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www.mathsrevision.com Higher Outcome 1 www.mathsrevision.com Higher Unit 1 Distance Formula The Midpoint Formula Gradients Collinearity Gradients of Perpendicular Lines The Equation of a Straight Line Median, Altitude & Perpendicular Bisector Exam Type Questions

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www.mathsrevision.com Higher Outcome 1 www.mathsrevision.com Starter Questions 1.Calculate the length of the length AC. (a)( 1, 2) and ( 5, 10)(b)( -4, -10) and ( -2,-6) 2. Calculate the coordinates that are halfway between. 6 m 8 m A CB

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www.mathsrevision.com Higher Outcome 1 Distance Formula Length of a straight line A(x 1,y 1 ) B(x 2, y 2 ) x 2 – x 1 y 2 – y 1 C x y O This is just Pythagoras’ Theorem

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www.mathsrevision.com Higher Outcome 1 Distance Formula The length (distance ) of ANY line can be given by the formula : Just Pythagoras Theorem in disguise

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www.mathsrevision.com Higher Outcome 1 Finding Mid-Point of a line A(x 1,y 1 ) B(x 2, y 2 ) x y O x 1 x 2 M y 1 y 2 The mid-point between 2 points is given by Simply add both x coordinates together and divide by 2. Then do the same with the y coordinates.

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www.mathsrevision.com Higher Outcome 1 Straight line Facts Y – axis Intercept y = mx + c Another version of the straight line general formula is: ax + by + c = 0

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www.mathsrevision.com Higher Outcome 1 19-May-15www.mathsrevision.com 7 Gradient Facts m < 0 m > 0 m = 0 x = a y = c Sloping left to right up has +ve gradient Sloping left to right down has -ve gradient Horizontal line has zero gradient. Vertical line has undefined gradient.

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www.mathsrevision.com Higher Outcome 1 19-May-15www.mathsrevision.com 8 Gradient Facts m = tan θ m > 0 Lines with the same gradient means lines are Parallel The gradient of a line is ALWAYS equal to the tangent of the angle made with the line and the positive x-axis θ

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www.mathsrevision.com Higher Outcome 1 Collinearity A C x y O x1x1 x2x2 B Points 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. D E F

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www.mathsrevision.com Higher Outcome 1 Gradient of perpendicular lines If 2 lines with gradients m 1 and m 2 are perpendicular then m 1 × m 2 = -1 When rotated through 90º about the origin A (a, b) → B (-b, a) -a B(-b,a) -b A(a,b) a O y x Conversely: If m 1 × m 2 = -1 then the two lines with gradients m 1 and m 2 are perpendicular. -b

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www.mathsrevision.com Higher Outcome 1 = The Equation of the Straight Line The Equation of the Straight Line y – b = m (x - a) The equation of any line can be found if we know the gradient and one point on the line. O y x x - a P (x, y) m A (a, b) y - b x – a m = y - b (x – a) m Gradient, m Point (a, b) y – b = m ( x – a ) Point on the line ( a, b ) ax y b

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www.mathsrevision.com Higher Outcome 1 A BC D 19-May-15www.mathsrevision.com 12 A Median means a line from vertex to midpoint of the base. Altitude means a perpendicular line from a vertex to the base. B D C

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www.mathsrevision.com Higher Outcome 1 19-May-15www.mathsrevision.com 13 A B D C Perpendicular bisector - a line that cuts another line into two equal parts at an angle of 90 o

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www.mathsrevision.com Higher Outcome 1 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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www.mathsrevision.com Higher Outcome 1 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) Typical Exam Questions

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www.mathsrevision.com Higher Outcome 1 Find gradient of the line: Use table of exact values Use 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. Exam Type Questions

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www.mathsrevision.com Higher Outcome 1 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 mid-point of AB Find equation 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 Exam Type Questions

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www.mathsrevision.com Higher Outcome 1 The line AB makes an angle of radians 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 Use (x and y axes are perpendicular.) Typical Exam Questions

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www.mathsrevision.com Higher Outcome 1 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 AM Find gradient of median AM Typical Exam Questions

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www.mathsrevision.com Higher Outcome 1 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 equation of altitude PS Find gradient of PS (perpendicular to QR) Typical Exam Questions

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www.mathsrevision.com Higher Outcome 1 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 supplement of b Find gradient of Find a° Find b° angle between two linesUse angle sum triangle = 180° 72° Typical Exam Questions 45 o 72 o 63 o 135 o

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www.mathsrevision.com Higher Outcome 1 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 p Find gradient of p (-2, 2) 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 p q

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www.mathsrevision.com Higher Outcome 1 Triangle ABC has vertices A(2, 2), B(12, 2) and C(8, 6). a) Write down the equation of l 1, the perpendicular bisector of AB b) Find the equation of l 2, the perpendicular bisector of AC. c) Find the point of intersection of lines l 1 and l 2. Mid-point AB Find mid-point AC (5, 4) Find gradient of AC Equation of perp. bisector ACGradient AC perp. Point of intersection (7, 1) Perpendicular bisector AB Exam Type Questions l 1 l 2

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www.mathsrevision.com Higher Outcome 1 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 Equation of median CM using y – b = m(x – a) Gradient of perpendicular AD Gradient BC Equation of AD using y – b = m(x – a) Gradient CM (median) Solve simultaneously for point of intersection (6, -4) Exam Type Questions

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www.mathsrevision.com Higher Outcome 1 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 Product of gradients Gradient of median ADMid-point BCEquation AD Gradient BC Solve simultaneously for M, point of intersection Hence AB is perpendicular to BC, so B = 90° Gradient of median BE Mid-point AC Equation AD Exam Type Questions M

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