Home / IB Mathematics SL 3.5 Equations of perpendicular bisectors AI HL Paper 1- Exam Style Questions

IB Mathematics SL 3.5 Equations of perpendicular bisectors AI HL Paper 1- Exam Style Questions- New Syllabus

IBDP Maths AI HL Paper 1 - All Topics

Question

The following Voronoi diagram displays sites located at \(A(3, 3)\), \(B(10, 5)\), \(C(7, 2)\), \(D(5, 9)\), \(E(1, 5)\) and \(F(6, 5)\). The boundaries and cells for each site are also shown. The vertex \(X\) is defined as being equidistant from sites \(B\), \(C\), and \(F\).
(a) (i) State the coordinates of vertex \(X\).
(ii) Given that the length of the segment \(BX\) is \(\sqrt{n}\), determine the value of \(n\).
Another vertex \(Y\), with coordinates \((a, b)\), is equidistant from sites \(B\), \(D\), and \(F\).
(b) (i) State the value of \(a\).
(ii) Calculate the exact value of \(b\).

Most-appropriate topic codes:

• SL 3.6: Voronoi diagrams — part (a) and (b)
• SL 3.5: Equations of perpendicular bisectors — part (b-ii)
▶️ Answer/Explanation
Detailed solution

(a)
(i) From the diagram, \(X\) is the intersection of the edges separating cells \(B\), \(C\), and \(F\).
Coordinates: \((8, 4)\).

(ii) Calculate the distance \(BX\) using \(B(10, 5)\) and \(X(8, 4)\).
\(BX = \sqrt{(10-8)^2 + (5-4)^2} = \sqrt{2^2 + 1^2} = \sqrt{4+1} = \sqrt{5}\).
Thus, \(n = 5\).

(b)
(i) Vertex \(Y\) is the intersection of cells \(B\), \(D\), and \(F\). From the diagram, \(a = 8\).

(ii) \(Y\) lies on the perpendicular bisector of \(B(10, 5)\) and \(D(5, 9)\).
Midpoint \(M_{BD} = (\frac{10+5}{2}, \frac{5+9}{2}) = (7.5, 7)\).
Gradient \(m_{BD} = \frac{9-5}{5-10} = \frac{4}{-5}\).
Gradient of perpendicular \(m_{\perp} = \frac{5}{4}\).
Equation: \(y – 7 = \frac{5}{4}(x – 7.5)\).
Since \(a=8\), substitute \(x=8\):
\(y – 7 = \frac{5}{4}(8 – 7.5) = \frac{5}{4}(0.5) = 0.625\).
\(y = 7 + 0.625 = 7.625\).
Value of \(b = \frac{61}{8}\) or \(7.625\).

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