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Higher Unit 1

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. Starter Questions. 1. Calculate the length of the length AC. A. 6 m. B. C.

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Higher Unit 1

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  1. 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 www.mathsrevision.com

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

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

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

  5. y x O Finding Mid-Point of a line The mid-point between 2 points is given by 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

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

  7. 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

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

  9. F y x O Collinearity E D Points are said to be collinear if they lie on the same straight. C The coordinates A,B C arecollinear since they lie on the same straight line. D,E,F are not collinear they do not lie on the same straight line. B A x1 x2

  10. Gradient of perpendicular lines 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 Conversely: If m1× m2 = -1 then the two lines with gradients m1 and m2 are perpendicular.

  11. y P (x, y) m = m y - b A (a, b) x - a O x 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 - b y = m (x – a) y - b b x – a Gradient, m a x Point (a, b) y – b = m ( x – a ) Point on the line ( a, b )

  12. Terminology A Median means a line from vertex to midpoint of the base. B C D A D Altitude means a perpendicular line from a vertex to the base. B C www.mathsrevision.com

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

  14. 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:

  15. 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)

  16. Use 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 table of exact values

  17. 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 equation of AB Find gradient of the AB: Gradient of AB (perp): Find mid-point of AB Use y – b = m(x – a) and point ( 1, 2) to obtain line of perpendicular bisector of AB we get

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

  19. 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

  20. 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

  21. Find gradient of Find gradient of Typical Exam Questions 72o 135o 63o 45o 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 a° Find b° Find supplement of b 72° Use angle sum triangle = 180° angle between two lines

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

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

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

  25. 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 Gradient of median BE Equation AD Mid-point AC Solve simultaneously for M, point of intersection

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