Question
Mai is at an amusement park. A map of part of the amusement park is represented on the following coordinate axes. Mai’s favourite three attractions are positioned at \( A(0, 16) \), \( B(12, 20) \), and \( C(12, 0) \). All measurements are in metres.
(a) Write down the distance between \( B \) and \( C \).
(b) Calculate the distance between \( A \) and \( B \).
Mai is standing at the attraction at \( B \) and wants to walk directly to the attraction at \( A \).
(c) Calculate the bearing of \( A \) from \( B \).
A drinking fountain is to be installed at a point that is an equal distance from each of the attractions at \( A \), \( B \), and \( C \).
(d) (i) Write down the gradient of \( [AC] \).
(ii) Write down the mid-point of \( [AC] \).
(iii) Hence calculate the coordinates of the drinking fountain.
▶️Answer/Explanation
Detailed solution
(a) Distance between \( B \) and \( C \):
Points \( B(12, 20) \) and \( C(12, 0) \) have the same \( x \)-coordinate, so the distance between them is the difference in their \( y \)-coordinates:
\[ BC = |20 – 0| = 20 \, \text{m} \]
So, the distance between \( B \) and \( C \) is 20 metres.
(b) Distance between \( A \) and \( B \):
Points \( A(0, 16) \) and \( B(12, 20) \) are not aligned horizontally or vertically, so we use the distance formula:
\[ AB = \sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2} \]
Substitute the coordinates of \( A \) and \( B \):
\[ AB = \sqrt{(12 – 0)^2 + (20 – 16)^2} = \sqrt{12^2 + 4^2} = \sqrt{144 + 16} = \sqrt{160} = 12.649 \, \text{m} \]
So, the distance between \( A \) and \( B \) is approximately 12.6 metres.
(c) Bearing of \( A \) from \( B \):
To find the bearing of \( A \) from \( B \), we calculate the angle that the line \( BA \) makes with the north direction. First, find the horizontal and vertical differences between \( B \) and \( A \):
\[ \Delta x = 0 – 12 = -12 \, \text{m} \]
\[ \Delta y = 16 – 20 = -4 \, \text{m} \]
The angle \( \theta \) that \( BA \) makes with the horizontal is:
\[ \tan \theta = \frac{\Delta y}{\Delta x} = \frac{-4}{-12} = \frac{1}{3} \]
\[ \theta = \tan^{-1}\left(\frac{1}{3}\right) = 18.4349^\circ \]
The bearing is measured clockwise from north. Since \( BA \) is in the southwest direction, the bearing is:
\[ 180^\circ + 71.565^\circ = 251.565^\circ \]
So, the bearing of \( A \) from \( B \) is approximately 252°.
(d) Drinking Fountain Coordinates:
(i) Gradient of \( [AC] \):
Points \( A(0, 16) \) and \( C(12, 0) \):
\[ \text{Gradient} = \frac{0 – 16}{12 – 0} = \frac{-16}{12} = -\frac{4}{3} \]
So, the gradient of \( [AC] \) is \(-\frac{4}{3}\).
(ii) Mid-point of \( [AC] \):
The mid-point \( M \) of \( [AC] \) is:
\[ M = \left( \frac{0 + 12}{2}, \frac{16 + 0}{2} \right) = (6, 8) \]
So, the mid-point of \( [AC] \) is \( (6, 8) \).
(iii) Coordinates of the Drinking Fountain:
The drinking fountain is the circumcenter of triangle \( ABC \), which is the intersection point of the perpendicular bisectors of the sides of the triangle. We already have the perpendicular bisector of \( [AC] \):
Gradient of perpendicular bisector of \( [AC] \):
\[ m_{\perp} = \frac{3}{4} \]
Equation of perpendicular bisector of \( [AC] \):
\[ y – 8 = \frac{3}{4}(x – 6) \implies y = \frac{3}{4}x + 3.5 \]
Next, find the perpendicular bisector of \( [BC] \). Since \( [BC] \) is vertical, its perpendicular bisector is horizontal. The mid-point of \( [BC] \) is:
\[ M_{BC} = \left( \frac{12 + 12}{2}, \frac{20 + 0}{2} \right) = (12, 10) \]
Thus, the equation of the perpendicular bisector of \( [BC] \) is:
\[ y = 10 \]
Now, find the intersection of \( y = \frac{3}{4}x + 3.5 \) and \( y = 10 \):
\[ 10 = \frac{3}{4}x + 3.5 \implies \frac{3}{4}x = 6.5 \implies x = \frac{26}{3} \approx 8.6667 \]
So, the coordinates of the drinking fountain are \( \left( 8.\overline{6}, 10 \right) \) or approximately \( (8.67, 10) \).
……………………………Markscheme……………………………….
(a) \( BC = 20 \, \text{m} \).
(b) \( AB = 12.6 \, \text{m} \).
(c) Bearing of \( A \) from \( B \): \( 252^\circ \).
(d) (i) Gradient of \( [AC] \): \( -\frac{4}{3} \).
(ii) Mid-point of \( [AC] \): \( (6, 8) \).
(iii) Drinking fountain coordinates: \( \left( 8.\overline{6}, 10 \right) \).