Mathbench

IEB practice · Paper II

IEB Analytical geometry practice

8 newly generated questions across all 13 analytical geometry topics the IEB examines in Paper II, 100 marks in all. Work each one, then open its answer underneath to check yourself. Nothing here is taken from a past paper.

Download the booklet as PDF · Sit this paper under time · Afrikaans

The equation of a circle

2 questions · 24 marks

112 marks

The circle with centre M has the equation x^2 + y^2 + 4x − 2y − 20 = 0.

xyOMP(1; 5)Q(−2; −5)
  1. (a)
    Determine the coordinates of the centre M and the radius of the circle.
    (3)
  2. (b)
    Show that P(1; 5) lies on the circle.
    (2)
  3. (c)
    Determine the equation of the tangent to the circle at P.
    (4)
  4. (d)
    A tangent is drawn from Q(−2; −5) to the circle, touching it at T. Determine the length of QT.
    (3)
Show the answers

(a) Answer: M(−2; 1); r = 5

  1. (x + 2)^2 + (y − 1)^2 = 20 + 4 + 1
  2. (x + 2)^2 + (y − 1)^2 = 25
  3. M(−2; 1) and r = √(25) = 5

(b) Answer: See the working below.

  1. (1)^2 + (5)^2 + 4(1) − 2(5) − 20 = 0
  2. so P lies on the circle

(c) Answer: y = −(3/4)x + 23/4

  1. m_MP = (5 − (1))/(1 − (−2)) = 4/3
  2. the tangent is perpendicular to the radius: m = −3/4
  3. 5 = (−3/4)(1) + c, so c = 23/4
  4. y = −(3/4)x + 23/4

(d) Answer: QT = √11 units

  1. QM^2 = (−2 − (−2))^2 + (−5 − (1))^2 = 36
  2. MT ⊥ QT (radius ⊥ tangent), so QT^2 = QM^2 − r^2
  3. QT^2 = 36 − 25 = 11
  4. QT = √11
212 marks

Circle P has centre P(−5; 3) and touches the x-axis. Circle Q has the equation (x − 3)^2 + (y + 3)^2 = 49.

xyOPQ
  1. (a)
    Write down the equation of circle P.
    (2)
  2. (b)
    Calculate the distance PQ between the centres.
    (2)
  3. (c)
    Show that the circles touch externally.
    (2)
  4. (d)
    Determine the coordinates of T, the point where the circles touch.
    (3)
  5. (e)
    Determine the equation of the common tangent to the circles at T.
    (3)
Show the answers

(a) Answer: (x + 5)^2 + (y − 3)^2 = 9

  1. it touches the x-axis, so r = |3| = 3
  2. (x + 5)^2 + (y − 3)^2 = 9

(b) Answer: PQ = 10 units

  1. PQ = √((3 − (−5))^2 + (−3 − (3))^2)
  2. = 10

(c) Answer: See the working below.

  1. r_P + r_Q = 3 + 7 = 10 = PQ
  2. ∴ the circles touch externally

(d) Answer: T(−13/5; 6/5)

  1. T lies on PQ, 3 units from P: 3/10 of the way to Q
  2. (−5 + (3/10)(8); 3 + (3/10)(−6)) = (−13/5; 6/5)

(e) Answer: y = (4/3)x + 14/3

  1. m_PQ = −3/4, so the tangent has m = 4/3
  2. 6/5 = (4/3)(−13/5) + c, so c = 14/3
  3. y = (4/3)x + 14/3

The distance between two points

1 questions · 14 marks

314 marks

In the diagram, A(3; −1), B(0; 0), D(4; 2) and E(1; 3) are given. AE and DB intersect at C(2; k).

xyOA(3; −1)B(0; 0)D(4; 2)E(1; 3)C
  1. (a)
    Calculate the length of AB, in simplest surd form.
    (2)
  2. (b)
    Determine the equation of the line DB.
    (3)
  3. (c)
    Determine the value of k.
    (2)
  4. (d)
    Prove that AE ⊥ DB.
    (3)
  5. (e)
    Determine the size of ABˆD, correct to one decimal place.
    (4)
Show the answers

(a) Answer: AB = √10 units

  1. AB = √((0 − (3))^2 + (0 − (−1))^2)
  2. = √(10) = √10

(b) Answer: y = (1/2)x

  1. m_DB = (0 − (2))/(0 − (4)) = 1/2
  2. 2 = (1/2)(4) + c, so c = 0
  3. y = (1/2)x

(c) Answer: k = 1

  1. C lies on DB: k = (1/2)(2) + (0)
  2. k = 1

(d) Answer: See the proof below.

  1. m_AE = (3 − (−1))/(1 − (3)) = −2
  2. m_AE × m_DB = (−2)(1/2) = −1
  3. ∴ AE ⊥ DB

(e) Answer: ABˆD = 45.0°

  1. the inclination of AB is 161.6° and of DB is 26.6°
  2. the lines make 135.0°
  3. ABˆD = 45.0° (its supplement)

Two circles: touching, intersecting or apart

1 questions · 10 marks

410 marks

Two circles are given: the first, with centre A, has the equation (x + 2)^2 + y^2 = 16; the second, with centre B, has the equation x^2 + y^2 − 2x + 8y + 8 = 0.

xyOAB
  1. (a)
    Determine the centre B and the radius of the second circle.
    (3)
  2. (b)
    Determine the distance AB between the centres.
    (2)
  3. (c)
    Do the circles cut, touch or not meet? Justify your answer.
    (3)
  4. (d)
    The second circle keeps its centre B. What must its radius be for it to touch the first circle externally?
    (2)
Show the answers

(a) Answer: B(1; −4); r = 3

  1. complete the square in x and in y
  2. (x − 1)^2 + (y + 4)^2 = 9
  3. B(1; −4) and r = 3

(b) Answer: AB = 5 units

  1. AB = √((1 − (−2))^2 + (−4 − (0))^2)
  2. = 5

(c) Answer: They cut in two points.

  1. sum of the radii: 4 + 3 = 7
  2. AB = 5 and 7 > 5

(d) Answer: r = 1

  1. touching externally: r1 + r = AB
  2. r = 5 − 4 = 1

The midpoint of a line segment

2 questions · 26 marks

514 marks

In the diagram, AB is a diameter of the circle with centre M, where A(−4; −3) and B(0; −5).

xyOA(−4; −3)B(0; −5)M
  1. (a)
    Determine the coordinates of M, the centre of the circle.
    (2)
  2. (b)
    Determine the equation of the circle.
    (3)
  3. (c)
    Show that P(−3; −6) lies on the circle.
    (2)
  4. (d)
    Determine the equation of the tangent to the circle at A.
    (4)
  5. (e)
    Determine the y-intercepts of the circle, in simplest surd form where needed.
    (3)
Show the answers

(a) Answer: M(−2; −4)

  1. M is the midpoint of the diameter AB
  2. ((−4 + (0))/2; (−3 + (−5))/2) = (−2; −4)

(b) Answer: (x + 2)^2 + (y + 4)^2 = 5

  1. r^2 = MA^2 = (−4 − (−2))^2 + (−3 − (−4))^2 = 5
  2. (x + 2)^2 + (y + 4)^2 = 5

(c) Answer: See the working below.

  1. (−3 − (−2))^2 + (−6 − (−4))^2 = 5 = r^2
  2. ∴ P lies on the circle

(d) Answer: y = 2x + 5

  1. m_MA = −1/2
  2. the tangent is perpendicular to the radius MA: m = 2
  3. −3 = (2)(−4) + c, so c = 5
  4. y = 2x + 5

(e) Answer: y = −5 or y = −3

  1. x = 0: (−2)^2 + (y − (−4))^2 = 5
  2. (y − (−4))^2 = 1
  3. y = −5 or y = −3
612 marks

In the diagram, ABCD is a parallelogram with A(−7; −4), B(−4; 1) and C(−9; 4). Its diagonals meet at M.

xyOA(−7; −4)B(−4; 1)C(−9; 4)D
  1. (a)
    Determine the coordinates of M, the point where the diagonals meet.
    (2)
  2. (b)
    Hence determine the coordinates of D.
    (2)
  3. (c)
    Determine the length of AB in simplest surd form.
    (2)
  4. (d)
    Determine whether ABCD is a rectangle. Show your working.
    (3)
  5. (e)
    Calculate the area of ΔABC.
    (3)
Show the answers

(a) Answer: M(−8; 0)

  1. the diagonals of a parallelogram bisect each other: M is the midpoint of AC
  2. M = ((−7 + −9)/2; (−4 + 4)/2) = (−8; 0)

(b) Answer: D(−12; −1)

  1. M is also the midpoint of BD
  2. D = (2(−8) − (−4); 2(0) − (1)) = (−12; −1)

(c) Answer: AB = √34 units

  1. AB = √((−7 − (−4))^2 + (−4 − (1))^2) = √(34)
  2. = √34

(d) Answer: Yes: ABCD is a rectangle

  1. m_AB × m_BC = (5/3)(−3/5) = −1
  2. so AB ⊥ BC and the parallelogram has a right angle

(e) Answer: 17 square units

  1. area = (1/2)|x_A(y_B − y_C) + x_B(y_C − y_A) + x_C(y_A − y_B)|
  2. = 17

Gradient

2 questions · 26 marks

714 marks

In the diagram, A(3; −4), B(5; 4) and C(0; −2) are the vertices of ΔABC.

xyOA(3; −4)B(5; 4)C(0; −2)
  1. (a)
    Determine the gradient of AB.
    (2)
  2. (b)
    Determine θ, the angle of inclination of AB, correct to one decimal place.
    (2)
  3. (c)
    Determine the equation of the line through C that is perpendicular to AB.
    (3)
  4. (d)
    D(4; k) lies on the same straight line as A and B. Determine k.
    (3)
  5. (e)
    Determine the size of angle BAˆC, correct to one decimal place.
    (4)
Show the answers

(a) Answer: m_AB = 4

  1. m_AB = (4 − (−4))/(5 − (3))
  2. = 4

(b) Answer: θ = 76.0°

  1. tan θ = 4
  2. θ = 76.0°

(c) Answer: y = −(1/4)x − 2

  1. m = −1/(4) = −1/4
  2. −2 = (−1/4)(0) + c, so c = −2
  3. y = −(1/4)x − 2

(d) Answer: k = 0

  1. m_AD = m_AB
  2. (k − (−4))/(4 − (3)) = 4
  3. k = 0

(e) Answer: BAˆC = 70.3°

  1. the inclination of AB is 76.0°
  2. m_AC = −2/3, so the inclination of AC is 146.3°
  3. BAˆC = 146.3° − 76.0°
  4. BAˆC = 70.3°
812 marks

In the diagram, ABCD is a trapezium with AD ∥ BC. A(2; 4), B(4; 6) and C(7; 3) are given, and D lies on the x-axis.

xyOA(2; 4)B(4; 6)C(7; 3)D
  1. (a)
    Determine the gradient of BC.
    (1)
  2. (b)
    Prove that ABˆC = 90°.
    (3)
  3. (c)
    Determine the equation of AD.
    (3)
  4. (d)
    Determine the coordinates of D.
    (2)
  5. (e)
    Calculate the area of ΔABC.
    (3)
Show the answers

(a) Answer: m_BC = −1

  1. m_BC = (3 − (6))/(7 − (4)) = −1

(b) Answer: See the proof below.

  1. m_AB = 1
  2. m_AB × m_BC = (1)(−1) = −1
  3. ∴ AB ⊥ BC and ABˆC = 90°

(c) Answer: y = −x + 6

  1. AD ∥ BC, so m_AD = −1
  2. 4 = (−1)(2) + c, so c = 6
  3. y = −x + 6

(d) Answer: D(6; 0)

  1. 0 = (−1)x + (6)
  2. x = 6, so D(6; 0)

(e) Answer: 6 square units

  1. AB = 2√2 and BC = 3√2
  2. area = ½ × AB × BC (the right angle is at B)
  3. = 6

Download this booklet

The same questions as a booklet to print, with the memorandum as a separate file so you can work without the answers in front of you. In English and Afrikaans.

Every question on this page was generated for this site by a computer engine, and its answer was confirmed by an independent method at the moment the question was made.

All IEB practice · Sit this paper under time · Afrikaans