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IEB practice · Paper II

IEB Analytical geometry practice

12 newly generated questions across all 13 analytical geometry topics the IEB examines in Paper II, 156 marks in all. Work each one, then open its answer underneath to check yourself. Nothing here is taken from a past paper.

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The equation of a straight line

1 questions · 15 marks

Paper II · Section A · 15 marks

QUESTION 1

In the diagram:

  • The straight line through A(−2; −3) and B(2; −1) is drawn.
  • P(−6; 0) is a point that does not lie on AB.
xyOA(−2; −3)B(2; −1)P(−6; 0)
  1. (a)
    Determine the equation of the line AB.
    (2)
  2. (b)
    The line AB cuts the x-axis at C and the y-axis at D. Calculate the area of ΔOCD, where O is the origin.
    (3)
  3. (c)
    F is the foot of the perpendicular from P to AB. Determine the equation of PF.
    (3)
  4. (d)
    Determine the coordinates of F.
    (3)
  5. (e)
    Determine the shortest distance from P to the line AB, in simplest surd form.
    (2)
  6. (f)
    Calculate the area of ΔPAB.
    (2)

[15]

Show the answers

(a) Answer: y = (1/2)x − 2

  1. m = (−1 − (−3))/(2 − (−2)) = 1/2
  2. −3 = (1/2)(−2) + c, so c = −2

(b) Answer: 4 square units

  1. C(4; 0) and D(0; −2)
  2. area = ½ × 4 × 2 = 4

(c) Answer: y = −2x − 12

  1. PF ⊥ AB: m_PF = −1/(1/2) = −2
  2. 0 = (−2)(−6) + c, so c = −12

(d) Answer: F(−4; −4)

  1. (1/2)x − 2 = −2x − 12
  2. x = −4, y = −4

(e) Answer: PF = 2√5 units

  1. PF^2 = (−6 − (−4))^2 + (0 − (−4))^2 = 20
  2. PF = 2√5

(f) Answer: 10 square units

  1. AB = 2√5 and the height is PF = 2√5
  2. area = ½ × AB × PF = 10

The equation of a circle

4 questions · 49 marks

Paper II · Section B · 12 marks

QUESTION 2

In the diagram:

  • The circle with centre M has the equation x^2 + y^2 + 6x + 4y − 13 = 0.
xyOMP(−8; −1)Q(−3; −10)
  1. (a)
    Determine the coordinates of the centre M and the radius of the circle.
    (3)
  2. (b)
    Show that P(−8; −1) 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(−3; −10) to the circle, touching it at T. Determine the length of QT.
    (3)

[12]

Show the answers

(a) Answer: M(−3; −2); r = √26

  1. (x + 3)^2 + (y + 2)^2 = 13 + 9 + 4
  2. (x + 3)^2 + (y + 2)^2 = 26
  3. M(−3; −2) and r = √(26) = √26

(b) Answer: See the working below.

  1. (−8)^2 + (−1)^2 + 6(−8) + 4(−1) − 13 = 0
  2. so P lies on the circle

(c) Answer: y = 5x + 39

  1. m_MP = (−1 − (−2))/(−8 − (−3)) = −1/5
  2. the tangent is perpendicular to the radius: m = 5
  3. −1 = (5)(−8) + c, so c = 39

(d) Answer: QT = √38 units

  1. QM^2 = (−3 − (−3))^2 + (−10 − (−2))^2 = 64
  2. MT ⊥ QT (radius ⊥ tangent), so QT^2 = QM^2 − r^2
  3. QT^2 = 64 − 26 = 38
  4. QT = √38

Paper II · Section B · 11 marks

QUESTION 3

In the diagram:

  • A circle with radius r lies in the first quadrant and touches both axes.
  • The circle passes through P(8; 9).
  • There are two such circles.
xyOP(8; 9)
  1. (a)
    Show that r^2 − 34r + 145 = 0.
    (3)
  2. (b)
    Hence determine the equations of the two circles.
    (3)
  3. (c)
    Calculate the distance between the centres of the two circles, in simplest surd form.
    (2)
  4. (d)
    Determine the equation of the tangent to the smaller circle at P.
    (3)

[11]

Show the answers

(a) Answer: See the working below.

  1. the circle touches both axes, so its centre is (r; r)
  2. it passes through P: (8 − r)^2 + (9 − r)^2 = r^2
  3. 64 − 16r + r^2 + 81 − 18r + r^2 = r^2
  4. r^2 − 34r + 145 = 0

(b) Answer: (x − 5)^2 + (y − 5)^2 = 25 and (x − 29)^2 + (y − 29)^2 = 841

  1. (r − 5)(r − 29) = 0, so r = 5 or r = 29
  2. (x − 5)^2 + (y − 5)^2 = 25
  3. (x − 29)^2 + (y − 29)^2 = 841

(c) Answer: 24√2 units

  1. the centres are (5; 5) and (29; 29)
  2. √((29 − 5)^2 + (29 − 5)^2) = √(1152) = 24√2

(d) Answer: y = −(3/4)x + 15

  1. the radius from (5; 5) to P has m = (9 − 5)/(8 − 5) = 4/3
  2. the tangent is perpendicular to it: m = −3/4
  3. 9 = (−3/4)(8) + c, so c = 15

Paper II · Section B · 14 marks

QUESTION 4

In the diagram:

  • The circle with centre M has the equation x^2 + y^2 + 8x − 6y − 15 = 0.
  • The line y = −x + 7 cuts the circle at A and B.
xyOMAB
  1. (a)
    Determine the coordinates of A and B.
    (5)
  2. (b)
    Determine the length of the chord AB, in simplest surd form.
    (2)
  3. (c)
    N is the midpoint of AB. Determine the coordinates of N, and show that MN ⊥ AB.
    (3)
  4. (d)
    Determine the values of k for which the line y = −x + k is a tangent to the circle.
    (4)

[14]

Show the answers

(a) Answer: A(−2; 9) and B(2; 5)

  1. substitute y = −x + 7 into the equation of the circle
  2. (x + 4)^2 + (−x + 7 − 3)^2 = 40
  3. x^2 − 4 = 0
  4. x = −2 or x = 2

(b) Answer: AB = 4√2 units

  1. AB^2 = (2 − (−2))^2 + (5 − 9)^2 = 32
  2. AB = 4√2

(c) Answer: N(0; 7)

  1. N = ((−2 + 2)/2; (9 + 5)/2) = (0; 7)
  2. m_MN × m_AB = (1)(−1) = −1
  3. so MN ⊥ AB (line from the centre to the midpoint of a chord)

(d) Answer: k = −1 ± 4√5

  1. substitute y = −x + k into the equation of the circle
  2. a tangent meets the circle once: Δ = 0 in the quadratic in x
  3. (k − (−1))^2 = 40(1 + (−1)^2) = 80

Paper II · Section B · 12 marks

QUESTION 5

In the diagram:

  • Circle P has centre P(4; −3) and passes through the origin.
  • Circle Q has the equation (x − 12)^2 + (y − 3)^2 = 25.
xyOPQ
  1. (a)
    Determine 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)

[12]

Show the answers

(a) Answer: (x − 4)^2 + (y + 3)^2 = 25

  1. it passes through O: r = PO = √((4)^2 + (−3)^2) = 5

(b) Answer: PQ = 10 units

  1. PQ = √((12 − 4)^2 + (3 − (−3))^2)
  2. = 10

(c) Answer: See the working below.

  1. r_P + r_Q = 5 + 5 = 10 = PQ
  2. ∴ the circles touch externally

(d) Answer: T(8; 0)

  1. T lies on PQ, 5 units from P: 5/10 of the way to Q
  2. (4 + (5/10)(8); −3 + (5/10)(6)) = (8; 0)

(e) Answer: y = −(4/3)x + 32/3

  1. m_PQ = 3/4, so the tangent has m = −4/3
  2. 0 = (−4/3)(8) + c, so c = 32/3

The midpoint of a line segment

4 questions · 54 marks

Paper II · Section B · 13 marks

QUESTION 6

In the diagram:

  • The circle with centre M passes through the origin O.
  • The circle cuts the x-axis at A(−8; 0) and the y-axis at B(0; 6).
xyOA(−8; 0)B(0; 6)M
  1. (a)
    Explain why AB is a diameter of the circle.
    (2)
  2. (b)
    Determine the coordinates of the centre M and the radius of the circle.
    (3)
  3. (c)
    Show that the equation of the circle can be written as x^2 + y^2 + 8x − 6y = 0.
    (2)
  4. (d)
    Determine the equation of the tangent to the circle at O.
    (3)
  5. (e)
    The line y = −x cuts the circle at O and at C. Determine the coordinates of C.
    (3)

[13]

Show the answers

(a) Answer: See the explanation below.

  1. AOˆB = 90°: the axes are perpendicular
  2. AB subtends a right angle at O, a point on the circle, so AB is a diameter (converse ∠ in semi-circle)

(b) Answer: M(−4; 3); r = 5

  1. M is the midpoint of AB: ((−8 + 0)/2; (0 + 6)/2) = (−4; 3)
  2. r = MO = √((−4)^2 + (3)^2) = 5

(c) Answer: See the working below.

  1. (x + 4)^2 + (y − 3)^2 = 25
  2. x^2 + 8x + 16 + y^2 − 6y + 9 = 25
  3. x^2 + y^2 + 8x − 6y = 0

(d) Answer: y = (4/3)x

  1. m_MO = (3 − 0)/(−4 − 0) = −3/4
  2. the tangent is perpendicular to the radius MO: m = 4/3
  3. it passes through O: y = (4/3)x

(e) Answer: C(−7; 7)

  1. substitute y = −x: x^2 + (−x)^2 + 8x + 6x = 0
  2. 2x(x − (−7)) = 0
  3. x = 0 (the point O) or x = −7

Paper II · Section A · 12 marks

QUESTION 7

In the diagram:

  • ABCD is a parallelogram with A(1; 6), B(1; 2) and C(−3; 6).
  • The diagonals of the parallelogram intersect at M.
xyOA(1; 6)B(1; 2)C(−3; 6)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)

[12]

Show the answers

(a) Answer: M(−1; 6)

  1. the diagonals of a parallelogram bisect each other: M is the midpoint of AC
  2. M = ((1 + (−3))/2; (6 + 6)/2) = (−1; 6)

(b) Answer: D(−3; 10)

  1. M is also the midpoint of BD
  2. D = (2(−1) − 1; 2(6) − 2) = (−3; 10)

(c) Answer: AB = 4 units

  1. AB = √((1 − 1)^2 + (6 − 2)^2) = √(16)
  2. = 4

(d) Answer: No: ABCD is not a rectangle

  1. one of AB and BC is vertical
  2. so AB is not perpendicular to BC

(e) Answer: 8 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. = 8

Paper II · Section A · 16 marks

QUESTION 8

In the diagram:

  • A(1; 2), B(−1; 6), C(3; 4) and D(5; 0) are the vertices of quadrilateral ABCD.
  • The diagonals AC and BD intersect at M.
xyOA(1; 2)B(−1; 6)C(3; 4)D(5; 0)M
  1. (a)
    Determine the coordinates of M, the midpoint of AC.
    (2)
  2. (b)
    Show that M is also the midpoint of BD.
    (2)
  3. (c)
    Show that AC ⊥ BD.
    (3)
  4. (d)
    Hence prove that ABCD is a rhombus.
    (2)
  5. (e)
    Determine the size of DAˆB, correct to one decimal place.
    (4)
  6. (f)
    Calculate the area of the rhombus ABCD.
    (3)

[16]

Show the answers

(a) Answer: M(2; 3)

  1. M = ((1 + 3)/2; (2 + 4)/2) = (2; 3)

(b) Answer: the midpoint of BD is (2; 3) = M

  1. midpoint of BD = ((−1 + 5)/2; (6 + 0)/2) = (2; 3)

(c) Answer: m_AC × m_BD = −1, so AC ⊥ BD

  1. m_AC = (4 − 2)/(3 − 1) = 1
  2. m_BD = (0 − 6)/(5 − (−1)) = −1
  3. m_AC × m_BD = (1)(−1) = −1

(d) Answer: ABCD is a rhombus

  1. the diagonals bisect each other, so ABCD is a parallelogram
  2. the diagonals of the parallelogram are perpendicular, so ABCD is a rhombus

(e) Answer: DAˆB = 143.1°

  1. m_AB = −2, so the inclination of AB is 116.57°
  2. m_AD = −1/2, so the inclination of AD is 153.43°
  3. the lines make 36.86°; DAˆB is its supplement: 180° − 36.86°

(f) Answer: 12 square units

  1. AC = √((3 − 1)^2 + (4 − 2)^2) = 2√2
  2. BD = √((5 − (−1))^2 + (0 − 6)^2) = 6√2
  3. area = ½ × AC × BD = ½ × 2√2 × 6√2 = 12

Paper II · Section A · 13 marks

QUESTION 9

In the diagram:

  • ABCD is a square.
  • A(−5; −5) and C(3; −1) are opposite vertices.
xyOA(−5; −5)C(3; −1)BD
  1. (a)
    Determine the coordinates of M, the point where the diagonals of the square meet.
    (2)
  2. (b)
    Determine the equation of the diagonal BD.
    (3)
  3. (c)
    Determine the coordinates of B and D, given that B has the greater x-coordinate.
    (4)
  4. (d)
    Calculate the area of the square ABCD.
    (2)
  5. (e)
    Determine the inclination of AC, correct to one decimal place.
    (2)

[13]

Show the answers

(a) Answer: M(−1; −3)

  1. the diagonals of a square bisect each other: M is the midpoint of AC
  2. M = ((−5 + 3)/2; (−5 + (−1))/2) = (−1; −3)

(b) Answer: y = −2x − 5

  1. m_AC = (−1 − (−5))/(3 − (−5)) = 1/2
  2. the diagonals of a square are perpendicular: m_BD = −2
  3. through M: −3 = (−2)(−1) + c, so c = −5

(c) Answer: B(1; −7) and D(−3; 1)

  1. the diagonals of a square are equal and bisect each other: MB = MD = MA
  2. MA^2 = (−5 − (−1))^2 + (−5 − (−3))^2 = 20
  3. on BD: (x + 1)^2 + (−2x − 2)^2 = 20
  4. x = 1 or x = −3

(d) Answer: 40 square units

  1. AC^2 = 4 × MA^2 = 80
  2. the diagonals are equal and perpendicular: area = ½ × AC × BD = ½ × AC^2 = 40

(e) Answer: θ = 26.6°

  1. tan θ = m_AC = 1/2

Two circles: touching, intersecting or apart

1 questions · 11 marks

Paper II · Section B · 11 marks

QUESTION 10

In the diagram:

  • The circle with centre A has the equation (x + 4)^2 + (y + 4)^2 = 4.
  • The circle with centre B has the equation x^2 + y^2 + 24x + 8y + 124 = 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)
    Determine the coordinates of the point where the circles touch.
    (3)

[11]

Show the answers

(a) Answer: B(−12; −4); r = 6

  1. complete the square in x and in y
  2. (x + 12)^2 + (y + 4)^2 = 36
  3. B(−12; −4) and r = 6

(b) Answer: AB = 8 units

  1. AB = √((−12 − (−4))^2 + (−4 − (−4))^2)
  2. = 8

(c) Answer: They touch externally.

  1. sum of the radii: 2 + 6 = 8
  2. AB = 8 and 8 = 8

(d) Answer: (−6; −4)

  1. the point lies on AB, 2 units from A: 2/8 of the way to B
  2. (−4 + (2/8)(−8); −4 + (2/8)(0)) = (−6; −4)

Gradient

2 questions · 27 marks

Paper II · Section B · 13 marks

QUESTION 11

In the diagram:

  • A(−3; −8), B(1; −4) and C(−5; 2) lie on the circle with centre M.
xyOA(−3; −8)B(1; −4)C(−5; 2)
  1. (a)
    Show that AB ⊥ BC.
    (3)
  2. (b)
    Hence explain why AC is a diameter of the circle through A, B and C, and determine the coordinates of its centre M.
    (3)
  3. (c)
    Determine the equation of the circle.
    (2)
  4. (d)
    Determine the equation of the tangent to the circle at C.
    (3)
  5. (e)
    Calculate the area of ΔABC.
    (2)

[13]

Show the answers

(a) Answer: See the working below.

  1. m_AB = (−4 − (−8))/(1 − (−3)) = 1
  2. m_BC = (2 − (−4))/(−5 − 1) = −1
  3. m_AB × m_BC = (1)(−1) = −1, so AB ⊥ BC

(b) Answer: M(−4; −3)

  1. ABˆC = 90°, so AC subtends a right angle at B: AC is a diameter (converse ∠ in semi-circle)
  2. M is the midpoint of AC: ((−3 + (−5))/2; (−8 + 2)/2) = (−4; −3)

(c) Answer: (x + 4)^2 + (y + 3)^2 = 26

  1. r^2 = MA^2 = (−3 − (−4))^2 + (−8 − (−3))^2 = 26

(d) Answer: y = (1/5)x + 3

  1. m_MC = −5
  2. the tangent is perpendicular to the radius MC: m = 1/5
  3. 2 = (1/5)(−5) + c, so c = 3

(e) Answer: 24 square units

  1. AB = 4√2 and BC = 6√2
  2. area = ½ × AB × BC (the right angle is at B) = 24

Paper II · Section A · 14 marks

QUESTION 12

In the diagram:

  • A(−2; −4), B(3; −2) and C(0; 0) are the vertices of ΔABC.
  • θ is the angle of inclination of AB.
xyOA(−2; −4)B(3; −2)C(0; 0)
  1. (a)
    Determine the gradient of AB.
    (2)
  2. (b)
    Determine the size of θ, 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(−7; 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)

[14]

Show the answers

(a) Answer: m_AB = 2/5

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

(b) Answer: θ = 21.8°

  1. tan θ = 2/5

(c) Answer: y = −(5/2)x

  1. m = −1/(2/5) = −5/2
  2. 0 = (−5/2)(0) + c, so c = 0

(d) Answer: k = −6

  1. m_AD = m_AB
  2. (k − (−4))/(−7 − (−2)) = 2/5

(e) Answer: BAˆC = 41.6°

  1. the inclination of AB is 21.80°
  2. m_AC = 2, so the inclination of AC is 63.43°
  3. BAˆC = 63.43° − 21.80°

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