International Baccalaureate · MYP 5 Extended

Coordinate Geometry

⌚ ~170 min ▢ 19 blocks ✓ 76 exercises ➣ Prerequisites : Linear Equations, Inequalities and Formulae, Coordinate Geometry and Straight Lines, Quadratic Equations, Pythagoras and Right-Angled Trigonometry

I Configurations in the Plane

I.1 Collinear Points

Discover

On a town plan, one unit stands for $100$ m. A bus route runs in a straight line from the station $S(-2, 1)$ through the market $M(4, 5)$. The school stands at $K(10, 9)$. On the plan the three points look aligned, but a drawing cannot tell a straight line from a slightly bent one. Does the straight bus route pass through the school? Decide with gradients.

Answer

With the gradient formula, from the station $S$: $$\begin{aligned}[t] m_{SM} &= \dfrac{5 - 1}{4 - (-2)} = \dfrac{4}{6} = \dfrac{2}{3}, \\ m_{SK} &= \dfrac{9 - 1}{10 - (-2)} = \dfrac{8}{12} = \dfrac{2}{3}. \end{aligned}$$ The line $\Line{SM}$ and the line $\Line{SK}$ have the same gradient and pass through the same point $S$: they are one and the same line. So $S$, $M$ and $K$ are collinear, and the bus route passes through the school.

Definition — Gradient

The gradient (also called the slope) of a non-vertical line is defined as the ratio of the change in the vertical direction ($\Delta y$) to the change in the horizontal direction ($\Delta x$), for any two distinct points on the line. This ratio is the same no matter which two points on the line we choose:$$\text{gradient} = \dfrac{\textcolor{colorprop}{\Delta y}}{\textcolor{colordef}{\Delta x}} = \dfrac{\textcolor{colorprop}{\text{vertical change}}}{\textcolor{colordef}{\text{horizontal change}}}, \quad \text{where } \textcolor{colordef}{\Delta x} \neq 0$$

For a vertical line, the horizontal change $\Delta x$ is always zero, so the gradient is undefined.
In the special case where $x$ increases by exactly $1$ — that is, $\Delta x = 1$ — the gradient is simply $\Delta y$: the gradient tells how much $y$ changes for one unit step in $x$.

Proposition — Gradient Formula

The gradient (also called the slope) of a non-vertical line passing through two distinct points $A\left(\textcolor{colordef}{x_A}, \textcolor{colorprop}{y_A}\right)$ and $B\left(\textcolor{colordef}{x_B}, \textcolor{colorprop}{y_B}\right)$ is given by the formula:$$\text{gradient} =\frac{\textcolor{colorprop}{y_B}-\textcolor{colorprop}{y_A}}{\textcolor{colordef}{x_B}-\textcolor{colordef}{x_A}}, \quad \text{where } \textcolor{colordef}{x_A} \neq \textcolor{colordef}{x_B}$$ The order of the points does not matter, as long as we subtract in the same order in the numerator and the denominator.

Definition — Gradient-Intercept Form

The gradient-intercept form (also called the slope-intercept form) of a non-vertical line's equation is:$$y = mx + c$$ where $m$ is the gradient (also called the slope) and $c$ is the $y$-intercept.

The gradient $m$ tells us how much $y$ changes when $x$ increases by $1$.
The $y$-intercept $c$ is the value of $y$ when $x = 0$: the line crosses the $y$-axis at the point $(0, c)$.

Proposition — Horizontal and Vertical Lines

Let $c$ and $k$ be two numbers.

The horizontal line through the point $(0, c)$ has the equation $y = c$: its points are exactly the points with $y$-coordinate $c$. It is the line $y = 0x + c$: its gradient is $0$ and its $y$-intercept is $c$.
The vertical line through the point $(k, 0)$ has the equation $x = k$: its points are exactly the points with $x$-coordinate $k$. It has no equation of the form $y = mx + c$, and its gradient is undefined.

In particular, the $x$-axis has the equation $y = 0$ and the $y$-axis has the equation $x = 0$.

Proposition — Collinear Points and Gradients

Let $A$, $B$ and $C$ be three distinct points of the coordinate plane, and write $m_{AB}$ for the gradient of the line $\Line{AB}$ when it is not vertical.

If $x_B \neq x_A$ and $x_C \neq x_A$, then $A$, $B$ and $C$ are collinear if and only if $m_{AB} = m_{AC}$.
If $x_A = x_B = x_C$, the three points lie on one vertical line: they are collinear.
If exactly one of the points $B$ and $C$ has the $x$-coordinate of $A$, the three points are not collinear.

Proof

Suppose $x_B \neq x_A$ and $x_C \neq x_A$. The line $\Line{AB}$ is not vertical; let $m = m_{AB}$. By the Definition “Gradient-Intercept Form”, it has an equation $y = mx + c$, and since $A$ lies on it, $y_A = m x_A + c$, that is $c = y_A - m x_A$.

If $m_{AC} = m$, then $y_C - y_A = m(x_C - x_A)$, so $$\begin{aligned}[t] y_C &= m x_C - m x_A + y_A \\ &= m x_C + c. \end{aligned}$$ The coordinates of $C$ satisfy the equation of $\Line{AB}$: the point $C$ lies on $\Line{AB}$, and $A$, $B$, $C$ are collinear.
Conversely, if $C$ lies on $\Line{AB}$, the gradient of this line is the same whichever two of its points are used (Definition “Gradient”), so $m_{AC} = m_{AB}$.

If $x_A = x_B = x_C$, the three points lie on the vertical line with equation $x = x_A$ (Proposition “Horizontal and Vertical Lines”). If $x_B = x_A$ and $x_C \neq x_A$, the line $\Line{AB}$ is the vertical line $x = x_A$, which does not contain $C$: the three points are not collinear; the case $x_C = x_A$, $x_B \neq x_A$ is the same with $B$ and $C$ exchanged.

Proposition — Parallel Lines

Two distinct, non-vertical lines$$L_1 : y = m_1 x + c_1 \quad \text{and} \quad L_2 : y = m_2 x + c_2$$ are parallel if and only if they have the same gradient (also called slope), i.e.\ $m_1 = m_2$. (If $m_1 = m_2$ and $c_1 = c_2$, the two equations represent the same line — which is why the proposition assumes the lines are distinct.)
All vertical lines (of the form $x = k$) are parallel to each other.

Method — Testing Whether Three Points Are Collinear

For three distinct points $A$, $B$ and $C$:

Compare the $x$-coordinates: if the three are equal, the points are collinear (a vertical line); if exactly two are equal, they are not collinear.
Otherwise, compute two gradients that share a point, for example $m_{AB}$ and $m_{AC}$, or $m_{AB}$ and $m_{BC}$.
Conclude: equal gradients through a common point give one line, so the points are collinear; different gradients, and they are not.

Two equal gradients $m_{AB} = m_{CD}$, computed from two pairs of points that share no point, only prove that the lines $\Line{AB}$ and $\Line{CD}$ are parallel: they may be two different lines, or one and the same line.

Example

Determine whether the points are collinear.

$A(-2, -1)$, $B(1, 1)$ and $C(7, 5)$.
$P(-1, 4)$, $Q(2, 2)$ and $R(5, -1)$.
$E(3, -2)$, $F(3, 1)$ and $G(3, 6)$.

Answer

The $x$-coordinates are different. With the gradient formula: $$\begin{aligned}[t] m_{AB} &= \dfrac{1 - (-1)}{1 - (-2)} = \dfrac{2}{3}, \\ m_{AC} &= \dfrac{5 - (-1)}{7 - (-2)} = \dfrac{6}{9} = \dfrac{2}{3}. \end{aligned}$$ $m_{AB} = m_{AC}$: $A$, $B$ and $C$ are collinear.
$$\begin{aligned}[t] m_{PQ} &= \dfrac{2 - 4}{2 - (-1)} = -\dfrac{2}{3}, \\ m_{PR} &= \dfrac{-1 - 4}{5 - (-1)} = -\dfrac{5}{6}. \end{aligned}$$ $m_{PQ} \neq m_{PR}$: $P$, $Q$ and $R$ are not collinear. On the figure, $Q$ lies half a unit above the line $\Line{PR}$.
$E$, $F$ and $G$ have the same $x$-coordinate $3$: they lie on the vertical line $x = 3$, so they are collinear — here no gradient exists, and none is needed.

Example

Let $A(-1, 3)$ and $B(2, 1)$.

Find $k$ so that $A$, $B$ and $C(8, k)$ are collinear.
The line $\Line{AB}$ crosses the $x$-axis at $D(d, 0)$. Find $d$.

Answer

First, $m_{AB} = \dfrac{1 - 3}{2 - (-1)} = -\dfrac{2}{3}$.

$C$ does not have the $x$-coordinate of $A$, so $A$, $B$, $C$ are collinear exactly when $m_{AC} = m_{AB}$: $$\begin{aligned}[t] \dfrac{k - 3}{8 - (-1)} &= -\dfrac{2}{3} && \text{(equal gradients)} \\ k - 3 &= -\dfrac{2}{3} \times 9 \\ k - 3 &= -6 \\ k &= -3. \end{aligned}$$ Check: $m_{BC} = \dfrac{-3 - 1}{8 - 2} = -\dfrac{4}{6} = -\dfrac{2}{3}$ $\checkmark$.
$D$ lies on $\Line{AB}$, so $m_{AD} = m_{AB}$: $$\begin{aligned}[t] \dfrac{0 - 3}{d - (-1)} &= -\dfrac{2}{3} && \text{(equal gradients)} \\ -9 &= -2(d + 1) \\ d + 1 &= 4.5 \\ d &= 3.5. \end{aligned}$$ The line $\Line{AB}$ crosses the $x$-axis at $D(3.5, 0)$.

Skills to practice

Recognising Collinear Points
Finding a Coordinate for Collinearity
Proving That Points Are Collinear

I.2 Points at a Given Distance

Discover

In a park, one unit stands for $1$ m. A dog is tied to a post $P(2, -1)$ by a rope $5$ m long, and a straight path runs along the line $y = 3$. At which points of the path can the dog, at full stretch, just reach the path?

Answer

A point of the path is $M(x, 3)$. The dog just reaches $M$ when $PM = 5$, that is, since both sides are non-negative, when $PM^2 = 25$: $$\begin{aligned}[t] (x - 2)^2 + (3 - (-1))^2 &= 25 && \text{(distance formula)} \\ (x - 2)^2 + 16 &= 25 \\ (x - 2)^2 &= 9 \\ x - 2 &= 3 \quad \text{or} \quad x - 2 = -3 && \text{(Solutions of } x^2 = a\text{)} \\ x &= 5 \quad \text{or} \quad x = -1. \end{aligned}$$ The dog just reaches the path at $M_1(5, 3)$ and $M_2(-1, 3)$: the two points where the circle of centre $P$ and radius $5$ meets the line $y = 3$.

Proposition — Distance Between Two Points in the Plane

In a coordinate plane whose axes are perpendicular and have the same unit, the distance between the points $A(x_A, y_A)$ and $B(x_B, y_B)$ is $$AB = \sqrt{(x_B - x_A)^2 + (y_B - y_A)^2}.$$

Proposition — Solutions of $x^2 \equal a$

Let $a$ be a number. The solutions of the equation $x^2 = a$ depend on the sign of $a$:

if $a > 0$, the equation has two solutions: $x = \sqrt{a}$ and $x = -\sqrt{a}$;
if $a = 0$, the equation has exactly one solution: $x = 0$;
if $a < 0$, the equation has no solution: the square of a number is never negative.

Method — Classifying a Triangle from Its Vertices

Compute the squares $AB^2$, $BC^2$ and $AC^2$ of the three sides with the distance formula. Lengths are positive, so two lengths are equal exactly when their squares are equal: the square roots are not needed.
Compare the sides: three equal sides, the triangle is equilateral (and so also isosceles); exactly two, it is isosceles and not equilateral; none, it is scalene.
Test a longest side, of length $c$, the other two being $a$ and $b$: if $a^2 + b^2 = c^2$, the triangle is right-angled, and the right angle is opposite the longest side (converse of Pythagoras' theorem); if $a^2 + b^2 \neq c^2$, it is not right-angled (contrapositive of Pythagoras' theorem).

Method — Finding a Coordinate from a Given Distance

Write the condition with the distance formula, the unknown coordinate in it, and square both sides: for $d > 0$, $AB = d$ exactly when $AB^2 = d^2$, since a length is never negative. In the same way, $PA = PB$ exactly when $PA^2 = PB^2$.
Solve the equation. It often comes down to $(k - a)^2 = c$: two solutions $k = a + \sqrt{c}$ and $k = a - \sqrt{c}$ if $c > 0$, one solution $k = a$ if $c = 0$, and none if $c < 0$ (Proposition “Solutions of $x^2 = a$”).
Interpret each solution as a point and check it on a sketch: the points at the distance $d$ from $A$ lie on the circle of centre $A$ and radius $d$, which meets a line in two, one or no points.

Example

Let $A(-1, 2)$.

Find $b$ so that the point $B(3, b)$ is at the distance $5$ from $A$.
Is there a point $B(3, b)$ at the distance $3$ from $A$?

Answer

$AB = 5$ exactly when $AB^2 = 25$: $$\begin{aligned}[t] (3 - (-1))^2 + (b - 2)^2 &= 25 && \text{(distance formula)} \\ 16 + (b - 2)^2 &= 25 \\ (b - 2)^2 &= 9 \\ b - 2 &= 3 \quad \text{or} \quad b - 2 = -3 && \text{(Solutions of } x^2 = a\text{)} \\ b &= 5 \quad \text{or} \quad b = -1. \end{aligned}$$ There are two such points: $B_1(3, 5)$ and $B_2(3, -1)$. Check: $AB_1^2 = 4^2 + 3^2 = 25$ $\checkmark$.
$AB = 3$ would need $16 + (b - 2)^2 = 9$, that is $(b - 2)^2 = -7$: a square is never negative, so there is no such point. The line $x = 3$ is $4$ units from $A$, too far for the circle of radius $3$ to meet it.

Example

Find the point $P$ of the $y$-axis that is at the same distance from $A(1, 4)$ as from $B(5, 2)$.

Answer

A point of the $y$-axis is $P(0, p)$, and $PA = PB$ exactly when $PA^2 = PB^2$: $$\begin{aligned}[t] (1 - 0)^2 + (4 - p)^2 &= (5 - 0)^2 + (2 - p)^2 && \text{(distance formula)} \\ 1 + 16 - 8p + p^2 &= 25 + 4 - 4p + p^2 && \text{(expanding)} \\ 17 - 8p &= 29 - 4p \\ -4p &= 12 \\ p &= -3. \end{aligned}$$ So $P(0, -3)$. Check: $PA^2 = 1^2 + 7^2 = 50$ and $PB^2 = 5^2 + 5^2 = 50$ $\checkmark$. The squares $p^2$ cancel: the equation is of the first degree, so there is exactly one such point.

Skills to practice

Counting the Points at a Given Distance
Finding a Coordinate from a Given Distance
Finding Points at a Given Distance

I.3 The Equation of a Perpendicular Bisector

Discover

On a map, one unit stands for $100$ m. Two farms stand at $A(-1, 1)$ and $B(3, 5)$, and a well is to be dug at the same distance from the two farms.

Where can the well be dug? For a possible position $W(x, y)$, write the condition $WA = WB$ with the distance formula, simplify it, and describe the set of all the possible positions.

Answer

Lengths are non-negative, so $WA = WB$ exactly when $WA^2 = WB^2$: $$\begin{aligned}[t] (x + 1)^2 + (y - 1)^2 &= (x - 3)^2 + (y - 5)^2 && \text{(distance formula)} \\ x^2 + 2x + 1 + y^2 - 2y + 1 &= x^2 - 6x + 9 + y^2 - 10y + 25 && \text{(expanding)} \\ 2x - 2y + 2 &= -6x - 10y + 34 && \text{($x^2$ and $y^2$ cancel)} \\ 8y &= -8x + 32 \\ y &= -x + 4. \end{aligned}$$ The squares cancel: the condition is an equation of the first degree, so the possible positions form a line, $y = -x + 4$. This line passes through the midpoint $M(1, 3)$ of $\Segment{AB}$, since $-1 + 4 = 3$, and the product of its gradient $-1$ and the gradient $m_{AB} = \dfrac{5 - 1}{3 - (-1)} = 1$ of $\Segment{AB}$ is $-1$: it is perpendicular to $\Segment{AB}$. It is the perpendicular bisector of $\Segment{AB}$. For instance, the well can be dug at $W_1(0, 4)$ or at $W_2(4, 0)$: $W_1A^2 = 1 + 9 = 10 = W_1B^2$.

Definition — Point-Gradient Form

A line with gradient $m$ passing through the point $(x_1, y_1)$ has the equation $$y - y_1 = m(x - x_1),$$ called its point-gradient form (also called the point-slope form). It is the gradient formula in disguise: a point $(x, y)$ with $x \neq x_1$ lies on the line exactly when $\dfrac{y - y_1}{x - x_1} = m$.

Proposition — Perpendicular Lines

In a coordinate plane whose axes are perpendicular and have the same unit, two non-vertical lines are perpendicular if and only if the product of their gradients $m_1$ and $m_2$ is $-1$, i.e.\ $m_1 \cdot m_2 = -1$. Equivalently, their gradients are negative reciprocals:$$m_2 = -\frac{1}{m_1}.$$ Additionally, a vertical line (undefined gradient) is always perpendicular to a horizontal line (gradient $0$).

Definition — Midpoint Formula

The midpoint $M$ of the segment $\Segment{AB}$, with $A(x_A, y_A)$ and $B(x_B, y_B)$, has coordinates $$M\left(\frac{x_A + x_B}{2},\ \frac{y_A + y_B}{2}\right):$$ each coordinate of $M$ is the mean of the corresponding coordinates of $A$ and $B$.

Definition — Perpendicular bisector

The perpendicular bisector of a line segment is the line that passes through the midpoint of the segment and is perpendicular to the segment.

Proposition — Perpendicular bisector and equidistance

The perpendicular bisector of a segment $\Segment{AB}$ is the set of all the points at the same distance from $A$ and from $B$. In other words, a point $M$ lies on the perpendicular bisector of $\Segment{AB}$ if and only if $MA = MB$.

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 $y = -\dfrac{1}{2}x + \dfrac{9}{2}$.

Skills to practice

Calculating the Equation of a Perpendicular Bisector
Finding the Equation of a Perpendicular Bisector

II Coordinates in Space

II.1 Points in Space

Discover

On the map of a bay, a grid locates every point of the sea surface by two numbers: the origin $O$ is a buoy, the first axis points east, the second north, and the unit is the metre. A seagull flies $12$ m above the point $(20, 40)$ of the map, a boat floats at the point $(40, 10)$, and a diver swims $8$ m under the surface, right below the point $(10, 20)$. How many numbers are needed to say exactly where the seagull, the boat and the diver are? What does the last number measure, and what is special about the diver's numbers?

Answer

Three numbers. The first two locate the point of the map right below or right above the object; the third is its height above the sea surface, measured on a vertical axis through $O$: $$\text{seagull } G(20, 40, 12), \qquad \text{boat } B(40, 10, 0), \qquad \text{diver } D(10, 20, -8).$$ The boat is on the surface: its height is $0$. The diver is under the surface: its third number is negative, $-8$.

Definition — Coordinate System in Space

A coordinate system in space is made of a point $O$, the origin, and three number lines through $O$, perpendicular in pairs and graduated with the same unit (such a system is called orthonormal): the $x$-axis, the $y$-axis and the $z$-axis. Every point $P$ of space is located by an ordered triple of numbers, its coordinates, written $P(x, y, z)$: starting from $O$, move $x$ units along the $x$-axis, then $y$ units parallel to the $y$-axis, then $z$ units parallel to the $z$-axis — in the direction of the arrow for a positive number, in the opposite direction for a negative one. The number $x$ is the $x$-coordinate of $P$, $y$ its $y$-coordinate and $z$ its $z$-coordinate: the signed height of $P$ relative to the plane of the $x$- and $y$-axes, positive above this plane and negative below it. The origin is $O(0, 0, 0)$.

Definition — Right-Handed Coordinate System

By convention, the axes of a coordinate system in space are placed right-handed: when the thumb of the right hand points along the $x$-axis and the index finger along the $y$-axis, the middle finger, held at right angles to both, points along the $z$-axis — each finger in the direction of the arrow. Seen from above, from a point of the $z$-axis with $z > 0$, the quarter turn that takes the $x$-axis onto the $y$-axis is then anticlockwise. With the middle finger pointing the other way, the system would be left-handed: the convention excludes it.

Definition — Coordinate Planes

The axes, taken two at a time, span the three coordinate planes:

the $xy$-plane, which contains the $x$- and $y$-axes: its points are those with $z = 0$;
the $xz$-plane: the points with $y = 0$;
the $yz$-plane: the points with $x = 0$.

Each axis is perpendicular to the plane of the other two: the $z$-axis is perpendicular to every line of the $xy$-plane through $O$, like a vertical post and the straight lines drawn on the ground from its foot. A point lies on the $x$-axis when $y = z = 0$, on the $y$-axis when $x = z = 0$, and on the $z$-axis when $x = y = 0$.

Method — Plotting a Point in Space

The axes are drawn in perspective: the $y$-axis horizontal, the $z$-axis vertical, and the $x$-axis, which points towards the reader, slanting down to the left with its unit shortened. To plot the point $P(x, y, z)$:

from $O$, move $x$ units along the $x$-axis — towards the reader when $x > 0$;
then $y$ units parallel to the $y$-axis — to the right when $y > 0$;
then $z$ units vertically — up when $z > 0$, down when $z < 0$.

Keep the three moves on the drawing, dashed: a lone dot on a flat drawing could stand for infinitely many points of space. The point $P'(x, y, 0)$ reached after the first two moves is the point of the $xy$-plane right below or right above $P$.

Example

Plot the points $A(4, 2, 3)$ and $B(3, -2, -2)$ on a perspective drawing of the axes.

Answer

$A(4, 2, 3)$: $4$ units towards the reader along the $x$-axis, then $2$ units to the right, then $3$ units up.
$B(3, -2, -2)$: $3$ units towards the reader, then $2$ units to the left, since $y = -2$, then $2$ units down, since $z = -2$: the point $B$ lies below the $xy$-plane.

Example

The cuboid $ABCDEFGH$ has its vertex $A$ at the origin and its edges $\Segment{AB}$, $\Segment{AD}$ and $\Segment{AE}$ along the axes, with $AB = 4$, $AD = 5$ and $AE = 3$. Give the coordinates of its eight vertices.
The pyramid $SOPQR$ has a square base $OPQR$ of side $4$ in the $xy$-plane, with $O$ at the origin, $P$ on the $x$-axis and $R$ on the $y$-axis; its apex $S$ is $3$ units above the centre $K$ of the base. Give the coordinates of $P$, $Q$, $R$, $K$ and $S$.

Answer

$A$ is the origin: $A(0, 0, 0)$. The vertices $B$, $D$ and $E$ lie on the axes, at the distances $AB = 4$, $AD = 5$ and $AE = 3$ from $A$: $B(4, 0, 0)$, $D(0, 5, 0)$, $E(0, 0, 3)$. The vertex $C$ is reached by moving $4$ units along the $x$-axis, then $5$ units parallel to the $y$-axis: $C(4, 5, 0)$. The top face $EFGH$ lies right above the base, with $z$-coordinate $3$: $F(4, 0, 3)$, $G(4, 5, 3)$, $H(0, 5, 3)$.
$P(4, 0, 0)$, $Q(4, 4, 0)$ and $R(0, 4, 0)$. The centre of the square is halfway along each side of length $4$: $K(2, 2, 0)$. The apex is right above $K$, $3$ units up: $S(2, 2, 3)$.

Skills to practice

Locating Points on the Axes and in the Planes
Reading Coordinates in a Solid
Plotting and Locating Points in Space

II.2 Distance in Space

Discover

On the map of a bay, the origin $O$ is a buoy at sea level, the first axis points east, the second north, the third up, and the unit is the metre. A seagull is at $G(20, 40, 12)$ and a diver at $D(10, 20, -8)$. How far is the seagull from the diver? Why is the distance between the points $(20, 40)$ and $(10, 20)$ of the map not the answer?

Answer

The map distance ignores the difference of heights, $12 - (-8) = 20$ m. Two right-angled triangles give the true distance. From $D$, move $10$ m east to $E(20, 20, -8)$, then $20$ m north to $C(20, 40, -8)$, right below $G$ at the diver's depth.

The triangle $DEC$ is horizontal and right-angled at $E$, since east and north are perpendicular; the triangle $DCG$ is vertical and right-angled at $C$, since $\Segment{CG}$ is vertical and $\Segment{DC}$ horizontal: $$\begin{aligned}[t] DC^2 &= DE^2 + EC^2 && \text{(Pythagoras' theorem in } DEC\text{)}\\ &= 10^2 + 20^2 \\ &= 500,\\ DG^2 &= DC^2 + CG^2 && \text{(Pythagoras' theorem in } DCG\text{)}\\ &= 500 + 20^2 \\ &= 900. \end{aligned}$$ So $DG = \sqrt{900} = 30$ m, while the map distance is only $\sqrt{500} \approx 22.4$ m.

Proposition — Distance Between Two Points

In an orthonormal coordinate system in space, the distance between the points $A(x_A, y_A, z_A)$ and $B(x_B, y_B, z_B)$ is $$AB = \sqrt{(x_B - x_A)^2 + (y_B - y_A)^2 + (z_B - z_A)^2}.$$

Proof

We apply Pythagoras' theorem twice: first in a horizontal right-angled triangle, then in a vertical one. Let $D(x_B, y_A, z_A)$ and $C(x_B, y_B, z_A)$: from $A$, the point $D$ is reached parallel to the $x$-axis, then $C$ parallel to the $y$-axis, and $B$ is right above or right below $C$.

The horizontal triangle $ADC$. The segment $\Segment{AD}$ is parallel to the $x$-axis and $\Segment{DC}$ to the $y$-axis. These axes are perpendicular, so $ADC$ is right-angled at $D$, with $AD = |x_B - x_A|$ and $DC = |y_B - y_A|$.
The vertical triangle $ACB$. The segment $\Segment{CB}$ is parallel to the $z$-axis, which is perpendicular to the $xy$-plane; the points $A$ and $C$ have the same $z$-coordinate $z_A$, so $\Line{AC}$ is horizontal. Hence $ACB$ is right-angled at $C$, with $CB = |z_B - z_A|$.

A square does not depend on the sign of the number squared: $|x_B - x_A|^2 = (x_B - x_A)^2$, and likewise for $y$ and $z$. So $$\begin{aligned}[t] AB^2 &= AC^2 + CB^2 && \text{(Pythagoras' theorem in } ACB\text{)}\\ &= AD^2 + DC^2 + CB^2 && \text{(Pythagoras' theorem in } ADC\text{)}\\ &= (x_B - x_A)^2 + (y_B - y_A)^2 + (z_B - z_A)^2. && \text{(the three lengths)} \end{aligned}$$ A distance is non-negative, so $AB$ is the square root of this sum. When two of the points $A$, $D$, $C$, $B$ coincide, a triangle is flat, the corresponding length is $0$, and the same equalities hold.

Method — Calculating a Distance in Space

To calculate the distance $AB$:

write the three differences of coordinates $x_B - x_A$, $y_B - y_A$ and $z_B - z_A$ — the order of the two points does not matter, since $(x_A - x_B)^2 = (x_B - x_A)^2$;
square them and add;
take the square root, and give the exact value, with the square root simplified, before any rounding.

Example

Let $A(2, -1, 5)$ and $B(5, 3, -7)$. Calculate the distances $AB$ and $OA$, where $O$ is the origin; give exact values.

Answer

$$\begin{aligned}[t] AB &= \sqrt{(x_B - x_A)^2 + (y_B - y_A)^2 + (z_B - z_A)^2} && \text{(distance formula)}\\ &= \sqrt{(5 - 2)^2 + (3 - (-1))^2 + (-7 - 5)^2} && \text{(coordinates of } A \text{ and } B\text{)}\\ &= \sqrt{3^2 + 4^2 + (-12)^2} \\ &= \sqrt{9 + 16 + 144} \\ &= \sqrt{169} \\ &= 13, \end{aligned}$$ $$\begin{aligned}[t] OA &= \sqrt{(2 - 0)^2 + (-1 - 0)^2 + (5 - 0)^2} && \text{(distance formula)}\\ &= \sqrt{4 + 1 + 25} \\ &= \sqrt{30}. \end{aligned}$$ The number $30 = 2 \times 3 \times 5$ has no square factor other than $1$: $\sqrt{30}$ does not simplify.

Skills to practice

Calculating a Distance in Space
Using the Distance Formula in Space

II.3 Midpoint in Space

Discover

The origin $O$ is a buoy at sea level; the first axis points east, the second north, the third up, and the unit is the metre. A boat floats at $B(40, 10, 0)$. Its anchor rests on the sea bed at $A(30, 0, -30)$, and the anchor chain is taut, in a straight line from $B$ to $A$. A float is fixed at the middle $M$ of the chain. Guess the coordinates of $M$, then check the guess with distances.

Answer

Halfway from $B$ to $A$, each coordinate has changed by half as much: the guess is $M(35, 5, -15)$, each coordinate being the mean of those of $B$ and $A$.

Check with the distance formula: $$BA = \sqrt{(-10)^2 + (-10)^2 + (-30)^2} = \sqrt{1100} = 10\sqrt{11},$$ $$BM = \sqrt{(-5)^2 + (-5)^2 + (-15)^2} = \sqrt{275} = 5\sqrt{11}, \qquad MA = \sqrt{(-5)^2 + (-5)^2 + (-15)^2} = 5\sqrt{11}.$$ So $BM + MA = BA$: the point $M$ lies on the chain; and $BM = MA$: it is its middle. The float is at $M(35, 5, -15)$, $15$ m under the surface.

Proposition — Coordinates of the Midpoint

In an orthonormal coordinate system in space, the midpoint $M$ of the segment $\Segment{AB}$, with $A(x_A, y_A, z_A)$ and $B(x_B, y_B, z_B)$, has coordinates $$M\left(\frac{x_A + x_B}{2}, \frac{y_A + y_B}{2}, \frac{z_A + z_B}{2}\right):$$ each coordinate of the midpoint is the mean of the corresponding coordinates of the endpoints.

Proof

Let $M$ be the point with these coordinates. For the first coordinate, $$x_M - x_A = \frac{x_A + x_B}{2} - x_A = \frac{x_B - x_A}{2} \quad\text{and}\quad x_B - x_M = x_B - \frac{x_A + x_B}{2} = \frac{x_B - x_A}{2},$$ and the same computation holds for the second and third coordinates. The distance formula then gives $$\begin{aligned}[t] AM &= \sqrt{\left(\frac{x_B - x_A}{2}\right)^2 + \left(\frac{y_B - y_A}{2}\right)^2 + \left(\frac{z_B - z_A}{2}\right)^2} && \text{(distance formula)}\\ &= \frac{1}{2}\sqrt{(x_B - x_A)^2 + (y_B - y_A)^2 + (z_B - z_A)^2} && \text{(factor } \tfrac{1}{4} = \left(\tfrac{1}{2}\right)^2\text{)}\\ &= \frac{AB}{2}, && \text{(distance formula)} \end{aligned}$$ and in the same way $MB = \dfrac{AB}{2}$. Hence $AM + MB = AB$. The three points $A$, $M$ and $B$ lie in one plane, and by the triangle inequality, $AM + MB = AB$ happens only when the triangle $AMB$ is flat, with $M$ on the segment $\Segment{AB}$. So $M$ is a point of $\Segment{AB}$ with $AM = MB$: it is the midpoint of $\Segment{AB}$.

Example

Calculate the coordinates of the midpoint $M$ of $\Segment{AB}$, with $A(2, -1, 5)$ and $B(5, 3, -7)$.

Answer

By the midpoint formula, each coordinate of $M$ is the mean of those of $A$ and $B$: $$\begin{aligned}[t] x_M &= \frac{x_A + x_B}{2} = \frac{2 + 5}{2} = 3.5,\\ y_M &= \frac{y_A + y_B}{2} = \frac{-1 + 3}{2} = 1,\\ z_M &= \frac{z_A + z_B}{2} = \frac{5 + (-7)}{2} = -1. \end{aligned}$$ So $M(3.5, 1, -1)$.

Example

The point $M(1, -2, 4)$ is the midpoint of $\Segment{AB}$, and $A(3, 1, 7)$. Calculate the coordinates of $B$.

Answer

Each coordinate of $M$ is a mean: $x_M = \dfrac{x_A + x_B}{2}$ gives $x_B = 2x_M - x_A$, and likewise for $y$ and $z$: $$\begin{aligned}[t] x_B &= 2x_M - x_A = 2 \times 1 - 3 = -1,\\ y_B &= 2y_M - y_A = 2 \times (-2) - 1 = -5,\\ z_B &= 2z_M - z_A = 2 \times 4 - 7 = 1. \end{aligned}$$ So $B(-1, -5, 1)$. Check: $\dfrac{3 + (-1)}{2} = 1$, $\dfrac{1 + (-5)}{2} = -2$ and $\dfrac{7 + 1}{2} = 4$ are the coordinates of $M$.

Skills to practice

Calculating the Coordinates of a Midpoint
Finding an Endpoint from the Midpoint
Using Midpoints in Space

III Going further

Skills to practice

Reasoning with Coordinates in Space
Investigating Patterns with Collinear Points and Distances
Solving Real-Life Problems with Collinear Points and Distances
Quiz