A University has three student residences which are located at points (−6, 12) (0,0) (3,9). The University wants to build a tennis court an

Question

A University has three student residences which are located at points (−6, 12) (0,0) (3,9). The University wants to build a tennis court an equal distance from all three residences. Determine the coordinates of the tennis court. What is the distance from a residence to the court to the nearest 10th of a metre? (1unit=40m).

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Mít Mít 3 years 2021-09-04T20:04:58+00:00 1 Answers 5 views 0

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  1. minhkhoi
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    2021-09-04T20:06:12+00:00
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    Answer:

    a.  (-3, 6)

    b. 268.3 m

    Step-by-step explanation:

    a. Determine the coordinates of the tennis court.

    Let (x,y) be the coordinates of the tennis court. Since we have equal distances from the tennis court to each residence, we have using the equation for distance between two points (x₁,y₁) and (x₂,y₂), d = √[(y₂ – y₁)² + (x₂ – x₁)²]. Using the first two residences, (-6, 12) and (0,0) as (x₂,y₂) and (x, y) = (x₁,y₁), So,

    d = √[(12 – y)² + (-6 – x)²]

    d = √[(12 – y)² + (6 + x)²]

    d = √[(144 – 24y + y² + 36 + 12x + x²]

    d = √x² + y² + 12x – 24y + 180

    d² = x² + y² + 12x – 24y + 180  (1)

    Also, with (x₂, y₂) = (0,0) we have

    d = √[(0 – y)² + (0 – x)²]

    d = √[( – y)² + (- x)²]

    d = √[y² + x²]

    d = √[x² + y²]

    d² = x² + y²  (2)

    Also, with (x₂, y₂) = (3,9) we have

    d = √[(9 – y)² + (3 – x)²]  

    d = √[(81 – 18y + y² + 9 – 6x + x²]

    d = √(x² + y² – 6x – 18y + 90)

    d² = x² + y² – 6x – 18y + 90 (3)

    Substituting (2) into (1), we have

    d² = x² + y² + 12x – 24y + 180   (1)

    d² = d² + 12x – 24y + 180  (1)

    d² – d² = 12x – 24y + 180

    12x – 24y + 180 = 0

    12x – 24y = -180

    dividing through by 12, we have

    x – 2y = -15  (4)

    Substituting (2) into (3), we have

    d² = x² + y² – 6x – 18y + 90 (3)

    d² = d² – 6x – 18y + 90 (3)

    d² – d² = – 6x – 18y + 90

    – 6x – 18y + 90 = 0

    – 6x – 18y = -90

    dividing through by -6, we have

    x + 3y = 15 (5)

    Subtracting (5) from (4), we have

    x – 2y = -15  (4)

    –

    x + 3y = 15 (5)

    -5y = -30

    y = -30/-5

    y = 6

    substituting y = 6 into (4), we have

    x – 2y = -15

    x – 2(6) = -15

    x – 12 = -15

    x = 12 – 15

    x = -3

    So, the coordinates of the tennis court is (-3, 6)

    b. What is the distance from a residence to the court to the nearest 10th of a metre? (1unit=40m).

    Using equation (2),

    d² = x² + y²  (2) where x and y are the coordinates of the tennis court, and x = -3 and y = 6

    So,

    d² = x² + y²  

    d² = (-3)² + (6)²  

    d² = 9 + 36  

    d² = 45

    d = √45 units

    Since 1 unit = 40 m,

    d = √45 × 40 m

    d = 6.708 × 40 m

    d = 268.33 m

    d ≅ 268.3 m to the nearest tenth of a metre

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