IB Mathematics SL 3.5 Equations of perpendicular bisectors AI SL Paper 2 - Exam Style Questions - New Syllabus
▶️ Answer / Explanation
Solution
(a) Midpoint of \(B(6,0)\) and \(D(2,2)\):
\[ M_{BD}=\left(\frac{6+2}{2},\,\frac{0+2}{2}\right)=(4,1). \] [2]
(b) Line \((XZ)\) is the perpendicular bisector of \([BD]\):
Slope of \(BD\): \(m_{BD}=\dfrac{2-0}{2-6}=\dfrac{2}{-4}=-\frac{1}{2}\).
Therefore slope of \((XZ)\): \(m_{\perp}=2\). Passing through \((4,1)\):
\(y-1=2(x-4)\ \Rightarrow\ \boxed{y=2x-7}.\) [4]
Therefore slope of \((XZ)\): \(m_{\perp}=2\). Passing through \((4,1)\):
\(y-1=2(x-4)\ \Rightarrow\ \boxed{y=2x-7}.\) [4]
(c) \(X\) is the intersection of \((XY)\) and \((XZ)\):
Solve \(\begin{cases}y=2-x\\ y=2x-7\end{cases}\Rightarrow 2-x=2x-7\Rightarrow 3x=9\Rightarrow x=3\).
Then \(y=2-3=-1\). Hence \(\boxed{X(3,-1)}\). [3]
Then \(y=2-3=-1\). Hence \(\boxed{X(3,-1)}\). [3]
(d) Exact length of \([YZ]\) with \(Y(-1,3)\), \(Z(7,7)\):
\[ YZ=\sqrt{(7-(-1))^2+(7-3)^2}=\sqrt{8^2+4^2}=\sqrt{80}= \boxed{4\sqrt{5}}. \] [2]
(e) Given \(|XY|=\sqrt{32}=4\sqrt{2}\), and from (c) & (d), \(|XZ|=\sqrt{(7-3)^2+(7-(-1))^2}=\sqrt{80}=4\sqrt{5}\).
Use the cosine rule at angle \( \angle XYZ \) (sides adjacent to \(Y\) are \(YX\) and \(YZ\), opposite is \(XZ\)):
\[ \cos(\angle XYZ)=\frac{XY^2+YZ^2-XZ^2}{2\cdot XY\cdot YZ} =\frac{32+80-80}{2\cdot (4\sqrt2)\cdot (4\sqrt5)} =\frac{32}{32\sqrt{10}} =\frac{\sqrt{10}}{10}. \] Hence \(\angle XYZ=\cos^{-1}\!\left(\frac{\sqrt{10}}{10}\right)\approx \boxed{71.6^\circ}\) (to 1 d.p.). [4]
\[ \cos(\angle XYZ)=\frac{XY^2+YZ^2-XZ^2}{2\cdot XY\cdot YZ} =\frac{32+80-80}{2\cdot (4\sqrt2)\cdot (4\sqrt5)} =\frac{32}{32\sqrt{10}} =\frac{\sqrt{10}}{10}. \] Hence \(\angle XYZ=\cos^{-1}\!\left(\frac{\sqrt{10}}{10}\right)\approx \boxed{71.6^\circ}\) (to 1 d.p.). [4]
(f) Area of \(\triangle XYZ\):
\[ \text{Area}=\tfrac12\,(XY)(YZ)\sin(\angle XYZ) =\tfrac12\,(4\sqrt2)(4\sqrt5)\,\sqrt{1-\left(\tfrac{\sqrt{10}}{10}\right)^2} =\tfrac12\cdot 16\sqrt{10}\cdot \sqrt{\tfrac{9}{10}} =\tfrac12\cdot 16\cdot 3 =\boxed{24\ \text{km}^2}. \] [2]
(g) Example criticism:
A larger Voronoi area does not account for population density, transport links, or store type/amenities; proximity alone may not determine where people shop. [1]
Total: 18 marks