Solutions

1. Length AB = √(x₂ − x₁)² + (y₂ − y₁)²
   =  √(6 − 3)² + (6 − 2)²
   =  √3² + 4²
   =  √25
   = 5
   The length of AB is 5 units. 3 (3)
2. PQ = √(x_Q – x_P)² + (y_Q − y_P)²
   5 =  √(1 − 5)² + (− 3 − t)²
   =  √(− 4)² + 9 + 6t + t²
   =  √16 + 9 + 6t + t²
   =  √t² + 6t  + 25
   25 = t² + 6t  +  25 (square both sides)
   0 =  t² + 6t
   0 = t (t + 6) (factorise by taking out the HCF)
   t = 0 or t = −6 (both solutions are correct – plot the points to see why!)(3)
   [6] 

Solutions

1. Midpoint of KL = ( x₁ + x₂;y₁ + y₂)
                                        2           2
   = (−1 + 5 ; − 6 + 4)
            2            2
   = (2 ; -1) 3 (2)
2. Let B have coordinates (x_B; y_B).
   (–1; 4) = ( 3 + x_B ;6 + y_B)
                       2          2
   –1 = 3 + x_B and 4 = 6 + y_B
              2                      2
   (–1)(2) = 3 + x_B     (4)(2) = 6 + y_B
   –2 = 3 + x_B            8 = 6 + yB
   –5 = x_B                  2 = yB
   ∴ the coordinates of B are(–5; 2).

We can use coordinate geometry to identify the properties of geometric
shapes on the Cartesian plane. (3)
[5] 

Solutions

1. (2)
2. A (–4; 7) and C (0; –1)
   Midpoint ( x₁ + x₂;y₁ + y₂) =  (−4 + 0; 7 − 1) = (–2; 3)
                       2           2                2         2
   So the midpoint of AC is (–2; 3) (2)
3. Diagonals of parallelogram ABCD bisect each other
   ∴ midpoint of DB is (–2; 3).
   So midpoint (–2; 3) = ( 4+2a; 5 +2b)
                                          2        2  
   –2 = 4 +2a and 3 = 5 +2b
              2                     2
   –4 = 4 + a and 6 = 5 + b
   –8 = a 3 and 1 = b 
   ∴ Point D has coordinates (–8; 1) (3)
   [7] 

Solutions

1. First calculate the gradient of PQ:
   m =  y₂ − y₁ =  8 − 2 =  6 = 3 
           x₂ - x₁     3 - 1     2
   Then use the form y − y1 = m (x − x1)
   y − y1 = 3(x − x₁) 
   Substituting P(1; 2)
   y − 2 = 3 (x − 1)
   y − 2 = 3x − 3
   ∴ The equation of linePQ is y = 3x − 1.  (3)
2. m_CD = –2 and CD ⊥AB.
   ∴ m_AB = ½
   So now we have y = ½ x + c
   Substitute (3; 4) to find the value of c.
   4 = ½ (3) + c 
   c = 4 – 1½
   ∴ c = 2½
   ∴ equation of line AB is y = ½ x + 2½ (2)
   [5] 

Solutions

1. m_CD = –4 and m_AB⊥m_CD, – 4 × ¼ = –1
   So mAB = ¼ 
   So tanθ = ¼  = 0,25 and θ = 14,04°. (2)
2. P (–6; 2) and Q (3; 10)
   So mPQ = y₂ − y₁ =  10 − 2 = 8 
                      x₂ - x₁      3-(-6)   9
   So tan θ = ⁸/₉ [To find θ, use 8 ÷ 9 = shift tan on your calculator]
   Angle of inclination: θ = 41,63°
   NOTE: (round off to 2 decimal places) (2)
3. Draw a rough sketch first. Place the triangle on the Cartesian plane.
   Use angles α and β
   
   mAB = tan α.
   ∴tan α = 6 + 1 =  7 = 1 
                 5 + 2     7
   ∴α = 45° 3 (special angles)
   mBC = tan
   ∴ tanβ = − 2 − 6 =  −8 = −4 
                    7 - 5      2
   ∴ β = −75,963°… + 180°
   = 104,04°
   ABC = β−α (ext angle of ∆)
   = 104,04°− 45° = 59,04°(6)
   [10] 

Solutions

1. 1.1 Length AB = √(x₂ − x₁)2 + (y₂ − y₁)² = √(6 − 3)² + (6 − 2)²
   √3² + 4² = √25 = 5 units
   1.2 Midpoint coordinates:
   x₁ + x₂ = 3 + 6 = 4 ½
       2           2
   y₁ + y₂  =  6 + 2 = 4. So the midpoint has the coordinates (4½ ;4)
        2            2 
   1.3 tan θ = m_AB = 2 − 6 = −4 = 4 ∴ θ = 53,13°
                                3 - 6     -3     3
   1.4 m_AB = ⁴/₃ and you know the coordinates of A and B.
   Use y – y₁ = m(x – x₁)
   y – y₁ = ⁴/₃ (x – x₁) now substitute either point A or point B 
   y – 2 = ⁴/₃ (x – 3) here point B has been substituted for (x₁; y₁)
   y – 2 = ⁴/₃ x – 4 ∴y = ⁴/₃ x – 2 
   1.5 AB⊥ GH ∴ mAB × mGH = –1 3 ∴ m_AB =  ⁴/₃ so mGH = ^{− 3}/₄
   The midpoint of AB is (4½; 4)
   y – y₁ = m(x – x₁)
   y – 4 = ^-3/₄ (x − ⁹/₂) 
   y – 4 = ^-3/₄ x + ²⁷/₈
   y = ^-3/₄ x + 3³/₈ + 4
   y = ^-3/₄ x + 7³/₈ (11)
2. m_FG = 6 – 5 = 1 
               4 - 1     3
   m_H1 = 2 – 0 = 2 = 1 
              8 - 2     6     3
   ∴ FG and HI are parallel.
   m_F1 = 0 – 5 = –5 = 5  and m_GH = 2 − 6 =  − 4 = − 1
              4 - 1     -1                           8 - 4       4
   so FI is not parallel to GH.
   ∴ FGHI is a trapezium (one pair of opp sides∥) 3 (5)
3. 3.1 m_AC =  y_C − y_Asubstitution
                      x_C - x_A
   answer
   =  − 5 − 1 (2)
         -1 - 5
   =  − 6 Answer only: full marks
        - 6
   = 1
   3.2 substitution
   y − y₁ = m(x − x₁)
   y − 1 = 1(x − 5)
   y = x − 4 equation (3)
   3.3 Midpoint of BD = (x₂ + x₁; y₂ + y₁)
                                           2          2 
   = (−3 + 9; 5 − 7)= (3; − 1) midpoint (3;-1)
            2        2
   line AC is y = x − 4
   y = 3 − 4
   y =  −1 3 substitution of M in the equation of line AC
   ∴ M lies on AC.
   conclusion (3)
   3.4 M_AM =  y₂ − y₁ gradient of AM
                      x₂ - x₁
   =  − 1 − 5
          3 + 3
   = −1
   and M_MB =  − 1 − 1  gradient of BM
                         3 - 5
   = 1
   M_AM × M_MB = −1
   M_AM × M_MB = −1
   ∴ AMB  = 90°. (2)
   3.5 BM =  √(5 + 1)² + (− 3 − 3)²  substitution into distance formula
   BM =  √72
   AC =  √(5 + 1)² + (1 + 5)²   BM = √72
   AC =  √72    AC = √72
   Area of ΔABC =  ½ ( √72 ) ( √72 )  formula for area of ∆
   = 36 square units  answer (5)
   [31] 

Solutions

1. 1. To get the equation in the form (x – a)² + (y – b)² = r², we need to add in
   numbers to complete the squares using x2 with –2x and y2 with 10y.
   (x²– 2x) + (y²+ 10y) = –14
   (x²– 2x + 1) + (y²+ 10y + 25) = –14 + 1 + 25
   (x – 1)² + (y + 5)²= 12
   So the centre is the point (1; –5) and the radius is √12 = √2².3 = 2√3  (3)
2. First find the value of r²:
   r² = (x – a)² + (y – b)²
   r² = (x + 1)² + (y + 2)² 
   Substitute B(1; –6)
   r² = (1+ 1)² + (–6 + 2)² 
   r² = (2)2 + (–4)²
   r² = 4 + 16 =20
   ∴20 = (x + 1)² + (y + 2)² (3)
   [6]