International Baccalaureate · DP 2: Applications and Interpretation HL

Revision: Mixed Practice

⌚ ~27 min ▢ 3 blocks ✓ 12 exercises ➣ Prerequisites : Revision: Statistics and Probability, Revision: Geometry and Trigonometry

I Toolbox: Perpendicular Bisectors and Voronoi Cells

Recappreviously studied — click to reread

Method — Finding the Equation of a Perpendicular Bisector

To find the equation of the perpendicular bisector of the segment joining $A(x_A, y_A)$ and $B(x_B, y_B)$ (with $A \neq B$):

Find the gradient of the segment $\Segment{AB}$ (when $x_A \neq x_B$): $$ m_{AB} = \frac{y_B - y_A}{x_B - x_A}. $$
Find the gradient of the perpendicular line ($m_\perp$) when $m_{AB} \neq 0$: $$ m_\perp = -\frac{1}{m_{AB}}. $$
Find the midpoint of $\Segment{AB}$: $$ M = \left(\frac{x_A+x_B}{2}, \frac{y_A+y_B}{2}\right). $$
Use the point-gradient form with the midpoint $M(x_M,y_M)$ and the perpendicular gradient $m_\perp$: $$ y - y_M = m_\perp(x - x_M). $$

Special cases:

If $\Segment{AB}$ is horizontal ($m_{AB}=0$), then the perpendicular bisector is a vertical line $x = x_M$.
If $\Segment{AB}$ is vertical ($x_A = x_B$), then the perpendicular bisector is a horizontal line $y = y_M$.

Example

Find the equation of the perpendicular bisector of the line segment with endpoints $A(2, 1)$ and $B(4, 5)$.

Answer

The gradient of the segment $\Segment{AB}$ is$$m_{AB} = \frac{5-1}{4-2} = \frac{4}{2} = 2.$$
The gradient of the perpendicular bisector is $m_\perp = -\dfrac{1}{2}$.
The midpoint of $\Segment{AB}$ is$$M = \left(\frac{2+4}{2}, \frac{1+5}{2}\right) = (3, 3).$$
Using the point-gradient form $y-y_1 = m(x-x_1)$ with the point $(3,3)$ and the gradient $-\dfrac{1}{2}$: $$ \begin{aligned} y-3 &= -\frac{1}{2}(x-3) \\ y-3 &= -\frac{1}{2}x + \frac{3}{2} \\ y &= -\frac{1}{2}x + \frac{3}{2} + 3 \\ y &= -\frac{1}{2}x + \frac{9}{2}. \end{aligned} $$

The equation of the perpendicular bisector is $y = -\dfrac{1}{2}x + \dfrac{9}{2}$.

Method — Constructing a Voronoi Diagram for 3 Sites

To construct the Voronoi diagram of three sites $A$, $B$ and $C$ that are not on one line:

Draw the triangle $ABC$ (in dashed lines, for clarity).
Draw the perpendicular bisector of each side of the triangle.
These three bisectors meet at a single point $V$, equidistant from $A$, $B$ and $C$: it is the only vertex of the diagram.
The point $V$ cuts each bisector into two rays. The edge between the cells of $A$ and $B$ is the ray of the bisector of $[AB]$ whose points are nearer to $A$ and $B$ than to $C$: it is the ray starting at $V$ that points towards the side of the line $(AB)$ where $C$ is not. Check it with one point of the ray. The two other edges are found in the same way.

When the triangle has an obtuse angle, $V$ lies outside the triangle, beyond its longest side, and two of the edges cross the triangle: “away from the triangle” is not a valid rule. When the three sites lie on one line, the two bisectors are parallel: the diagram has no vertex, and its two edges are whole lines.

Example

Construct the Voronoi diagram for sites $A$, $B$, and $C$.

The vertex is the centre of the circumcircle of triangle $ABC$. It is equidistant from $A$, $B$, and $C$.

Skills to practice

Finding the Area of a Voronoi Cell

II Toolbox: Diagonals and Angles in a Cuboid

Recappreviously studied — click to reread

Proposition — Space Diagonals of a Cuboid

In a cuboid whose three edges from one vertex measure $a$, $b$ and $c$, the four space diagonals have the same length $$d = \sqrt{a^2 + b^2 + c^2}$$ and the same midpoint, the centre of the cuboid. In a cube of edge $a$, the space diagonals measure $a\sqrt{3}$.

Method — Finding the Angle Between a Line and a Plane

To find the angle $\theta$ between a line $\Line{QM}$ and a plane $\mathcal{P}$ that it meets at $Q$ — the line being neither perpendicular to the plane, otherwise $\theta = 90^\circ$, nor contained in it, otherwise $\theta = 0^\circ$:

project $M$ on the plane: choose a point $M$ of the line outside the plane and drop the perpendicular from $M$ to the plane; its foot $M'$ is the projection of $M$, and the line $\Line{QM'}$ is the projection of the line on the plane. With coordinates, on the $xy$-plane, the projection of $M(x, y, z)$ is $M'(x, y, 0)$, right below or right above $M$; on a plane parallel to a coordinate plane — a face of a cuboid placed along the axes, for instance — the coordinate that is constant on the plane takes its value, the other two are kept;
draw the right-angled triangle $QM'M$, formed by the segment $\Segment{QM}$ of the line, its projection $\Segment{QM'}$ and the perpendicular $\Segment{MM'}$, right-angled at $M'$;
compute two of its sides and use a trigonometric ratio at $Q$ (SOH-CAH-TOA), for instance $\tan\theta = \dfrac{MM'}{QM'}$: the result is the acute angle $\theta$ between the line and its projection.

In the figure, the angle between the line $L$ and the plane $\mathcal{P}$ is the acute angle $\theta$ at $Q$ of the right-angled triangle $QM'M$.

Example

The pyramid $SABCD$ has a square base $ABCD$ in the $xy$-plane, with $A(0, 0, 0)$, $B(6, 0, 0)$, $C(6, 6, 0)$ and $D(0, 6, 0)$, and its apex is $S(3, 3, 4)$. Calculate the angle between the edge $\Segment{SC}$ and the base, to the nearest tenth of a degree.

Answer

The edge meets the base at $C$. The projection of $S(3, 3, 4)$ on the base, the $xy$-plane, is $K(3, 3, 0)$, the centre of the square. The angle sought is $\theta = \Angle{SCK}$, in the triangle $CKS$ right-angled at $K$.

Its sides: $SK = 4$ and $CK = \sqrt{(3 - 6)^2 + (3 - 6)^2 + 0^2} = \sqrt{18} = 3\sqrt{2}$. The side $\Segment{SK}$ is opposite $\theta$ and $\Segment{CK}$ adjacent: $$\begin{aligned}[t] \tan\theta &= \frac{SK}{CK} && \text{(opposite over adjacent)}\\ &= \frac{4}{3\sqrt{2}}, \end{aligned}$$ so $\theta = \tan^{-1}\left(\dfrac{4}{3\sqrt{2}}\right) \approx 43.3^\circ$.

Skills to practice

Finding the Length of a Space Diagonal

III Exam-Style Questions

Skills to practice

Working with Technology: Mixed Practice