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Q3 Calculate the size of the angle between y = ½ x + 2 & y = 3x - 1

Q3 Calculate the size of the angle between y = ½ x + 2 & y = 3x - 1. Answer: (Let angles be α and β for each gradient) If m = tan θ :- tan α = ½ & tan β = 3 α = tan -1 ( ½) β = tan -1 ( 3) α = 71.6 o β = 26.6 o Thus angle between 2 lines is

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Q3 Calculate the size of the angle between y = ½ x + 2 & y = 3x - 1

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  1. Q3 Calculate the size of the angle between y = ½x + 2 & y = 3x - 1 Answer: (Let angles beα and β for each gradient) If m = tanθ:- tan α = ½ & tanβ = 3 α = tan-1(½) β = tan-1(3) α = 71.6o β = 26.6o Thus angle between 2 lines is α – β =θ = 71.6 – 26.6 = 45o

  2. Q4 Find the equation of the line passing thro’ (1, 2) perpendicular to y – 2x = -5 y – 2x = -5 must change to y = mx + c y = 2x – 5  m = 2 As m1x m2 = -1 perp gradient is m = -½ Thus if passes thro (1, 2) and m = -½ (y – 2) = -½(x – 1) 2y – 4 = -x + 1 x + 2y – 5 = 0 [or an alternative equation]

  3. Q5. Where do y = 2x + 7 & y = -3x - 3 intersect? If y = …. & y = …  y = y 2x + 7 = -3x – 3 5x = - 10 x = -2 If x = -2 subst to find y:(either equation is fine) Eq1 y = 2x + 7 or Eq2 y = -3x – 3 = -4 + 7 = 6 – 3 = 3 = 3 Thus point of intersection is at (-2, 3)

  4. Q6 Where does 3y – 2x – 12 = 0 cut the x & y-axis? Cuts x-axis when y = 0:3y – 2x – 12 = 0 0 – 2x – 12 = 0 Thus cuts – 2x = 12 x-axis at x = -6 (-6 , 0) Cuts y-axis when x = 0 3y – 2x – 12 = 0 3y – 0 – 12 = 0 Thus cuts 3y = 12 y-axis at y = 4 (0, 4)

  5. Q7 (a) Find altitude from C to ABGiven A(3 , 1); B(11 , 5) & C(2 , 8)? C mAB = 5 – 1 = 4 = 1 11 – 3 8 2 If mAB = ½  mc = -2 Altitude from C(2, 8) with mc = -2:- y – 8 = -2(x – 2) y – 8 = -2x + 4 2x + y – 12 = 0 A B

  6. Q7 (b) Find altitude from A to BCGiven A(3 , 1); B(11 , 5) & C(2 , 8)? A mBC = 8 – 5 = 3 = -1 2 – 11 -9 3 If mBC = -1 mA = 3 3 Altitude from A(3, 1) with mc = 3:- y – 1 = 3(x – 3) y – 1 = 3x - 9 y = 3x - 8 C B

  7. Q7 (c) Find the coordinates of T, the point of intersection of the 2 altitudes. A Altitude from A  y = 3x – 8 Altitude from C  2x + y – 12 = 0 Substituting Equation 1 into 2 gives: 2x + y – 12 = 0 2x + (3x – 8) – 12 = 0 5x – 20 = 0 5x = 20 x = 4 Using y = 3x – 8: y = 12 – 8 = 4  T is ( 4 , 4 ) C B T( 4 , 4 )

  8. Q8 (a) Find line perpendicular to y = ⅓x + 1 which passes thro P(4 , 10) ? From the above equation the gradient is m = ⅓  mperp = -3 Perp line thro P(4, 10) with mp = -3:- y – 10 = -3(x – 4) y – 10 = -3x + 12 3x + y – 22 = 0(3x + y – 21 = 0 if used (4 , 9))

  9. Q8 (b) Find the coordinates where both lines meet? y = ⅓x + 1 & 3x + y – 22 = 0 3x + y – 22 = 0 y = ⅓x + 1 3x + (⅓x + 1) - 22 = 0 y = ⅓(6.3) + 1 3x + ⅓x + 1 - 22 = 0 y = 2.1 + 1 3⅓x = 21 y = 3.1 10x = 21 3 Thus both lines meet at 10x = 63 (6.3 , 3.1) x = 6.3(If used 3x + y – 21 = 0 meet at (6 , 3))

  10. Q9 (a) Find altitude from C to ABGiven A(2 , -4); B(14 , 2) & C(10 , 10)? C(10,10) B(14 , 2) A(2, -4)

  11. Q9 (b) Find median thro A given A(2 , -4); B(14 , 2) & C(10 , 10)? C Midpoint of BC = (10+14 , 2+10) = (12, 6) 2 2 If mAD = 6 –(-4)= 10 = 1 12 – 2 10 Median from A with mAD = 1:- y – ( -4)= 1(x – 2) y + 4 = x - 2 y = x - 6 A B

  12. Q9 (c) Find the perpendicular bisector of AB? C Midpoint of AB, say E = (2+14 , -4+2) = (8 , -1) 2 2 If mAB = 2–(-4)= 6= 1 mperp = -2 14 – 2 12 2 Perpendicular Bisector from AB with mAD = -2:- y – ( -1)= -2(x – 8) y + 1 = -2x + 16 2x +y = 15 A B

  13. Q9 (d) Find the coordinates of W, the point of intersection of the 2 lines? C 2x + y = 15 -----1 y = x - 6 -----2 2x + (y) = 15 2x + (x – 6) = 15 3x = 21 x = 7 If y = x – 6 = 7 – 6 y = 1  W( 7 , 1) A B

  14. In Q10(a) depending on which 2 of the 3 equations you find will result in your working being laid out differently. So to make it easier I have called each possibility “Options 1, 2 and 3” for you to view your own particular choice.

  15. Q10(a) Find the coordinates of the Centroid P, where medians meet. If D(1 , 2) ; E(-1 , 1) & F(3 , -1). Median from D Midpoint of EF = (-1 + 3 , 1+(-1)) =(1 , 0 ) 2 2 If mD = 0 – 2 = -2 = ∞  vertical line 1 - 1 0 Median from D ( 1 , 2 ) with undefined gradient is therefore x =1 D E F

  16. Q10(a) Find the coordinates of the Centroid P, where medians meet. If D(1 , 2) ; E(-1 , 1) & F(3 , -1). Median from E Midpoint of DF, = (1 + 3 , 2+(-1)) = (2 , ½) 2 2 If mE = ½ – 1 = - ½ = -1 2 –(-1) 3 6 Median from E ( -1 , 1 ) with mE = -1/6:- y – 1 = -1/6(x – (-1)) 6y - 6 = -x - 1 x + 6y – 5 = 0 D E F

  17. Q10(a) Find the coordinates of the Centroid P, where medians meet. If D(1 , 2) ; E(-1 , 1) & F(3 , -1). Median from F Midpoint of DE, = (1 + (-1) , 2 + 1) = (0 , 1½) 2 2 If mF = -1 – 1½= -2½ = -5 3 - 0 3 6 Median from F ( 3 , -1 ) with mF = -5/6:- y – (- 1) = -5/6(x - 3) 6(y + 1) = -5x + 15 6y + 6 = -5x + 15 5x + 6y – 9 = 0 D E F

  18. Q10(a) Find the coordinates of the Centroid P (where medians meet) of triangle DEF? x = 1 -----1 x + 6y – 5 = 0-----2 5x + 6y – 9 = 0 -----3 Choose any 2 of 3 possible median equations to solve for the point P D E F

  19. Q10(a)OPTION 1Find the coordinates of the Centroid P (where medians meet) of triangle DEF? x = 1 -----1 Choose any x + 6y – 5 = 0-----2 of 2 median 5x + 6y – 9 = 0 -----3 eqns to solve x = 1 -----1 x + 6y – 5 = 0 -----2 1 + 6y – 5 = 0 6y = 4 y = ⅔ If x = 1 and y = ⅔  Centroid P must be ( 1 , ⅔) D E F

  20. Q10(a)OPTION 2Find the coordinates of the Centroid P (where medians meet) of triangle DEF? x = 1 -----1 Choose any x + 6y – 5 = 0 -----2 of 2 median 5x + 6y – 9 = 0 -----3 eqns to solve x = 1 -----1 5x + 6y – 9 = 0 -----3 5 + 6y – 9 = 0 6y = 4 y = ⅔ If x = 1 and y = ⅔  Centroid P must be ( 1 , ⅔) D E F

  21. Q10(a) OPTION 3 Find the coordinates of the Centroid P (where medians meet) of triangle DEF? x = 1 -----1 Choose any x + 6y – 5 = 0 -----2 of 2 median 5x + 6y – 9 = 0 -----3 eqns to solve x + 6y – 5 = 0 -----2 5x + 6y – 9 = 0 -----3 4x – 4 = 0 4x = 4 x = 1 If x = 1 then 5x + 6y – 9 = 0 5 + 6y – 9 = 0 6y = 4 y = ⅔ Centroid P must be ( 1 , ⅔) D E F

  22. Q10(b) Find the coordinates of the Orthocentre Q of triangle DEF (where altitudes meet)? Given D(1 , 2) ; E(-1 , 1) & F(3 , -1). Altitude from D If mEF = 1 – (-1)= 2 = -½  mPerpD = 2 -1 – 3 -4 Altitude from D( 1 , 2 ) with mPerpD= 2 (y – 2) = 2(x – 1) y – 2 = 2x – 2 y = 2x D E F

  23. Q10(b) Find the coordinates of the Orthocentre Q of triangle DEF (where altitudes meet)? Given D(1 , 2) ; E(-1 , 1) & F(3 , -1). Altitude from E If mDF = 2 – (-1)= 3  mPerpE = ⅔ 1 – 3 -2 Altitude from E( -1 , 1 ) with mPerpE= ⅔ (y – 1) = ⅔(x – (-1)) 3y – 3 = 2x + 2 2x – 3y + 5 = 0 D E F

  24. Q10(b) Find the coordinates of the Orthocentre Q of triangle DEF (where altitudes meet)? Given D(1 , 2) ; E(-1 , 1) & F(3 , -1). Altitude from F If mDE= 2 – 1 = 1 mPerpF = -2 1 – (-1) 2 Altitude from F( 3 , -1 ) with mPerpF= -2 (y – (-1)) = -2(x –3) y + 1 = -2x + 6 2x + y – 5 = 0 D E F

  25. Again in Q10(b) depending on which 2 of the 3 equations you find will result in your working being laid out differently. So to make it easier I have called each possibility “Options 1, 2 and 3” for you to view your own particular choice.

  26. Q10(b) Find the coordinates of the Orthocentre Q (where altitudes meet) of triangle DEF? y = 2x -----1 2x - 3y + 5 = 0-----2 2x + y – 5 = 0 -----3 Choose any 2 of 3 possible altitude equations to solve for the Orthocentre Q

  27. Q10(b)OPTION 1Find the coordinates of the Orthocentre Q (where altitudes meet) of triangle DEF? y = 2x -----1 Choose any 2x - 3y + 5 = 0-----2 of 2 altitude 2x + y – 5 = 0 -----3 eqns to solve Substituting 1 into 2 gives: 2x - 3y + 5 = 0 2x – 3(2x) + 5 = 0 2x – 6x = -5 -4x = -5 x = 5/4 (or 1.25) If x = 5/4 and y = 2x = 2(5/4) y = 5/2  Orthocentre Q must be ( 5/4 , 5/2) D E F

  28. Q10(b)OPTION 2Find the coordinates of the Orthocentre Q (where altitudes meet) of triangle DEF? y = 2x -----1 2x + y – 5 = 0 -----3 Substituting y = 2x into eqn 3 gives: 2x + y - 5 = 0 2x + (2x) - 5 = 0 4x = 5 x = 5/4 (or 1.25) If x = 5/4 and y = 2x = 2(5/4) = 5/2 Orthocentre Q must be ( 5/4 , 5/2) D E F

  29. Q10(b)OPTION 3 Find the coordinates of the Orthocentre Q (where altitudes meet) of triangle DEF? 2x - 3y + 5 = 0-----2 2x + y – 5= 0-----3 Equation 3 - 2 gives: 4y - 10 = 0 4x = 10 x = 10/4 x = 5/4 (or 1.25) If x = 5/4 and 2x + y - 5 = 0 5/2 + y = 5 y = 5/2  Orthocentre Q must be ( 5/4 , 5/2) D E F

  30. Q10(c) Find the coordinates of the Circumcentre R (where perpendicular bisectors meet) of triangle DEF? Perp Bisector of DE Midpoint of DE = (1 + (-1) , 2 + 1) = (0 , 3/2)or (0,1.5) 2 2 If mDE = 2 - 1= 1 mperp = -2 1 – (-1) 2 Perpendicular Bisector from DE & mperp = -2 y – 1.5 = -2(x – 0) y – 1.5 = -2x 4x + 2y – 3 = 0 D E F

  31. Q10(c) Find the coordinates of the Circumcentre R (where perpendicular bisectors meet) of triangle DEF? Perp Bisector of DF Midpoint of DF = (1 + 3 , 2 + (-1)) = (2 , ½) or (2,0.5) 2 2 If mDF = 2 – (-1)= 3 mperp = ⅔ 1 – 3 -2 Perpendicular Bisector from DF & mperp = ⅔ y – 0.5 = ⅔(x – 2) 3y – 1.5 = 2x - 4 2x - 3y – 2.5 = 0 4x – 6y – 5 = 0 D E F

  32. Q10(c) Find the coordinates of the Circumcentre R (where perpendicular bisectors meet) of triangle DEF? Perp Bisector of EF Midpoint of EF = (-1 + 3 , 1+ (-1)) = (1 , 0) 2 2 If mEF = 1 – (-1)= 2 = - ½ mperp = 2 -1 – 3 -4 Perpendicular Bisector from EF & mperp = 2 y – 0 = 2(x – 1) y = 2x - 2 D E F

  33. Again in Q10(c) depending on which 2 of the 3 equations you find will result in your working being laid out differently. So to make it easier I have called each possibility “Options 1, 2 and 3” for you to view your own particular choice.

  34. Q10(c) Find the coordinates of the Circumcentre R where perpendicular bisectors meet. 4x + 2y – 3 = 0 -----1 4x – 6y – 5 = 0-----2 y = 2x – 2 -----3 Choose any 2 of 3 possible altitude equations to solve for the Circumcentre R

  35. Q10(c)OPTION 1Find the coordinates of the Circumcentre R, where perpendicular bisectors meet. 4x + 2y – 3 = 0 -----1 4x – 6y – 5 = 0-----2 y = 2x – 2 -----3 Subtracting Equation 1 - 2 gives: 8y + 2 = 0 8y = -2 y = -¼ If y = -¼ and 4x +2y – 3 = 0 4x + 2(-¼)- 3 = 0 4x - ½ - 3 = 0 4x = 3 ½ 8x = 7 x = 7/8  Circumcentre R must be ( 7/8 , -¼) D E F

  36. Q10(c)OPTION 2Find the coordinates of the Circumcentre R, where perpendicular bisectors meet. 4x + 2y – 3 = 0 -----1 4x – 6y – 5 = 0 -----2 y = 2x – 2 -----3 Substituting Equation 3 into Equation 1 gives: 4x + 2y – 3 = 0 4x + 2(2x – 2) – 3 = 0 4x + 4x – 4 – 3 = 0 8x = 7 x = 7/8 If x = 7/8 and y = 2x – 2 y = 2(7/8) – 2 y = 7/4 - 2 y = -¼  Circumcentre R must be ( 7/8 , -¼) D E F

  37. Q10(c)OPTION 2Find the coordinates of the Circumcentre R, where perpendicular bisectors meet. 4x + 2y – 3 = 0 -----1 4x – 6y – 5 = 0 -----2 y = 2x – 2 -----3 Substituting Equation 3 into Equation 2 gives: 4x - 6y – 5 = 0 4x - 6(2x – 2) – 5 = 0 4x - 12x + 12 – 5 = 0 -8x = -7 x = 7/8 If x = 7/8 and y = 2x – 2 y = 2(7/8) – 2 y = 7/4 - 2 y = -¼  Circumcentre R must be ( 7/8 , -¼) D E F

  38. Q10(d) Show that P, Q, R are Collinear. Centroid  P ( 1 , ⅔) Orthocentre  Q ( 5/4 , 5/2) Circumcentre  R ( 7/8 , -¼) mpq = 5/2 - ⅔ = 15/6 – 4/6 = 11/6 = 44 = 22 5/4 – 1 ¼ ¼ 6 3 mqr = 5/2-( - ¼) = 10/4 + ¼ = 11/4 = 88 = 22 5/4 – 7/8 10/8 – 7/8 3/8 12 3 As mpq & mqr have equal gradients and a common point exists at Q => points P, Q & R are collinear

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