Mathematics Grade 12 Questions and AnswersAnalytic geometry Questions and Answers Mathematics Grade 12 Pdf Download

Analytic geometry Questions and Answers Mathematics Grade 12 Pdf Download

Analytic geometry Questions and Answers Mathematics Grade 12. In mathematics, analytic geometry, also known as coordinate geometry or Cartesian geometry, is the study of geometry using a coordinate system. This contrasts with synthetic geometry. Analytic geometry is used in physics and engineering, and also in aviation, rocketry, space science, and spaceflight.

ANALYTICAL GEOMETRY QUESTIONS AND ANSWERS GRADE 12

Activity 1

  1. For a line passing through the two points A(6; 6) and B(3; 2), calculate the length of AB. (3)
  2. If PQ = 5 units; P (5; t) and Q (1; –3) determine the possible value(s) of t. (3)
    [6]
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 = √(xQ – xP)2 + (yQ − yP)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] 

3. The midpoint of a line
5
If you know the coordinates of the two endpoints of a line, you can find the point that is halfway between them. This is called the midpoint.
The midpoint of a line has the coordinates
( x1 + x2;y1 + y2)
2           2
where (x1; y1) and (x2; y2) are the endpoints of the line.

e.g.3
For a line passing through the two points A(6; 6) and B(3; 2), find the
coordinates of the midpoint of AB.
Midpoint of AB = ( x1 + x2;y1 + y2)
2           2
= (6 + 3 ; 6 + 2) = (4½ ;4)
2         2
So the midpoint has the coordinates (4½ ;4)

e.g.4
The coordinates of the midpoint of the line AB are (1;-4). Find the coordinates of A if the coordinates of B are (4;-3).
When midpoint is given:
x as the midpoint of AB = xA + xB and y as midpoint of AB = yA + yB
2                                                     2
1 = xA + 4 and –4 = yA − 3
2                       2
2 = xA + 4 and –8 = yA − 3
–2 = xA and –5 = yA
Coordinates of A are (-2;-5)

Activity 2
K (–1; –6) and L (5; 4) are two coordinates on the same straight line. Determine the coordinates of the midpoint. (2)
If M (–1; 4) is the midpoint of line segment AB, and the coordinates of A (3;6) are given, find the coordinates of the endpoint B. (3)
[5]

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 (xB; yB).
    (–1; 4) = ( 3 + xB ;6 + yB)
    2          2
    –1 = 3 + xB and 4 = 6 + yB
               2                      2
    (–1)(2) = 3 + xB     (4)(2) = 6 + yB
    –2 = 3 + xB            8 = 6 + yB
    –5 = xB                  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]

Activity 3
A. (–4; 7), B (4; 5), C (0; –1) and D (a; b) are the vertices of parallelogram ABCD.

  1. Draw the parallelogram on squared paper. (2)
  2. Find the midpoint of the diagonal AC. (2)
  3. Use the information that you have to find the coordinates of point D. (3)
    [7]
Solutions

  1. 6(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] 

Activity 4

  1. Determine the equation of the straight line that passes through the points P(1; 2) and Q(3; 8) in the form y = …. (3)
  2. Line AB is perpendicular to CD, which has a gradient of –2. The point (3; 4) lies on AB. Determine the equation of line AB. (2)
    [5]
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. mCD = –2 and CD ⊥AB.
    ∴ mAB = 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] 

Activity 5
Give your answers correct to two decimal places.

  1. Line AB is perpendicular to CD, which has a gradient of –4.
    Find the inclination θ of AB. (2)
  2. Determine the inclination of the straight line that passes through the points
    P (–6; 2) and Q (3; 10). (2)
  3. Given the points A (–2; –1), B (5; 6) and C (7; –2), calculate the size of ABC . (6)
    [10]
Solutions

  1. mCD = –4 and mAB⊥mCD, – 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
    8
    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] 

Activity 6

  1. For a line passing through the two points A(6; 6) and B(3; 2):
    1.1 Calculate the length of AB.
    1.2 Find the coordinates of the midpoint of AB.
    1.3 Calculate the angle of inclination of the line.
    1.4 Determine the equation of the line passing through A and B.
    1.5 Determine the equation of a line GH perpendicular to AB passing through the midpoint of AB. (11)
  2. F, G, H and I are the vertices of a quadrilateral shown below.
    What kind of quadrilateral is FGHI? (5)
    9
  3. ABCD is a quadrilateral with vertices A(5 ; 1), B(−3 ; 5), C(−1 ; −5) and D(9 ; −7).
    10
    3.1 Calculate the gradient of AC. (2)
    3.2 Determine the equation of AC in the form y = … (3)
    3.3 Hence, or otherwise, show that the midpoint M of BD lies on AC. (3)
    3.4 Show that AMB  = 90°. (2)
    3.5 Calculate the area of ∆ ABC. (5)
    [31]
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 = mAB = 2 − 6 = −4 = 4 ∴ θ = 53,13°
    3 – 6     -3     3
    1.4 mAB = 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 ∴ mAB =  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. mFG = 6 – 5 = 1
    4 – 1     3
    mH1 = 2 – 0 = 2 = 1
    8 – 2     6     3
    ∴ FG and HI are parallel.
    mF1 = 0 – 5 = –5 = 5  and mGH = 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 mAC =  yC − yAsubstitution
    xC – xA
    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 MAM =  y2 − y1 gradient of AM
    x2 – x1
    =  − 1 − 5
    3 + 3
    = −1
    and MMB =  − 1 − 1  gradient of BM
    3 – 5
    = 1
    MAM × MMB = −1
    MAM × MMB = −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] 

Activity 7

  1. Determine the coordinates of the centre and the length of the radius if a circle has the equation: x2– 2x + y2+ 10y = –14 (3)
  2. Determine the equation of a circle with centre C(–1; –2) and passing through the point B(1; –6). (3)
    [6]
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] 

Activity 8

  1. ABCD is a quadrilateral with vertices A(–3; 0), B(–1; –3), C(2; –1) and D(0; 2).
    16
    1.1 Determine the coordinates of M, the midpoint of AC. (2)
    1.2 Show that AC and BD bisect each other. (3)
    1.3 Prove that A| D^ C = 90°. (4)
    1.4 Show that ABCD is a square. (4)
    1.5 Determine the size of θ, the angle of inclination of DC, correct to ONE decimal place. (3)
    1.6 Does C lie inside or outside the circle with centre (0; 0) and radius 2? Justify your answer. (2)
    [18]
Solutions
1.1 Midpoint M of AC: 2 − 3; − 1 + 0 = ( − ½ ; -½) (2)
2         2
1.2 Midpoint M of BD: ( −1 + 0; − 3 + 2) = ( −½;-½)
2          2
∴ So the midpoint of AC and the midpoint of BD are the same point, so they bisect each other.(3)
1.3 mAD= 2 − 0 =  2 and mDC=  − 1 − 2 =  −3
0-(0-3)  3                       2-0         2
mAD × mDC = 2 × −3 = –1
3      2      1
∴AD ⊥ DC
∴ADC = 90° (4)
1.4Here is one possible answer:
The diagonals AC and BD bisect each other (proved in 1.2)
ADC = 90° (proved in 1.3)
AD2 = (2 – 0)2 + (0 – (–3))2 = 4 + 9 = 13

HINT: There are several ways to prove that ABCD is a square:

  • Prove that diagonals are equal and bisect each other at 90°
  • Prove that ABCD is a rectangle and has a pair of adjacent sides equal.
  • Prove that all four sides are equal and that one internal angle is 90°.

AD = √13
CD2 = (– 1 – 2)2 + (2 – 0)2 = 9 + 4 = 13
CD = √13
So adjacent sides are equal in length
∴ ABCD is a square. (4)
1.5 tan θ = mDC =  − 1 − 2 =  −3
2 – 0       2
θ = –56,3099324… + 180°
θ = 123,7° (3)
1.6 OC2 = (2 – 0)2 + (– 1 – 0)2
OC2= 4 + 1 = 5
OC = √5
This is longer than the radius of 2 of the given circle, so C(2; –1) lies outside the circle 3. (2)
[18] 

2. O is the centre of the circle in the figure below. P(x; y) and Q(12; 5) are two points on the circle. POQ is a straight line. The point R(t; –1) lies on the tangent to the circle at Q.
17
2.1 Determine the equation of the circle. (3)
2.2 Determine the equation of the straight line through P and Q. (2)
2.3 Determine x and y, the coordinates of P. (4)
2.4 Show that the gradient of QR is -12/5 (2)
2.5 Determine the equation of the tangent QR in the form y = … (3)
2.6 Calculate the value of t. (2)
2.7 Determine an equation of the circle with centre Q(12; 5) and passing through the origin. (3)
[19]

Solutions
The centre is at the origin, so x2 + y2 = r2.
2.1 OQ2 = (5)2 + (12)2 = 25 + 144 = 169
So the equation of the circle is x2 + y2 = 169 (3)
2.2 mPQ = mOQ = 0 – 5 = 5
0 – 12  12
PQ has y-intercept of 0. 3 (2)
y = 5/12 x
2.3 By symmetry, P is the point (–12; –5).  OR
Substitute y = 5/12 x into x2 + y2 = 169
x2 + (5/12 x )2 = 169
x2 + 25/144 x2 = 169
144x2 + 25x2 = 169 × 144
169x2 = 24 336
x2 = 144
x = 12 or x = –12 x = –12 according to given diagram
y = 5/12 x = 5/12 × (–12) = –5  (4)
So P is the point (–12; –5).
2.4 tangent ⊥ radius so QR ⊥ PQ
mPQ = 0 − 5 = 5
0 -12   12
∴mQR =  − 12/5 (2)
2.5 y = − 12/5 x + c  OR y – y1 = − 12/5 (x – x1)
Substitute Q(12; 5) into equation to find c:
5 = − 12/5 (12) + c y – 5 = − 12/5 (x – 12)
5 + 144/5 = c     y =− 12/5 x + 144 + 5
c = 169/5             y = − 12/5 x + 169/5
y = − 12/5 x + 169/5(3)
2.6 R(t; –1) lies on line with equation y = − 12/5 x + 169/5
∴ –1 = − 12/5 t + 169/5
–5 = –12t + 169
12t = 174
t = 14,5 (2)
2.7 OQ2 = (x – 12)2 + (y – 5)2  Q(12; 5) is centre of circle
Substitute (0; 0) into equation:
OQ2 = (0 – 12)2 + (0 – 5)2
OQ2 = 144 + 25 = 169
∴ (x – 12)2 + (y – 5)2 = 169 (3)
[19] 

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