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Q3 Calculate the size of the angle between y = ½x + 2 & y = 3x - 1 Answer: (Let angles be α and β for each gradient) If m = tanθ:- tan α = ½&tanβ = 3 α

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Presentation on theme: "Q3 Calculate the size of the angle between y = ½x + 2 & y = 3x - 1 Answer: (Let angles be α and β for each gradient) If m = tanθ:- tan α = ½&tanβ = 3 α"— Presentation transcript:

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

2 Q4Find 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 m 1 x m 2 = -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) Eq1y = 2x + 7orEq2y = -3x – 3 = = 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 AB Given A(3, 1); B(11, 5) & C(2, 8)? m AB = 5 – 1 = 4 = 1 11 – If m AB = ½ m c = -2 Altitude from C(2, 8) with m c = -2:- y – 8 = -2(x – 2) y – 8 = -2x + 4 2x + y – 12 = 0 C AB

6 Q7 (b) Find altitude from A to BC Given A(3, 1); B(11, 5) & C(2, 8)? m BC = 8 – 5 = 3 = -1 2 – If m BC = -1 m A = 3 3 Altitude from A(3, 1) with m c = 3:- y – 1 = 3(x – 3) y – 1 = 3x - 9 y = 3x - 8 A CB

7 Q7 (c) Find the coordinates of T, the point of intersection of the 2 altitudes. 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 ) A CB T( 4, 4 )

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

9 Q8 (b) Find the coordinates where both lines meet? y = x + 1 & 3x + y – 22 = 0 3x + y – 22 = 0 y = x + 1 3x + (x + 1) - 22 = 0 y = (6.3) + 1 3x + x = 0 y = x = 21 y = x = 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 AB Given A(2, -4); B(14, 2) & C(10, 10)? B(14, 2) A(2, -4) C(10,10)

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

12 Q9 (c) Find the perpendicular bisector of AB? Midpoint of AB, say E = (2+14, -4+2) = (8, -1) 22 If m AB = 2–(-4) = 6 = 1 m perp = – Perpendicular Bisector from AB with m AD = -2:- y – ( -1)= -2(x – 8) y + 1 = -2x x + y = 15 C AB

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

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 m D = 0 – 2 = -2 = vertical line 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 m E = ½ – 1 = - ½ = -1 2 –(-1) 3 6 Median from E ( -1, 1 ) with m E = -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 m F = -1 – 1½= -2½ = Median from F ( 3, -1 ) with m F = -5/6:- y – (- 1) = -5/6(x - 3) 6(y + 1) = -5x y + 6 = -5x x + 6y – 9 = 0 D E F

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

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

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

21 Q10(a) OPTION 3 Find the coordinates of the Centroid P (where medians meet) of triangle DEF? x = Choose any x + 6y – 5 = of 2 median 5x + 6y – 9 = eqns to solve x + 6y – 5 = x + 6y – 9 = x – 4 = 0 4x = 4 x = 1 If x = 1 then 5x + 6y – 9 = y – 9 = 0 6y = 4 y = Centroid P must be ( 1, ) 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 m EF = 1 – (-1)= 2 = -½ m PerpD = 2 -1 – 3 -4 Altitude from D( 1, 2 ) with m PerpD = 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 m DF = 2 – (-1)= 3 m PerpE = 1 – 3 -2 Altitude from E( -1, 1 ) with m PerpE = (y – 1) = (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 m DE = 2 – 1 = 1 m PerpF = – (-1) 2 Altitude from F( 3, -1 ) with m PerpF = - 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 x - 3y + 5 = x + y – 5 = Choose any 2 of 3 possible altitude equations to solve for the Orthocentre Q

27 Q10(b) OPTION 1 Find the coordinates of the Orthocentre Q (where altitudes meet) of triangle DEF? y = 2x Choose any 2x - 3y + 5 = of 2 altitude 2x + y – 5 = 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 2 Find the coordinates of the Orthocentre Q (where altitudes meet) of triangle DEF? y = 2x x + y – 5 = 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 = x + y – 5 = Equation 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? D E F Perp Bisector of DE Midpoint of DE = (1 + (-1), 2 + 1) = (0, 3/2)or (0,1.5) 2 2 If m DE = = 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

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

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

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 = x – 6y – 5 = y = 2x – Choose any 2 of 3 possible altitude equations to solve for the Circumcentre R

35 Q10(c) OPTION 1 Find the coordinates of the Circumcentre R, where perpendicular bisectors meet. 4x + 2y – 3 = x – 6y – 5 = y = 2x – Subtracting Equation gives: 8y + 2 = 0 8y = -2 y = -¼ If y = -¼and 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 2 Find the coordinates of the Circumcentre R, where perpendicular bisectors meet. 4x + 2y – 3 = x – 6y – 5 = y = 2x – 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 2 Find the coordinates of the Circumcentre R, where perpendicular bisectors meet. 4x + 2y – 3 = x – 6y – 5 = y = 2x – 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, ) Orthocentre Q ( 5/4, 5/2) Circumcentre R ( 7/8, -¼) m pq = 5/2 - = 15/6 – 4/6 = 11/6 = 44 = 22 5/4 – 1¼ ¼ 6 3 m qr = 5/2-( - ¼) = 10/4 + ¼ = 11/4 = 88 = 22 5/4 – 7/8 10/8 – 7/8 3/ As m pq & m qr have equal gradients and a common point exists at Q => points P, Q & R are collinear


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