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A computer game shows the location of four flowers A(1, 7), B(1, 2), C(6, 2), and D(5, 6) on a grid - Junior Cycle Mathematics - Question 16 - 2014

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Question 16

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A computer game shows the location of four flowers A(1, 7), B(1, 2), C(6, 2), and D(5, 6) on a grid. The object of the game is to collect all the nectar from the flo... show full transcript

Worked Solution & Example Answer:A computer game shows the location of four flowers A(1, 7), B(1, 2), C(6, 2), and D(5, 6) on a grid - Junior Cycle Mathematics - Question 16 - 2014

Step 1

A bee found a hidden flower halfway between flower B and flower D. Find the co-ordinates of this hidden flower.

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Answer

To find the coordinates of the midpoint between flower B (1, 2) and flower D (5, 6), we can use the midpoint formula:

extMidpoint=(x1+x22,y1+y22) ext{Midpoint} = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)

Plugging in the coordinates:

Midpoint=(1+52,2+62)=(62,82)=(3,4)\text{Midpoint} = \left( \frac{1 + 5}{2}, \frac{2 + 6}{2} \right) = \left( \frac{6}{2}, \frac{8}{2} \right) = (3, 4)

Therefore, the coordinates of the hidden flower are (3, 4).

Step 2

Another flower E can be located by completing the square ABCE. Write down the coordinates of the point E.

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Answer

To locate point E, we can move horizontally or vertically from points A and B. Starting from B (1, 2), if we move 5 squares to the right, we land at E (6, 2) which is 5 units to the right of B.

Alternatively, we can also find E by moving 5 squares upward from C (6, 2), getting us to E (6, 7).

Thus, the coordinates of point E are (6, 7).

Step 3

Bee 1 flies directly from flower A to B and then on to C. Bee 2 flies from flower A directly to D and then on to C. Write down which bee has travelled the shortest total distance. Give a reason for your answer.

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Answer

To determine which bee has the shortest distance, we will calculate the total distance travelled by each bee:

  • Bee 1:

    • A to B:
      AB=(11)2+(27)2=0+25=5 unitsAB = \sqrt{(1-1)^2 + (2-7)^2} = \sqrt{0 + 25} = 5 \text{ units}
    • B to C:
      BC=(61)2+(22)2=25+0=5 unitsBC = \sqrt{(6-1)^2 + (2-2)^2} = \sqrt{25 + 0} = 5 \text{ units}
    • Total distance = 5 + 5 = 10 units.
  • Bee 2:

    • A to D:
      AD=(51)2+(67)2=16+1=174.12 unitsAD = \sqrt{(5-1)^2 + (6-7)^2} = \sqrt{16 + 1} = \sqrt{17} \approx 4.12 \text{ units}
    • D to C:
      DC=(65)2+(26)2=1+16=174.12 unitsDC = \sqrt{(6-5)^2 + (2-6)^2} = \sqrt{1 + 16} = \sqrt{17} \approx 4.12 \text{ units}
    • Total distance = ≈ 4.12 + 4.12 = ≈ 8.24 units.

Therefore, Bee 2 travelled the shortest total distance.

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