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Higher Unit 1.1 Straight Line

Higher Unit 1.1 Straight Line. Drawing a sketch is always worth the time and effort involved. Distance Formula Length of a straight line. y. x. O. AB = √ ( x 2 – x 1 ) 2 + ( y 2 – y 1 ) 2. Pythagoras’ Theorem. B ( x 2, y 2 ). AB 2 = AC 2 + CB 2. y 2 – y 1.

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Higher Unit 1.1 Straight Line

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  1. Higher Unit 1.1 Straight Line Drawing a sketch is always worth the time and effort involved

  2. Distance FormulaLength of a straight line y x O AB = √(x2 – x1)2 + (y2 – y1) 2 Pythagoras’ Theorem B(x2, y2) AB 2 = AC 2 + CB 2 y2 – y1 A(x1,y1) x2 – x1 C AB 2 = (x2 – x1)2 + (y2 – y1) 2

  3. Finding Mid-Point of a line y x O B(x2, y2) The mid-point of AB ● M A(x1,y1)

  4. What You Should Know About GRADIENTS

  5. All graphs are read from left to right m> 0 m is +ve, when line slopes up m is –ve, when line slopes down m< 0

  6. All graphs are read from left to right m is 0, when line is horizontal m= 0 Equation of line is y = a constant m is undefined or infinite, when line is vertical Equation of line is x = a constant m= ∞

  7. More Gradient facts Parallel lines have equal gradients m1 = m2 The gradient = the tangent of the angle the line makes with the positive x - axis θ m = tan θ

  8. 30º 2 45º 3 2 1 60º 45º 1 1 x 0º 30º 45º 60º 90º sin xº 1 0 cos xº 1 0 tan xº 0 Exact Values 1

  9. All Sinners Take Care CAST 90° 90° to 180° sin positive 0° to 90° all positive S A 180 - 180° 0°,360° 180 + 360 - C T 180° to 270° tan positive 270° to 360° cos positive 270°

  10. Find the size of the angle between the positive x-axis and the lines with gradient: 60o 150o x x m = tan θ 90o  A S 0o 180o 360o  T C 270o Angle must be between 90 and 180

  11. Collinearity C y ● B ● A ● x O Points are said to be collinear if they lie on the same straight line To prove: show mAB = mBC Then state A, B and C are collinear since AB and BC are parallelthrough the common point B

  12. Collinearity Example Prove that the points A(– 2 , – 3), B(– 1 , 0) and C(1 , 6) Show mAB = mBC mAB = mBC  AB is parallel to BC So A, B and C are collinear since AB and BC are parallelthrough the common point B

  13. Gradient of perpendicular lines y x Draw a line from O to (3 , 2) Now rotate the line 90o anticlockwise Work out the gradient of each line What do you notice about the gradients? m1× m2 = – 1

  14. Gradient of perpendicular lines y x O If two lines with gradients m1 and m2 are perpendicular then m1×m2 = –1 Conversely: If m1×m2 = –1 then the two lines with gradients m1 and m2 are perpendicular m1 m2 The converse is used to prove lines perpendicular

  15. Gradient of perpendicular lines If two lines with gradients m1 and m2 are perpendicular then m1×m2 = -1 You need to be able to work out NEGATIVE INVERSES m1m2 m1m2

  16. Perpendicular and Parallel Lines Exercise 3 Page 7 Numbers 1, 2, 3, 6, 7, 9, 10 and 11

  17. Straight line Facts y =mx +c y- intercept

  18. ax + by + c = 0 ax + by + c = 0 is an alternative version of the straight line equation. When given the equation in this form it is often useful to rearrange it into the formy = mx + c, especially if you need to know the gradient of the line. Step 1: isolate y term on left hand side. by = – ax – c Step 2: divide by y term coefficient

  19. Points on a line A point will only lie on a line if its coordinates satisfy the equation of the line. In other words if substituting the x and y coordinates makes the equation work out. Given the line with equation 3x – 2y = 4 (2 , 1) lies on the line since 3×2 – 2×1 = 4 As does (0 , – 2) since 3×0 – 2×(– 2) = 4 But (3 , 2) does NOT since 3×3 – 2×2 = 5

  20. Points on a line Given the line with equation2x – y = 9, find the value of k, so that (2k , k) lies on the line. (2k , k) lies on the line  2×2k – k = 9 3k = 9 k = 3

  21. Angle between lines Find the angle between the lines with equations 6x – 2y + 6 = 0 and 2x + 5y – 10 = 0 Remember the connection between angles and lines is m = tan θ, where θ is the angle between the line and the positive x-axis However, when m is negative θ becomes the angle between the line and the negative x-axis. Positive direction x-axis Negative direction x-axis

  22. Angle between lines Find the angle between the lines with equations 6x – 2y + 6 = 0 and 2x + 5y – 10 = 0 m = 0∙75 m = – 0∙75 tanθ = 0∙75 tanθ = – 0∙75 tan-1 0∙75 = 36∙9o tan-1 0∙75 = 36∙9o Positive direction x-axis Negative direction x-axis 36∙9o 36∙9o

  23. Angle between lines y x Find the angle between the lines with equations 6x – 2y + 6 = 0 and 2x + 5y – 10 = 0 5y = – 2x + 10 – 2y = – 6x – 6 y = – 2/5x + 2 y = 3x + 2 tanθ = – 2/5 tanθ = 3 tan-1 0∙4 = 21∙8o tan-1 3 = 71∙6o Angle between lines = 180o – (21∙8o +71∙6o) = 86∙6o 21∙8o 71∙6o

  24. Another Equation of the Straight Line y x O The equation of the line through the point (a , b) with gradient m is (a , b) ● m y – b = m(x – a)

  25. Using y – b = m(x – a) All questions require you to find the gradient, m and the coordinates (a , b) The gradient may be given or found from two points on the line found from angle line makes with x-axis found by changing given line equation into y = mx + c then parallel  m1 = m2 perpendicular  m1×m2= – 1

  26. Using y – b = m(x – a) All questions require you to find the gradient, m and the coordinates (a , b) Find the equation of the line through the points (– 5 , 3) and ( 7 , 7) Use one of the given points as (a , b) × 3

  27. Using y – b = m(x – a) All questions require you to find the gradient, m and the coordinates (a , b) Find the equation of the line parallel to 2x + 3y = 12 passing through the point ( 1 , 5) Rearrange 2x + 3y = 12 to find m. × 3 Parallel m1 = m2

  28. Using y – b = m(x – a) All questions require you to find the gradient, m and the coordinates (a , b) Find the equation of the line perpendicular to 2x + 3y = 12 passing through the point ( 1 , 5) Rearrange 2x + 3y = 12 to find m. × 2 Perpendicular m1 × m2 = – 1

  29. Past Paper Multiple Choice Questions

  30. The line with equation y = ax + 4 is perpendicular to the line with equation 3x + y + 1 = 0. What is the value of a? A. – 3 B. – 1/3 C. 1/3 D. 3 Perpendicular m1 × m2 = – 1

  31. The line 2y = 3x + 6 meets the y-axis at C. The gradient of the line joining C to A (4,-3) is: A. 9/4 B. 2/3 C. -2/3 D. -3/2 E. -9/4 2y = 3x + 6 cuts y-axis at x = 0  2y = 6  y = 3 C(0 , 3)

  32. The line 3x - 2y – 6 = 0 cuts the x-axis at P and the y-axis at R. The mid-point of PR is: A.(3/2,1) B.(3/2,-1) C.(1,3/2) D.(1,-3/2) 3x - 2y – 6 = 0cuts x-axis at y = 0  3x – 6 = 0 P(2 , 0)  x = 2 3x - 2y – 6 = 0cuts y-axis at x = 0  – 2y – 6 = 0 R(0 , – 3)  y = – 3

  33. Given that the line joining the points (2 , 3) and (8 , k) is perpendicular to the line 2y – 3x + 5 = 0, then the value of k is: A. –1 B. –2 C. –7 D. 1 2y – 3x + 5 = 0,  2y = 3x – 5  y = 3/2 x – 5/2  m = 3/2 Perpendicular m1 × m2 = – 1  m = – 2/3

  34. For varying values of p the equation y – 1 = p(x – 1) is the equation of a straight line. All such lines • A. have the same gradient B. cut the x axis at the same point. C. pass through a fixed point not on the axis. D. cut the y axis at the same point.  From y – b = m(x – a) y – 1 = p(x – 1) Varies with p Line passes through (1 , 1) m = p Cuts x-axisat y = 0 Cuts y-axisat x = 0 Varies with p Varies with p

  35. A straight line passes through the points P(−5, −2) and Q(−2, −1). What is the equation of the straight line which passes through P and is perpendicular to PQ ? Perpendicular m1 × m2 = – 1

  36. Median ● ● ● A median is a line joining a vertex to the midpoint of the opposite side The medians are CONCURRENT, they all pass through the same point

  37. Altitude An altitude is a line from a vertex to meet the opposite side at right angles The altitudes are CONCURRENT, they all pass through the same point

  38. Perpendicular Bisector ● A perpendicular bisector is a line which cuts the given line in half at right angles

  39. Intersecting Lines The point of intersection of two lines is found by solving simultaneous equations Three or more lines are said to be concurrent if they all pass through a common point. To prove concurrency solve one pair of simultaneous equations and show that the point of intersection lies on the other line.

  40. Show that the lines x + y + 3 = 0, 2x – y = 0 and 3x – 2y = 1 are concurrent. Add Put x = – 1 and y = – 2 in 3x – 2y Put x = – 1 in x + y = – 3 Lines are concurrent since they all pass through (-1 , -2)

  41. Questions from SQA Past Papers

  42. Find the equation of the line which passes through the point (-1, 3) and is perpendicular to the line with equation Use y = mx + c to find gradient of given line: Use m1× m2 = -1 to find gradient of perpendicular: Use y – b = m(x – a) to find equation:

  43. Find the equation of the straight line which is parallel to the line with equation and which passes through the point (2, –1). Usey = mx + cto find gradient of given line: Parallel lines have equal gradients

  44. B(3√3, 2) ● 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. y Find gradient of the line: ao Use m = tan θ x O A(0, -1) a = 30 Use table of exact values

  45. 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 (5, 5) (-3, -1) ● ● y To find equation of AB (a , b) O Usingy – b = m(x – a)

  46. 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 (5, 5) (-3, -1) ● ● y To find perpendicular bisector (a , b) O Usingy – b = m(x – a)

  47. The line AB makes an angle of π/3 radians with the y-axis, as shown in the diagram. Find the exact value of the gradient of AB. y x O The angle between AB and x-axis is equal to the angle between the line and a parallel line B π/3 π/6 From m = tan θ A m = tan π/6 m = 1/√3

  48. y A(4 , 3) B(6 , 1) O ● x C(-2 , -3) 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 A. Find mid-point of BC: M(2 , -1) Find gradient of median AM M Equation of AM

  49. 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. y P(-4 , 5) R(4 , 1) O x Q(-2 , -2) Find gradient of QR: Find gradient of PS (perpendicular to QR) S Equation of altitude PS

  50. The lines y = 2x + 4 and x + y = 13 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, and the acute angle between the lines. y y = 2x+4 x + y = 13 bo ao x O Gradient of y = 2x + 4 is 2 tan ao = 2  a = tan-1 2 = 63o 135o Gradient of x + y = 13 is -1 tan bo = -1  a = tan-1 -1 = 135o 45o 63o Acute angle = 180 – (63 + 45) Acute angle = 72o

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