Straight Line Applications 1.1 Prior Knowledge Distance Formula The Midpoint Formula m = tan θ Gradients of Perpendicular Lines Median, Altitude & Perpendicular Bisector Collinearity Mindmap Exam Type Questions www.mathsrevision.com

Circle Vector Logs & Exponentials Differentiation Recurrence Relations Vector Where is Straight Line Theory Used in Higher Logs & Exponentials Differentiation

Parallel lines have same gradient Distance between 2 points m < 0 m = undefined Terminology Median – midpoint Bisector – midpoint Perpendicular – Right Angled Altitude – right angled m1.m2 = -1 m > 0 m = 0 Possible values for gradient Form for finding line equation y – b = m(x - a) (a,b) = point on line Straight Line y = mx + c Parallel lines have same gradient m = gradient c = y intercept (0,c) For Perpendicular lines the following is true. m1.m2 = -1 θ m = tan θ

Another version of the straight line formula is Straight Line Facts y = mx + c Y – axis Intercept Another version of the straight line formula is ax + by + c = 0

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

The Equation of the Straight Line y – b = m (x - a) Demo The equation of any line can be found if we know the gradient and one point on the line. O y x P (x, y) m A (a, b) y - b y m m = = y - b (x – a) y - b b x - a x – a Gradient, m a x Point (a, b) y – b = m ( x – a ) Point on the line ( a, b )

Gradient Facts Sloping left to right up has +ve gradient m > 0 Sloping left to right down has -ve gradient m < 0 Horizontal line has zero gradient. m = 0 y = c Vertical line has undefined gradient. x = a www.mathsrevision.com 24-Apr-17

Gradient Facts Lines with the same gradient means lines are Parallel www.mathsrevision.com 24-Apr-17

Straight Line Theory

Straight Line Theory

Straight Line Theory

Straight Line Theory

Typical Exam Questions 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 gradient Find equation: Using y – b =m(x - a)

HHM Book Practice HHM Ex1A HHM Ex1B Q4 , Q5 HHM Ex1E HHM Ex1G Q1 , Q2

Distance Formula Length of a straight line x y O B(x2,y2) This is just Pythagoras’ Theorem y2 – y1 A(x1,y1) C x2 – x1

Distance Formula The length (distance ) of ANY line can be given by the formula : Just Pythagoras Theorem in disguise Demo

2√26 2√26 Isosceles Triangle ! 8√2 Demo

HHM Book Practice HHM Ex12B

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

HHM Book Practice Use Show me boards

Straight Line Facts m = tan θ m = Demo tan θ = The gradient of a line is ALWAYS equal to the tangent of the angle made with the line and the positive x-axis θ m = tan θ 0o ≤ θ < 180o θ O A H m = V H O A = O A tan θ = Demo www.mathsrevision.com 24-Apr-17

m = tan θ 60o m = tan 60o = √3

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

Exam Type Questions Find the size of the angle a° that the line joining the points A(0, -1) and B(33, 2) makes with the positive direction of the x-axis. Find gradient of the line: Use Use table of exact values

Typical Exam Questions 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. 60o Find angle between AB and x-axis: (x and y axes are perpendicular.) Use Use table of exact values

Typical Exam Questions 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 b b) Hence find the acute angle between the two given lines. Find gradient of Find a° Find gradient of Find b° Find supplement of b 72° Use angle sum triangle = 180° angle between two lines

HHM Book Practice HHM Ex1A Q9 , Q10 HHM Ex1B Q6 , Q7 , Q10 HHM Ex1D Q7

Gradient of perpendicular lines Investigation If 2 lines with gradients m1 and m2 are perpendicular then m1 × m2 = -1 When rotated through 90º about the origin A (a, b) → B (-b, a) y B(-b,a) -a A(a,b) -b O a x -b Demo Conversely: If m1 × m2 = -1 then the two lines with gradients m1 and m2 are perpendicular.

HHM Book Practice HHM Ex1D HHM Ex1F HHM Ex1G Q3 – Q5

Terminology A B C D A B C D Demo Demo Median means a line from a vertex to the midpoint of the base. B C D A B C D Demo Altitude means a perpendicular line from a vertex to the base. Demo www.mathsrevision.com 24-Apr-17

Terminology Perpendicular bisector - a line that cuts another line into two equal parts at an angle of 90o A B C D Demo www.mathsrevision.com 24-Apr-17

Common Straight Strategies for Exam Questions Finding the Equation of an Altitude To find the equation of an altitude: B • Find the gradient of the side it is perpendicular to ( ). mAB C • To find the gradient of the altitude, flip the gradient of AB and change from positive to negative: maltitude = –1 mAB • Substitute the gradient and the point C into Important Write final equation in the form y – b = m ( x – a ) A x + B y + C = 0 with A x positive.

Common Straight Strategies for Exam Questions Finding the Equation of a Median M = To find the equation of a median: = P • Find the midpoint of the side it bisects, i.e. x2 x1 + y2 y1 + O ( ) , M = 2 2 • Calculate the gradient of the median OM. Important • Substitute the gradient and either point on the line (O or M) into Write answer in the form A x + B y + C = 0 y – b = m ( x – a ) with A x positive.

Demo 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

HHM Book Practice HHM Ex1I HHM Ex1K HHM Ex1M HHM Ex1N

x Collinearity y C A Points are said to be collinear if they lie on the same straight. A C x y O x1 x2 B 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. D E F

they have a point in common Q Ratio DPQ:DPQ 1:2 Straight Line Theory DPQ=2√2 DPQ=√2 Since mPQ = mQR and they have a point in common Q PQR are collinear.

HHM Book Practice HHM Ex1B Q1 – Q3 HHM Ex1O

Typical Exam Questions 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:

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

Typical Exam Questions 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 gradient of median AM Find equation of median AM

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

Solve p and q simultaneously for intersection Exam Type Questions p 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. q Find mid-point of AB (-2, 2) Find gradient of p Find equation of p Find mid-point of BC (1, 0) Find gradient of BC Find gradient of q Find equation of q Solve p and q simultaneously for intersection (0, 2)

Exam Type Questions l2 l1 (5, 4) (7, 1) 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 AB Perpendicular bisector AB Find mid-point AC (5, 4) Find gradient of AC Gradient AC perp. Equation of perp. bisector AC Point of intersection (7, 1)

Gradient of perpendicular AD Exam Type Questions 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 AB Gradient CM (median) Equation of median CM using y – b = m(x – a) Gradient BC Gradient of perpendicular AD Equation of AD using y – b = m(x – a) Solve simultaneously for point of intersection (6, -4)

Hence AB is perpendicular to BC, so B = 90° Exam Type Questions M A triangle ABC has vertices A(–3, –3), B(–1, 1) and C(7,–3). a) Show that the triangle ABC is right angled at B. b) The medians AD and BE intersect at M. i) Find the equations of AD and BE. ii) Find find the co-ordinates of M. Gradient AB Gradient BC Product of gradients Hence AB is perpendicular to BC, so B = 90° Mid-point BC Gradient of median AD Equation AD Mid-point AC Gradient of median BE Equation AD Solve simultaneously for M, point of intersection

Parallel lines have same gradient Distance between 2 points m < 0 m = undefined Terminology Median – midpoint Bisector – midpoint Perpendicular – Right Angled Altitude – right angled m1.m2 = -1 m > 0 m = 0 Possible values for gradient Form for finding line equation y – b = m(x - a) (a,b) = point on line Straight Line y = mx + c Parallel lines have same gradient m = gradient c = y intercept (0,c) For Perpendicular lines the following is true. m1.m2 = -1 θ m = tan θ

Are you on Target ! Update you log book Make sure you complete and correct ALL of the Straight Line questions in the past paper booklet.