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 = 71.6o 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 = must change to y = mx + c
y = 2x – 5  m = 2
As m1 x 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
= = 6 – 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 = Thus cuts
– 2x = x-axis at x = (-6 , 0)
Cuts y-axis when x = 0
3y – 2x – 12 = 0
3y – 0 – 12 = Thus cuts
3y = y-axis at
y = (0, 4)

5. Q7 (a) Find altitude from C to AB Given A(3 , 1); B(11 , 5) & C(2 , 8)?
mAB = 5 – 1 = 4 = 1
11 –
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 BC Given A(3 , 1); B(11 , 5) & C(2 , 8)?
mBC = 8 – 5 = 3 = -1
2 –
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.
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 =  T is ( 4 , 4 ) A C B
T( 4 , 4 )

8. Q8 (a) Find line perpendicular to y = ⅓x + 1
Q8 (a) Find line perpendicular to y = ⅓x 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 & 3x + y – 22 = 0
3x + y – 22 = y = ⅓x + 1
3x + (⅓x + 1) - 22 = 0 y = ⅓(6.3) + 1
3x + ⅓x = 0 y =
3⅓x = y = 3.1
10x = 21
3 Thus both lines meet at
10x = (6.3 , 3.1)
x = (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)?

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)
If mAD = 6 –(-4) = 10 = 1
12 –
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?
Midpoint of AB, say E = (2+14 , -4+2) = (8 , -1)
2 2
If mAB = 2–(-4) = 6 = 1  mperp = -2
14 –
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
Q9 (d) Find the coordinates of W, the point of intersection of the 2 lines? C
2x + y =
y = x
2x + (y) = 15
2x + (x – 6) = 15
3x = 21
x = 7
If y = x – 6
= 7 – 6
y =  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
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+(-1)) =(1 , 0 )
If mD = 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
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)
If mE = ½ – 1 = - ½ = -1
2 –(-1)
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
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)
If mF = -1 – 1½= -2½ = -5
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. Choose any 2 of 3 possible median equations to solve for the point P
Q10(a) Find the coordinates of the Centroid P (where medians meet) of triangle DEF?
x =
x + 6y – 5 =
5x + 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 =
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 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 =
5x + 6y – 9 =
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 = Choose any
x + 6y – 5 = of 2 median
5x + 6y – 9 = eqns to solve
x + 6y – 5 =
5x + 6y – 9 =
4x – 4 = 0
4x = 4
x = 1
If x = 1 then x + 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
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 –
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
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 –
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
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. Choose any 2 of 3 possible altitude
Q10(b) Find the coordinates of the Orthocentre Q (where altitudes meet) of triangle DEF?
y = 2x
2x - 3y + 5 =
2x + 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
2x + 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 =
2x + 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?
Perp Bisector of DE
Midpoint of DE = (1 + (-1) , 2 + 1) = (0 , 3/2)or (0,1.5)
If mDE = = 1  mperp = -2
1 – (-1)
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 = ( , 2 + (-1)) = (2 , ½) or (2,0.5)
If mDF = 2 – (-1) = 3  mperp = ⅔
1 –
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+ (-1)) = (1 , 0)
If mEF = 1 – (-1) = 2 = - ½  mperp = 2
-1 –
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. Choose any 2 of 3 possible altitude
Q10(c) Find the coordinates of the Circumcentre R where perpendicular bisectors meet.
4x + 2y – 3 =
4x – 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 =
4x – 6y – 5 =
y = 2x –
Subtracting Equation 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 2 Find the coordinates of the Circumcentre R, where perpendicular bisectors meet.
4x + 2y – 3 =
4x – 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 =
4x – 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 , 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
mqr = 5/2-( - ¼) = 10/4 + ¼ = 11/4 = 88 = 22
5/4 – 7/ /8 – 7/ /
As mpq & mqr have equal gradients and a common point exists at Q => points P, Q & R are collinear