a and bb)Hence find the acute angle between the two given lines. Find supplement of bFind aFind bangle between two linesUse angle sum triangle = 18072Typical Exam Questions45o72o63o135o

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    HHM Book PracticeHHMEx1A Q9 , Q10HHMEx1B Q6 , Q7 , Q10HHMEx1D Q7

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    Gradient of perpendicular lines

    If 2 lines with gradients m1 and m2 are perpendicular then m1 m2 = -1 When rotated through 90 about the origin A (a, b) B (-b, a)-aB(-b,a)-bA(a,b)aOyxConversely:If m1 m2 = -1 then the two lines with gradients m1 and m2 are perpendicular.-bInvestigationDemo

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    HHM Book PracticeHHMEx1DHHMEx1FHHMEx1G Q3 Q5

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    www.mathsrevision.comHigherOutcome 1

    **www.mathsrevision.comAMedian means a line from a vertex tothe midpoint of the base.Altitude means a perpendicular linefrom a vertex to the base.BDCDemoDemo

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    **www.mathsrevision.comABDCPerpendicular bisector - a line that cuts another lineinto two equal parts at an angle of 90oDemo

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  • Finding the Equation of an AltitudeABTo find the equation of an altitude:Find the gradient of the side it is perpendicular to ( ).mABCTo find the gradient of the altitude, flip the gradientof AB and change from positive to negative:maltitude = mAB1Substitute the gradientand the point C intoy b = m ( x a ) ImportantWrite final equation in the formA x + B y + C = 0with A x positive.Common Straight Strategies for Exam Questions

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  • Finding the Equation of a MedianPQTo find the equation of a median:Find the midpoint of the side itbisects, i.e.OCalculate the gradient of the median OM.Substitute the gradient and eitherpoint on the line (O or M) intoy b = m ( x a ) ImportantWrite answer in the formA x + B y + C = 0with A x positive.==M( )M = 2y2 y1 2x2 x1 ,++Common Straight Strategies for Exam Questions

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  • Any number of lines are said to be concurrent if there is a point through which they all pass.For three lines to be concurrent, they must all pass through a single point.Demo

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    HHM Book PracticeHHMEx1IHHMEx1KHHMEx1MHHMEx1N

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    CollinearityPoints are said to be collinear if they lie on the same straight.The coordinates A,B C are collinear since they lie on the same straight line.D,E,F are not collinear they do not lie on the same straight line.

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    Straight Line TheorySince mPQ = mQR and they have a point in common QPQR are collinear.DPQ=2DPQ=22RatioDPQ:DPQ1:2

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    HHM Book PracticeHHMEx1B Q1 Q3HHMEx1O

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    Find the equation of the line which passes through the point (-1, 3) and is perpendicular to the line with equation Find gradient of given line:Find gradient of perpendicular:Find equation:Typical Exam Questions

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    A and B are the points (3, 1) and (5, 5).Find the equation ofa)the line AB.b)the perpendicular bisector of AB Find gradient of the AB:Find mid-point of ABFind equation of ABGradient of AB (perp):Use y b = m(x a) and point ( 1, 2) to obtain line of perpendicular bisector of AB we get Exam Type Questions

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    A triangle ABC has vertices A(4, 3), B(6, 1)and C(2, 3) as shown in the diagram. Find the equation of AM, the median from B to C Find mid-point of BC:Find equation of median AMFind gradient of median AMTypical Exam Questions

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    P(4, 5), Q(2, 2) and R(4, 1) are the verticesof triangle PQR as shown in the diagram. Find the equation of PS, the altitude from P. Find gradient of QR:Find equation of altitude PSFind gradient of PS (perpendicular to QR)Typical Exam Questions

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    Triangle ABC has vertices A(1, 6), B(3, 2) and C(5, 2)Find: a) the equation of the line p, the median from C of triangle ABC. b) the equation of the line q, the perpendicular bisector of BC. c) the co-ordinates of the point of intersection of the lines p and q. Find mid-point of AB Find equation of pFind gradient of p(-2, 2)Find mid-point of BC (1, 0)Find gradient of BCFind gradient of qFind equation of qSolve p and q simultaneously for intersection(0, 2)Exam Type Questionspq

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    Triangle ABC has vertices A(2, 2), B(12, 2) and C(8, 6). a) Write down the equation of l1, the perpendicular bisector of AB b) Find the equation of l2, the perpendicular bisector of AC. c) Find the point of intersection of lines l1 and l2. Mid-point ABFind mid-point AC (5, 4)Find gradient of ACEquation of perp. bisector ACGradient AC perp.Point of intersection(7, 1)Perpendicular bisector ABExam Type Questionsl1l2

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    A triangle ABC has vertices A(4, 1), B(12,3) and C(7, 7). a) Find the equation of the median CM. b) Find the equation of the altitude AD. c) Find the co-ordinates of the point of intersection of CM and AD Mid-point ABEquation of median CM using y b = m(x a) Gradient of perpendicular ADGradient BCEquation o