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A TY maths class has created a game involving a co-ordinate treasure map, as shown below - Leaving Cert Mathematics - Question 10 - 2021
Question 10
A TY maths class has created a game involving a co-ordinate treasure map, as shown below. The game consists of tasks that involve directions, distances, and location... show full transcript
Worked Solution & Example Answer:A TY maths class has created a game involving a co-ordinate treasure map, as shown below - Leaving Cert Mathematics - Question 10 - 2021
Step 1
Step 2
Food is located at a point F on the map. The point F is on a line which contains the point A(–3, –5) and has a slope of –1. The point F is also on a line which contains B(6, 4) and has a slope of 0. By drawing appropriate lines on the map above, or otherwise, find the co-ordinates of F.
Answer
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To find the equation of the line through A(–3, –5) with a slope of –1:
Using the point-slope form:
y - y_1 = m(x - x_1),
we get:
y - (–5) = –1(x - (–3))
or:
y = –x – 2.
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For line B(6, 4) with slope 0 (a horizontal line):
The equation is simply:
y = 4.
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To find F, set the y-values equal:
–x – 2 = 4
Solving this gives:
–x = 6
x = –6.
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Thus, point F is at F(–6, 4).
Step 3
A clue to another treasure is hidden in a locked box at point B(6, 4). The 4-digit code to open the box is d^4, where d is the distance from B to C(–3, 2), and d ∈ N. Find the 4-digit code (d^4).
Answer
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To calculate the distance d between points B(6, 4) and C(–3, 2), use the distance formula:
d = extup{sqrt}((x_2 - x_1)^2 + (y_2 - y_1)^2).
Substituting in the coordinates:
d = extup{sqrt}((-3 - 6)^2 + (2 - 4)^2) = extup{sqrt}((-9)^2 + (-2)^2) = extup{sqrt}(81 + 4) = extup{sqrt}(85).
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Now calculate d^4:
(d = 85^{1/2} ext{ so } d^4 = 85^2 = 7225).
Therefore, the 4-digit code is 7225.
Step 4
A meeting point P(2, m) is below the x-axis. P is a distance of sqrt(41) from point B(6, 4). Find the value of m and plot P on the map above.
Answer
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The distance from P(2, m) to B(6, 4) is given by:
d = extup{sqrt}((x_2 - x_1)^2 + (y_2 - y_1)^2).
Set d = extup{sqrt}(41):
extup{sqrt}((6 - 2)^2 + (4 - m)^2) = extup{sqrt}(41)
4 + (4 - m)^2 = 41.
Simplifying gives:
(4 - m)^2 = 37.
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Taking the square root:
4 - m = extup{sqrt}(37) ext{ or } 4 - m = - extup{sqrt}(37).
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Solving these equations:
m = 4 - extup{sqrt}(37) ext{ (valid as it's above x-axis)}
ext{ or } m = 4 + extup{sqrt}(37) ext{ (reject as it must be below x-axis)}.
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Thus, the valid value is m = 4 - extup{sqrt}(37). Location P can then be plotted on the map.
Step 5
The line k has equation x – y – 3 = 0. Verify, using substitution, that the point T(–2, –5) is on k.
Answer
To verify that T(–2, –5) lies on the line k, substitute the coordinates into the equation:
x - y - 3 = 0 ightarrow -2 - (–5) - 3 = 0.
Simplifying gives:
-2 + 5 - 3 = 0 ightarrow 0 = 0,
which confirms that T(–2, –5) is indeed on line k.
Step 6
Another treasure also needs to be somewhere on the line k. You must pick a spot along k to contain this treasure. Use algebra to find another point on k, other than T.
Answer
To find another point on line k: Start with the equation of line k:
x - y - 3 = 0 ightarrow y = x - 3.
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Choose a value for x; for example, let x = 0:
y = 0 - 3 = -3.
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Thus, another point on line k is (0, –3).
Step 7
A spade for digging is hidden on line l which is parallel to k. The line l contains the point C(–3, 2). Find the equation of line l.
Answer
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Since line l is parallel to line k, it has the same slope as k, which is 1 (found from y = x - 3).
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Using point C(–3, 2) to find the y-intercept:
y - 2 = 1(x + 3).
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This simplification yields:
y - 2 = x + 3 ightarrow y = x + 5.
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Therefore, the equation of line l is:
y = x + 5.