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A point P(2, 3) is mapped onto P(-7, 0) under an enlargement with scale factor of -2. Without drawing, find the centre of enlargement.

Mathematics

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9. A point P(2, 3) is mapped onto P'(-7, 0) under an enlargement with scale factor of -2. Without drawing, find the centre of enlargement.
Let the centre of enlargement be C(a,b)C(a, b).
The formula for enlargement is P′=C+k(P−C)P' = C + k(P - C), where kk is the scale factor.
In coordinate form, this means:
x′=a+k(x−a)x' = a + k(x - a)
y′=b+k(y−b)y' = b + k(y - b)

  1. Substitute the given values into the x-coordinate equation.

    Given P(x,y)=(2,3)P(x, y) = (2, 3), P′(x′,y′)=(−7,0)P'(x', y') = (-7, 0), and k=−2k = -2.

    −7=a+(−2)(2−a)-7 = a + (-2)(2 - a) −7=a−4+2a-7 = a - 4 + 2a −7=3a−4-7 = 3a - 4
  2. Solve for aa.

    −7+4=3a-7 + 4 = 3a −3=3a-3 = 3a a=−1a = -1
  3. Substitute the given values into the y-coordinate equation.

    0=b+(−2)(3−b)0 = b + (-2)(3 - b) 0=b−6+2b0 = b - 6 + 2b 0=3b−60 = 3b - 6
  4. Solve for bb.

    6=3b6 = 3b b=2b = 2

    The centre of enlargement is (−1,2)\boxed{(-1, 2)}.

    10. The present age of a father is 10 years more than twice the present age of his son. In how many years time will the age of the father be twice the age of the son.
    Let FF be the father's current age and SS be the son's current age.

  5. Formulate the first equation based on the present ages.

    "The present age of a father is 10 years more than twice the present age of his son."

    F=2S+10(1)F = 2S + 10 \quad (1)
  6. Formulate the second equation based on future ages.

    Let tt be the number of years from now when the father's age will be twice the son's age.
    In tt years, the father's age will be F+tF+t.
    In tt years, the son's age will be S+tS+t.
    "In how many years time will the age of the father be twice the age of the son."

    F+t=2(S+t)(2)F+t = 2(S+t) \quad (2)
  7. Substitute equation (1) into equation (2) and solve for tt.

    (2S+10)+t=2(S+t)(2S + 10) + t = 2(S+t) 2S+10+t=2S+2t2S + 10 + t = 2S + 2t

    Subtract 2S2S from both sides:

    10+t=2t10 + t = 2t

    Subtract tt from both sides:

    10=t10 = t

    The age of the father will be twice the age of the son in 10 years\boxed{10 \text{ years}}.

    11. A triangular prism ABCDEF with the cross section ABF is such that AB = 3cm, BF = 5.5cm and AF = 4cm. The length of the prism is 6cm. Draw the net of the solid.
    As an AI, I cannot physically draw the net. However, I can describe the components and their dimensions, which would allow you to draw it.

    A triangular prism consists of two identical triangular bases and three rectangular lateral faces.
    The cross-section is triangle ABF, so the two triangular bases are △ABF\triangle ABF and △CDE\triangle CDE.
    The dimensions of the triangular bases are:

    • Side AB=3 cmAB = 3 \text{ cm}
    • Side BF=5.5 cmBF = 5.5 \text{ cm}
    • Side AF=4 cmAF = 4 \text{ cm}

    The length of the prism is 6 cm6 \text{ cm}, which is the height of the rectangular faces.
    The three rectangular lateral faces are:

    • Rectangle ABCDABCD: with dimensions AB×length of prism=3 cm×6 cmAB \times \text{length of prism} = 3 \text{ cm} \times 6 \text{ cm}.
    • Rectangle BCFEBCFE: with dimensions BF×length of prism=5.5 cm×6 cmBF \times \text{length of prism} = 5.5 \text{ cm} \times 6 \text{ cm}.
    • Rectangle AFEDAFED: with dimensions AF×length of prism=4 cm×6 cmAF \times \text{length of prism} = 4 \text{ cm} \times 6 \text{ cm}.

    To draw the net, you would typically lay out the three rectangular faces side-by-side, sharing edges. For example, you could place the 3 cm×6 cm3 \text{ cm} \times 6 \text{ cm} rectangle, then attach the 5.5 cm×6 cm5.5 \text{ cm} \times 6 \text{ cm} rectangle to its 6 cm6 \text{ cm} side, and then attach the 4 cm×6 cm4 \text{ cm} \times 6 \text{ cm} rectangle to the 6 cm6 \text{ cm} side of the second rectangle. Finally, attach the two triangular bases (with sides 3 cm3 \text{ cm}, 5.5 cm5.5 \text{ cm}, 4 cm4 \text{ cm}) to the 3 cm3 \text{ cm} side of the first rectangle and the 4 cm4 \text{ cm} side of the third rectangle, respectively.

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