Introduction
The D’Shire Island Investigation involves finding the equations of lines on the map of D’Shire
Island along with volume, perpendicular and parallel lines, gradient and midpoints. I will use a
variety of formulas, these are listed below.
-
The gradient formula 𝑚 = (𝑦2 − 𝑦1)/(𝑥2 − 𝑥1)
-
The distance formula 𝑑 =
-
The midpoint formula (𝑥𝑚, 𝑦𝑚) = ((𝑥1 + 𝑥2)/2, (𝑦1 + 𝑦2)/2)
-
Equation of a line (gradient-intercept) 𝑦 = 𝑚𝑥 + 𝑐
Negative reciprocal (of a gradient) 𝑥/𝑦 is the negative reciprocal of 𝑦/𝑥
The volume of a rectangle V = Length * Height * Width
2
2
(𝑥2 − 𝑥1) + (𝑦2 − 𝑦1)
By showing both the working out, and explaining my methodology in paragraph form, I have
displayed basic algebraic understanding, along with understanding of the use of the formulas
listed above. My DI includes comprehensive explanations of the methods used and breaks
down each step in my working out.
Part 1
Part 1 asks us to find the equations of lines: North St, East Ave, South Rd, and West St.
Question A.
This question requires us to find the line of North St. We are given the coordinates of (1, 7) and
(7, 6) which are on North St. First, find the gradient using the formula ((y2–y1)/(x2-x1)). To
complete the equation, we must find the y-intercept, C. To find C, we substitute the y and x
values from one of the sets of coordinates that lies on the North St. Next, solve the equation
with substitute x and y values. Finally, substitute the C value into the original equation, ending
up with the final answer.
Given coordinates: (1 ,7) (7, 6)
m = (6-7)/(7-1)
∴ m = -⅙
∴ y = -⅙x + c
∴ 7 = -⅙ + c
∴c=7⅙
∴ y = (-x+43)/6 or -⅙x + 7 ⅙
Question B.
We can repeat the steps in Question A for Questions B - D, changing the coordinates used from
the ones that lie on North St to the ones for the other respective streets.
Given coordinates: (7, 6) (6, -5)
m = (-5-6)/6-7
∴ m = 11
∴ y = 11x + c
∴ -5 = 66 + c
∴ -71 = c
∴ y = 11x - 71
Question C.
Given coordinates: (6, -5) (-5, -6)
m = (-5 + 6)/(6 + 5)
∴ m = 1/11
∴ y = 1/11x + c
∴ -5 = 1/11(6) + c
∴ -61/11 = c
∴ y = (x-61)/11
Question D.
Given coordinates: (-5, -6) (-5, -1)
m = (-1+6)/(-5+5)
∴ m = (5)/(0)
Since the gradient is undefined, we must look at the given coordinates again. Delta x is zero.
This means that the line is x = n. In this case, x = -5.
∴ x = -5
Part 2
Part 2 requires the use of equations of a line and basic algebra. The question requires us to find
where North St and West St intersect. First, review the equations of North St and West St. Since
West St is represented as x = -5, this means we can substitute x = -5 into the North St line, then
solve algebraically for y.
North St line: (-x+43)/6
West St line: x = -5
y = (-1(-5)+43)/6
∴ y = 48/6
⊥∴ y = 8
∴ North St and West St intersect at (-5, 8)
Part 3
Question A.
Question A requires use of the gradient formula to determine whether the railway line is parallel
to North St. First, calculate the gradients of North St and the railway line. Compare the resulting
gradients. If they are equal, North St is parallel to the railway line. Otherwise, they are not
parallel.
M[North St] = (6-7)/(7-1)
∴ M[North St] = -⅙
M[Railway Line] = (-3 + 2)/(9+2)
∴ M[Railway Line] = -1/11
-1/11 ≠ -1/6
∴ North St ∦ Railway Line
Question B.
This question also uses the gradient formula. Additionally, it requires use of negative reciprocals
to determine whether East Ave and the railway line are perpendicular. First, calculate the
gradient of East Ave. Then, take the negative reciprocal of East Ave and compare it to the
gradient of the railway line. If the negative reciprocal of one equals the gradient of the other,
they are perpendicular.
m[East Ave] = (-5-6)/(6-7)
∴ m[East Ave] = 11
∴ m[East Ave]^-1 = 1/11
∴ -m[East Ave]^-1 = -1/11
m[Railway Line] = -1/11
∴ -1/11 = -m[East Ave]^-1
∴ East Ave ⊥ Railway Line
Part 4
Question A.
This question requires use of the distance formula to calculate the length of North St between
the intersections of Town St ∩ North St and East Ave ∩ North St. The points of intersection are
already given, being (1, 7) and (7, 6). Because of this, all that is required is to plug in these
points into the distance formula.
𝑑 =
2
2
(7 − 1) + (6 − 7)
∴ 𝑑 = 36 + 1
∴𝑑 =
37
Question B.
Question B is similar to the question above. However, the length being calculated is coincident
with South Rd, between South Rd ∩ West St and South Rd ∩ East Ave. The same method is
used for this solution, that being the distance formula.
𝑑 =
2
2
(6 + 5) + (− 5 + 6)
∴𝑑 = 121 + 1
∴𝑑 = 122
Part 5
Part 5 requires finding the point halfway between the church and the hotel. Another term for
halfway between two points is the midpoint. The midpoint formula can be used to find the
coordinates halfway between the church and the hotel.
Church = (-4, 3)
Hotel = (6, 5)
(𝑥𝑚, 𝑦𝑚) = ((6 − 4)/2, (5 + 3)/2)
∴(𝑥𝑚, 𝑦𝑚) = (1, 4)
Part 6
Part 6 requires the use of the distance formula to find the distances between D’Shire Castle,
which can also be referred to as the origin, and points on the D’Shire plane.
Question A.
Question A requires the distance between the origin and the church. The church’s coordinates
have been given: (-4, 3). The calculation of this distance requires the distance formula.
𝑑 =
2
2
(0 + 4) + (0 − 3)
∴ 𝑑 = 16 + 9
∴ 𝑑 = 25
∴𝑑 = 5
Question B.
Using the same methodology as Question A, we can calculate the distance between the origin
and the police station. The coordinates of the police station have been given: (1, -3)
𝑑 =
2
2
(0 − 1) + (0 + 3)
∴𝑑 = 1 + 9
∴𝑑 =
10
Question C.
Substituting (0, 0) and (9, -3) into the distance formula allows us to find the distance between
the origin and the railway station.
𝑑 =
2
2
(0 − 9) + (0 + 3)
∴ 𝑑 = 81 + 9
∴𝑑 =
90
∴ 𝑑 = 3 10
Part 7
Question A.
We can use the volume formula and substitute the values given of the pond and lake.
Pond: L(length) = 20, W(width) = 30, H(depth) = 3; Lake: L = 50, W = 10, H = 2. After using the
volume formula, whichever total volume is greater is the water source that provides more water.
River:
50 × 10 × 2 = 1000
Pond:
20 × 30 × 3 = 1800
1800 > 1000
∴ The river holds more water
Question B.
Using basic algebra, dividing the volume of the container by the rate of water will yield the time it
takes to fill the container.
120/15 = 8
∴The container will take 8 hours to fill
Part 8
Question A.
Question A requires us to find the x-value of point C. It is known that the y-value is 7 through the
diagram. Letting CA = BC due to that being the goal of this question is the first step. After this,
we plug in C(x, 7) and A(5, 4) into the distance formula along with C(x, 7) and B(2, 3). Finally,
use either substitution or elimination methods to find the x value. Simplify the fraction to get the
final coordinates: (7/3, 7).
Let CA = BC
2
CA:
2
(𝑥 − 5) + (7 − 4)
2
∴ (𝑥 − 10𝑥 + 25) + (9)
2
∴ 𝑥 − 10𝑥 + 34
2
2
(𝑥 − 2) + (7 − 3)
2
∴ (𝑥 − 4𝑥 + 4) + (16)
2
∴ (𝑥 − 4𝑥 + 20)
CA = BC
2
∴ 𝑥 − 10𝑥 + 34 =
∴6𝑥 = 14
∴𝑥 = 14/6
∴𝑥 = 7/3
∴C = (7/3, 7)
2
(𝑥 − 4𝑥 + 20)
Question B.
This question requires the total of CA and BC. First, use the distance formula and substitute the
values C(7/3, 7), A(5, 4) or C and B (2, 3). Lastly, multiply the result by 2 - as CA should equal
BC - to get the total and final answer.
2
2
(15/3 − 7/3) + (4 − 7)
2
∴ (8/3) + 9
∴ 64/9 + 81/9
∴ 145/9
∴2( 145)/3
Question C.
This question asks whether the sum of CA and BC is larger or smaller than the sum of DB and
BA. CA + BC is already given from Question B, so all that is required to work out is DB + BA.
The distance formula is used, by substituting D and B, and B and A into the formula and adding
these values together the final answer can be found.
2
2
2
2
(5 − 5) + (7 − 4)
∴ 0+9
∴ 9
∴3
(5 − 2) + (4 − 3)
∴ 9+1
∴ 64/9 + 81/9
∴ 10
3+ 10 < (2 145)/2
∴Routing a pipeline from D to B to A is cheaper than going from C to B and C to A.
Conclusion
In this Directed Investigation, I thoroughly explained my methodology and reasoning for each
step of each question. I used the formulas listed in the introduction along with basic algebraic
knowledge to investigate and find the solutions to each part of the D’Shire Island investigation.
Each question has a unique, simple explanation and uses correct terminology and
demonstrates my understanding of the topic.