Home / IB Mathematics SL 3.5 Equations of perpendicular bisectors AI SL Paper 2 – Exam Style Questions

IB Mathematics SL 3.5 Equations of perpendicular bisectors AI SL Paper 2 - Exam Style Questions - New Syllabus

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Question

Six Eatery locations (labelled A, B, C, D, E and F) are shown, together with their Voronoi diagram. All distances are measured in kilometres. 
(a) Elena wants to eat at the closest Eatery to her. Write down the Eatery she should go to, if she is at
(i) \((2,\,7)\).
(ii) \((0,\,1)\), when Eatery \(A\) is closed. [2]
Eatery \(C\) is at \((7,\,8)\) and Eatery \(D\) is at \((7,\,5)\).
(b) Find the equation of the perpendicular bisector of \(\overline{CD}\). [2]
Eatery \(B\) is at \((3,\,6)\).
(c) Find the equation of the perpendicular bisector of \(\overline{BC}\). [5]
(d) Hence find
(i) the coordinates of the point which is of equal distance from \(B\), \(C\) and \(D\);
(ii) the distance of this point from \(D\). [4]
▶️ Answer / Explanation
Key facts from the diagram/data
  • \(C(7,8),\ D(7,5)\) so \(\overline{CD}\) is a vertical segment with midpoint \(\left(7,\tfrac{8+5}{2}\right)=(7,6.5)\).
  • \(B(3,6),\ C(7,8)\) so the midpoint of \(\overline{BC}\) is \(\left(\tfrac{3+7}{2},\tfrac{6+8}{2}\right)=(5,7)\).

(a)

From the Voronoi diagram:
(i) The point \((2,7)\) lies in the region for B ⟹ closest Eatery is B. A1
(ii) The point \((0,1)\) lies in the region for A, but with \(A\) closed the nearest open site is F. A1
[2 marks]

(b)

For \(\overline{CD}\): midpoint \(M_{CD}=(7,6.5)\). Since \(CD\) is vertical, its perpendicular bisector is horizontal through \(M_{CD}\):
\[ \boxed{y=6.5}. \] M1 A1 [2 marks]

(c)

For \(\overline{BC}\): midpoint \(M_{BC}=(5,7)\). Gradient of \(BC\): \[ m_{BC}=\frac{8-6}{7-3}=\frac{2}{4}=\tfrac12. \] Perpendicular gradient \(m_\perp=-\dfrac{1}{m_{BC}}=-2.\)
Equation through \(M_{BC}\): \[ y-7=-2(x-5)\;\Rightarrow\; \boxed{y=-2x+17}\quad\text{(equivalently }y-7=-2(x-5)\text{).} \] A1 M1 A1 M1 A1 [5 marks]

(d)

(i) Intersection of the two bisectors from (b) and (c): solve \[ \begin{cases} y=6.5,\\ y=-2x+17 \end{cases} \Rightarrow -2x+17=6.5\Rightarrow x=\dfrac{10.5}{2}=5.25,\ y=6.5. \] Therefore the point equidistant from \(B,C,D\) is \[ \boxed{(5.25,\ 6.5)}. \] M1 A1
(ii) Distance from this point to \(D(7,5)\): \[ \begin{aligned} \text{dist} &= \sqrt{(5.25-7)^2+(6.5-5)^2} =\sqrt{(-1.75)^2+(1.5)^2}\\ &=\sqrt{3.0625+2.25}=\sqrt{5.3125} =\sqrt{\frac{85}{16}}=\frac{\sqrt{85}}{4} \approx \boxed{2.305\ \text{km}}\ (\text{to }3\ \text{sf}). \end{aligned} \] M1 A1 [4 marks]
Total: 13 marks
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