IB Mathematics SL 3.5 Equations of perpendicular bisectors AI HL Paper 2- Exam Style Questions- New Syllabus
▶️ Answer/Explanation
(a)
Midpoint formula: \(\left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)\)
Points B(6, 0) and D(2, 2)
\(\left( \frac{6 + 2}{2}, \frac{0 + 2}{2} \right) = \left( \frac{8}{2}, \frac{2}{2} \right) = (4, 1)\)
Result:
(4, 1)
(b)
(XZ) is perpendicular bisector of [BD]
Slope of BD: \(\frac{0 – 2}{6 – 2} = -\frac{2}{4} = -\frac{1}{2}\)
Perpendicular slope: 2
Midpoint from (a): (4, 1)
Equation: \(y – 1 = 2 (x – 4)\)
\(y – 1 = 2x – 8\)
\(y = 2x – 7\)
Result:
\(y = 2x – 7\)
(c)
X is intersection of (XY) and (XZ)
(XY): \(y = 2 – x\)
(XZ): \(y = 2x – 7\)
Solve: \(2 – x = 2x – 7\)
\(2 + 7 = 2x + x\)
\(9 = 3x\)
\(x = 3\)
\(y = 2 – 3 = -1\)
Result:
(3, -1)
(d)
Points Y(-1, 3) and Z(7, 7)
Distance formula: \(\sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2}\)
\(\sqrt{(7 – (-1))^2 + (7 – 3)^2} = \sqrt{8^2 + 4^2}\)
\(\sqrt{64 + 16} = \sqrt{80}\)
\(\sqrt{80} = 4\sqrt{5}\)
Result:
\(4\sqrt{5}\) km
(e)
Sides: XY = \(\sqrt{32}\), YZ = \(\sqrt{80}\), XZ = \(\sqrt{(7-3)^2 + (7-(-1))^2} = \sqrt{16 + 64} = \sqrt{80}\)
Cosine rule: \(\cos \angle XYZ = \frac{YZ^2 + XY^2 – XZ^2}{2 \times YZ \times XY}\)
\(\cos \angle XYZ = \frac{80 + 32 – 80}{2 \times \sqrt{80} \times \sqrt{32}}\)
\(\cos \angle XYZ = \frac{32}{2 \times 4\sqrt{5} \times 4\sqrt{2}}\)
\(\cos \angle XYZ = \frac{32}{32 \sqrt{10}} = \frac{1}{\sqrt{10}}\)
\(\angle XYZ = \cos^{-1}\left(\frac{1}{\sqrt{10}}\right) \approx 71.6^\circ\)
Result:
\(71.6^\circ\)
(f)
Area formula: \(\frac{1}{2} \times XY \times YZ \times \sin \angle XYZ\)
XY = \(\sqrt{32}\), YZ = \(\sqrt{80}\), \(\sin 71.6^\circ \approx 0.948\)
\(\frac{1}{2} \times 4\sqrt{2} \times 4\sqrt{5} \times 0.948 \approx 8 \sqrt{10} \times 0.948 \approx 24\)
Or: \(\frac{1}{2} \times \sqrt{80} \times \sqrt{72} = 24\) (using height method)
Result:
24 km2
(g)
A larger area does not necessarily result in an increase in population.
Result:
A larger area does not necessarily result in an increase in population.