Geometry

The points in the activity all obey one rule, “y is one more than x”, and together they make a straight line. A set of points that obey a given rule, and only those points, is called a locus (plural loci, Latin for “place”). Here the locus is the line, and its equation of the locus is y = x + 1: a point is on the line exactly when its coordinates make the equation true.

So a locus can be given in three ways that say the same thing: as a rule in words, as a shape, or as an equation.

A circle as a locus

Take a fixed point C and a length r. The points P that are exactly r from C make a circle: C is its center and r its radius. In words, the circle is the locus of a point that moves so that it is always r from C.

Let C be (h, k) and P any point (x, y) on the circle. By the distance formula,
$$r = \sqrt{(x - h)^2 + (y - k)^2}$$
Squaring both sides,
$$(x - h)^2 + (y - k)^2 = r^2$$
This is the equation of the circle, the equation of the locus. A point is on the circle exactly when its coordinates make this true.

Examples

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Two points A(3, 2) and B(7, −4) are given, and P(x, y) moves. Find the equation of the locus of P when:

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Question 15

A straight path runs from a school at A to a bus stop at B, 5 km long. A tea shop stands 2 km from A and 3 km from B. We say the tea shop divides AB in the ratio 2 : 3. If you know the coordinates of A and B, can you work out the tea shop’s coordinates, without measuring anything? That is what this section’s formula does.

A point P divides a segment AB in the ratio m₁:m₂ when AP:PB = m₁:m₂, that is, AP = km₁ and PB = km₂ for some length k. There are two ways it can happen.

• Internally: P lies between A and B. In the tea shop’s case AP = 2 km, PB = 3 km and AB = 5 km.
• Externally: P lies on the line AB but outside the segment, beyond one end. For example, X, R and Y on a line with XR = 10 cm, RY = 15 cm and XY = 25 cm: Y divides XR externally in the ratio XY:RY = 25:15 = 5:3.

The section formula

Let A(x₁, y₁) and B(x₂, y₂) be two points, and let P(x, y) divide AB internally in the ratio m₁:m₂. We want x and y.

Draw $AD \perp OX$, $PF \perp OX$, $BE \perp OX$, $AM \perp PF$ and $PN \perp BE$. Read the lengths of the small triangles’ sides from the coordinates:

• AM = DF = OF − OD = x − x₁ and PN = FE = OE − OF = x₂ − x
• PM = PF − MF = y − y₁ and BN = BE − NE = y₂ − y

The triangles AMP and PNB are similar: each has a right angle, and their slanting sides AP and PB lie on the same line, so they make the same angle with the horizontal. In similar triangles, corresponding sides are in the same ratio, here $\frac{AP}{PB} = \frac{m_1}{m_2}$. So

$$\frac{AM}{PN} = \frac{m_1}{m_2}: \quad \frac{x - x_1}{x_2 - x} = \frac{m_1}{m_2}$$

Cross-multiply: m₂x − m₂x₁ = m₁x₂ − m₁x, so m₁x + m₂x = m₁x₂ + m₂x₁ and
$$x = \frac{m_1x_2 + m_2x_1}{m_1 + m_2}$$
In the same way $\frac{PM}{BN} = \frac{m_1}{m_2}$ gives $\frac{y - y_1}{y_2 - y} = \frac{m_1}{m_2}$, and
$$y = \frac{m_1y_2 + m_2y_1}{m_1 + m_2}$$
This is the section formula:
$$P(x, y) = \left(\frac{m_1x_2 + m_2x_1}{m_1 + m_2}, \frac{m_1y_2 + m_2y_1}{m_1 + m_2}\right)$$

Special case A: the midpoint

When P is the midpoint, AP = PB, so m₁:m₂ = 1:1. The formula becomes
$$P(x, y) = \left(\frac{1 \cdot x_2 + 1 \cdot x_1}{1 + 1}, \frac{1 \cdot y_2 + 1 \cdot y_1}{1 + 1}\right)$$
$$= \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)$$
the midpoint formula: the midpoint’s coordinates are the averages of the ends’.

Special case B: dividing externally

Now let P(x, y) be on the line AB beyond B, with AP:BP = m₁:m₂ (here m₁ > m₂, since P is further from A).

Draw $AL \perp OX$, $BM \perp OX$, $PN \perp OX$, $AR \perp PN$ and $BS \perp PN$. Then

• AR = LN = ON − OL = x − x₁ and BS = MN = ON − OM = x − x₂
• PR = PN − RN = y − y₁ and PS = PN − SN = y − y₂

The right-angled triangles ARP and BSP are similar (they share the angle at P), so $\frac{AR}{BS} = \frac{AP}{BP} = \frac{m_1}{m_2}$:
$$\frac{x - x_1}{x - x_2} = \frac{m_1}{m_2}$$
m₂x − m₂x₁ = m₁x − m₁x₂, so m₁x₂ − m₂x₁ = x(m₁ − m₂), and $x = \frac{m_1x_2 - m_2x_1}{m_1 - m_2}$. In the same way, $\frac{PR}{PS} = \frac{m_1}{m_2}$ gives y. So:
$$P(x, y) = \left(\frac{m_1x_2 - m_2x_1}{m_1 - m_2}, \frac{m_1y_2 - m_2y_1}{m_1 - m_2}\right)$$
It is the section formula with m₂ replaced by −m₂. So when you work out a ratio and it comes out negative, as −3:2 or 1: − 2, the point divides the segment externally.

Examples

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Find the point dividing the segment joining the given points internally in the given ratio.

Question 4

Find the point dividing the segment joining the given points externally in the given ratio.

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Question 12

Find the points that divide the segment joining the given points into three equal parts.

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Question 17

Prove that the given points are the vertices of a parallelogram.

Question 18

Three consecutive vertices of a parallelogram are given. Find the fourth.

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Straight lines

Straight lines come in three kinds: lines parallel to the X-axis (horizontal), lines parallel to the Y-axis (vertical), and lines that cross the X-axis at an angle.

A. Slope or gradient

Look at a ladder leaning on a wall at home. It isn’t upright and it isn’t lying flat: it is tilted. How much it is tilted is its inclination, an angle θ at its foot between the ground and the ladder. Treat the ground as the X-axis and the wall as the Y-axis. Then the ladder is a piece of a line AB, and:

• OA, the part of the X-axis from the origin to where the line meets it, is the line’s x-intercept;
• OB, the part of the Y-axis from the origin to where the line meets it, is its y-intercept.

The angle θ that a line makes with the positive direction of the X-axis, measured anticlockwise from OX, is its inclination, and tan θ is its slope or gradient, written m:
$$m = \tan\theta$$
The slope says how many units the line goes up for every one unit to the right. A slope of 2 climbs 2 for each 1 across; a slope of $\frac{1}{4}$ climbs 1 for each 4 across; a negative slope goes down as you go right.

B. The slope of the line through two points

Let a line pass through P(x₁, y₁) and Q(x₂, y₂) and meet the X-axis at A, with inclination ∠PAX = θ.

Draw $PM \perp OX$, $QN \perp OX$ and $PR \perp QN$. Then

• PR = MN = ON − OM = x₂ − x₁ (the run, how far across)
• QR = QN − RN = QN − PM = y₂ − y₁ (the rise, how far up)

PR is parallel to the X-axis, so ∠QPR = ∠PAX = θ (corresponding angles). In the right-angled triangle PRQ,
$$m = \tan\theta = \frac{QR}{PR} = \frac{y_2 - y_1}{x_2 - x_1}$$
The slope is rise ÷ run, the change in y divided by the change in x, between any two points of the line.

C. The x-intercept and the y-intercept

If a line AB meets the X-axis at A(a, 0) and the Y-axis at B(0, b), then OA = a and OB = b: x-intercept = a and y-intercept = b. Each is a number with a sign: a line meeting the Y-axis below the origin has a negative y-intercept.

To find the intercepts from an equation, put y = 0 to find where it meets the X-axis, and x = 0 for the Y-axis. For 3x − 2y − 4 = 0: y = 0 gives $x = \frac{4}{3}$, and x = 0 gives y = −2.

Examples

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Find the slopes of these lines.

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A straight line is a locus too: its points obey one rule. Depending on what you know about a line (a point it passes through, its slope, where it cuts the axes, or how far it is from the origin), the rule is easiest to write in a different form. Here are the five you need.

1. A line parallel to the Y-axis

Let AB be parallel to the Y-axis, a units from it, to its right. Every point of AB, wherever it is up or down the line, has the same x-coordinate, a, and no other point has. So the equation of AB is
$$x = a$$
The equation says nothing about y, because y can be anything.

2. A line parallel to the X-axis

In the same way, if AB is parallel to the X-axis and b units above it, every point of AB has y-coordinate b, and its equation is
$$y = b$$
A line b units below the X-axis is y = −b. The X-axis itself is y = 0, and the Y-axis x = 0.

3. Slope-intercept form

Let the line AB make an angle θ with the positive X-axis, so its slope is m =  tan θ, and let it cut the Y-axis at C(0, c), so its y-intercept is c. Take any point P(x, y) on it. Draw $PM \perp OX$ and $CN \perp PM$. Then ∠PCN = ∠BAX = θ (CN is parallel to the X-axis), CN = OM = x and PN = PM − NM = y − c.

In the right-angled triangle PNC:
$$\tan\theta = \frac{PN}{CN}: \quad m = \frac{y - c}{x}$$
so mx = y − c, and
$$y = mx + c$$
This is the slope-intercept form: m is the slope and c the y-intercept, both read straight off the equation.

Another way: C(0, c) and P(x, y) are both on AB, and a line has the same slope between any two of its points (why? the triangles of rise and run between any two points of one line are similar). So $m = \frac{y - c}{x - 0}$, which gives y = mx + c again.

Two special cases:

• a. If the line passes through the origin, c = 0: y = mx.
• b. If the line is parallel to the X-axis, $m = \tan 0^\circ = 0$: y = 0 · x + c, so y = c.

4. Double-intercept form

Let a line AB cut the X-axis at A(a, 0) and the Y-axis at B(0, b), so its intercepts are a and b, and let P(x, y) be any point on it. A, P and B are collinear (on one line), so the slope from A to P equals the slope from P to B:
$$\frac{y - 0}{x - a} = \frac{b - y}{0 - x}$$
Cross-multiply: −xy = (b − y)(x − a) = bx − ab − xy + ay. The −xy cancels: bx + ay = ab. Divide both sides by ab:
$$\frac{x}{a} + \frac{y}{b} = 1$$
This is the double-intercept form (or intercept form). Its use: the two intercepts are read straight off it, so the line is easy to draw by marking (a, 0) and (0, b) and joining them. A line through the origin, or parallel to an axis, has no such form.

5. Normal (perpendicular) form

Let the line AB cut the axes at A(a, 0) and B(0, b), and draw the perpendicular OD from the origin to it. Let OD = p and ∠DOA = α.

In the right-angled triangle ODA, $\cos\alpha = \frac{OD}{OA} = \frac{p}{a}$, so $a = \frac{p}{\cos\alpha}$.
In the right-angled triangle ODB, $\angle BOD = 90^\circ - \alpha$, so $\angle OBD = 180^\circ - 90^\circ - (90^\circ - \alpha) = \alpha$, and $\sin\alpha = \frac{OD}{OB} = \frac{p}{b}$, so $b = \frac{p}{\sin\alpha}$.

Put these into the intercept form $\frac{x}{a} + \frac{y}{b} = 1$:
$$\frac{x\cos\alpha}{p} + \frac{y\sin\alpha}{p} = 1$$
so
$$x\cos\alpha + y\sin\alpha = p$$
This is the normal form (perpendicular form), where p, the length of the perpendicular from the origin, is never negative, and α is the angle from the positive X-axis to that perpendicular (0∘° to 360∘°).

Go back to the widget “One line, three ways to see it” in 3.1.3 and try its “Intercepts” and “Normal” buttons. Drag A and B so the line passes through the origin: what happens to the intercepts, and to p? And make it vertical: which forms still work?

Examples

Questions 1 to 7

Question 8

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Question 12

Find the slope and the y-intercept of each line.

Question 13

Find the equation of the line:

Question 14

Find the equation of the line with:

Question 15

Find the x-intercept and y-intercept of:

Question 16

Find the equation of the line that:

Question 17

Question 18

The line AB meets the X-axis at A(6, 0) and the Y-axis at B(0, 8).

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Question 21

Find the equation of the line, where p is the length of the perpendicular from the origin and α the angle it makes with the positive X-axis:

Every straight line has an equation of the first degree,
$$Ax + By + C = 0$$
where A, B and C are numbers (A and B not both 0). This is the general form; it is also written ax + by + c = 0. It covers every line, even vertical ones (B = 0) and lines through the origin (C = 0). But the slope, the intercepts and p are hidden in it. Here is how to bring each out.

A. To slope-intercept form

By = −Ax − C, so $y = -\frac{A}{B}x - \frac{C}{B}$ (when B ≠ 0). Compare with y = mx + c:
$$m = -\frac{A}{B} = -\frac{\text{coefficient of } x}{\text{coefficient of } y}$$
$$c = -\frac{C}{B} = -\frac{\text{constant}}{\text{coefficient of } y}$$

B. To double-intercept form

Ax + By = −C. Divide both sides by −C (when C ≠ 0): $\frac{Ax}{-C} + \frac{By}{-C} = 1$, that is
$$\frac{x}{-C/A} + \frac{y}{-C/B} = 1$$
$$a = -\frac{C}{A} = -\frac{\text{constant}}{\text{coefficient of } x}$$
$$b = -\frac{C}{B} = -\frac{\text{constant}}{\text{coefficient of } y}$$

C. To normal form

We want a number k so that multiplying Ax + By + C = 0 by k gives exactly x cos α + y sin α − p = 0:
$$x\cos\alpha + y\sin\alpha - p = kAx + kBy + kC$$
So cos α = kA …(ii), sin α = kB …(iii) and −p = kC. Square (ii) and (iii) and add: $\cos^2\alpha + \sin^2\alpha = k^2(A^2 + B^2)$. The left side is 1, so
$$k^2 = \frac{1}{A^2 + B^2}, \qquad k = \pm\frac{1}{\sqrt{A^2 + B^2}}$$
Then
$$\cos\alpha = \frac{\pm A}{\sqrt{A^2 + B^2}}, \quad \sin\alpha = \frac{\pm B}{\sqrt{A^2 + B^2}}$$
$$p = \frac{\mp C}{\sqrt{A^2 + B^2}}$$
p is a length, so take the sign that makes p positive: move the constant to the right side, make it positive, then divide everything by $\sqrt{A^2 + B^2}$.

Question 1

Write in slope-intercept form.

Question 2

Write in double-intercept form.

Question 3

Write in normal form.

Question 4

The equation of a line is x − y + 4 = 0.

Question 5

The equation of a line is 4x + 3y − 24 = 0.

Question 6

The equation of a line is 6x + 8y = 24.

Question 7

The equation of a line is 3x + 4y = 12.

Now the health post. Thulogaun is at A(1, 5) and Sanogaun at B(7, 1), in kilometers, and the highway is the X-axis.

Every point equally far from both villages. Let P(x, y) be such a point. $PA^2 = PB^2$:
$$(x - 1)^2 + (y - 5)^2 = (x - 7)^2 + (y - 1)^2$$
−2x + 1 − 10y + 25 = −14x + 49 − 2y + 1, so 12x − 8y − 24 = 0, and dividing by 4,
$$3x - 2y - 6 = 0$$
This is the perpendicular bisector of AB: its slope is $\frac{3}{2}$, and it passes through the midpoint of AB, (4, 3) (12 − 6 − 6 = 0 ✓).

The point on the highway. On the highway y = 0: 3x − 6 = 0, x = 2. The health post goes at (2, 0), 2 km along the highway from the origin. Check: $PA = \sqrt{1^2 + 5^2} = \sqrt{26}$ and $PB = \sqrt{5^2 + 1^2} = \sqrt{26}$, about 5.1 km each. ✓

In fact (2, 0) is the x-intercept of the locus: in intercept form, 3x − 2y = 6 is $\frac{x}{2} + \frac{y}{-3} = 1$. Every tool of 3.1 has been used once: a locus, a midpoint, a slope and the intercepts.

Look in a mirror and your image looks back from behind the glass. The minute hand of a clock turns 90∘° clockwise in 15 minutes. You slide a book from one corner of a table to another. Each time, something keeps its shape but moves by a fixed rule. A transformation is a change in the position, the size, or both, of a shape, made by a definite rule. The shape you start with is the object (or pre-image), and what it becomes is its image.

A photo app on a phone does transformations all day. “Flip” is a reflection, “Rotate” a turn of 90∘°, and pinching to zoom an enlargement, each applied to every one of the photo’s millions of points by one rule.

Types of transformation

When a closed door is opened, its position changes. Does its size change? When a balloon is blown up, its size changes. So a transformation may change only the position of a shape, or its size too. There are two types.

• Isometric transformation (“same measure”): only the position changes; the shape and size stay the same, so the object and its image are congruent. Reflection, rotation and translation are isometric.
• Non-isometric transformation: the size of the image changes. Enlargement (and reduction) is non-isometric; the image has the same shape, so it is similar to the object, but not the same size.

A translation slides every point the same distance in the same direction, as the book slid across the table: P(x, y) → P′(x + a, y + b). For example, the translation “5 to the right and 3 down” sends P(x, y) to P′(x + 5, y − 3), so A(1, 2) goes to A′(6, −1). Grade 10 studies translation fully, with vectors: Translation of geometrical shapes.

In the figure, △ABC with A(4, 4), B(2, 1) and C(6, 1) is reflected in the X-axis. Measure AQ and A′Q, BP and B′P, CR and C′R, and the angles AQX, BPX and CRX. Do you and your friends get the same?

AQ = A′Q = 4 units, BP = B′P = 1 unit and CR = C′R = 1 unit: each point and its image are the same distance from the mirror line, the axis of reflection. And $\angle AQX = \angle BPX = \angle CRX = 90^\circ$: the line joining each point to its image is perpendicular to the axis. The corresponding sides and angles of △ABC and △A′B′C′ are equal, so $\triangle ABC \cong \triangle A'B'C'$.

Compare the coordinates:

The x-coordinates stay the same and the y-coordinates change sign. A point on the X-axis, such as (5, 0), stays where it is.

From this, the properties of reflection:

1. The object and its image are the same distance from the axis of reflection.
2. The line joining a point and its image is perpendicular to the axis of reflection.
3. The object and its image are congruent.
4. The image is inverted (flipped over): going round the object A → B → C one way, the image A′ → B′ → C′ goes round the other way, as your right hand in a mirror looks like a left hand.
5. Points on the axis of reflection are invariant (they don’t move).

a. Reflection in the line y = x

What kind of line is y = x? Every point on it has equal coordinates: (0, 0), (1, 1), (−2, −2)… It passes through the origin and through the first and third quadrants, at 45∘° to both axes.

b. Reflection in the line y = −x

The line y = −x is made of points whose coordinates are equal in size but opposite in sign: (0, 0), (1, −1), (−2, 2)… It passes through the origin and the second and fourth quadrants.

c. Reflection in the line x = a

The line x = a is parallel to the Y-axis, a units from it: every point on it has x-coordinate a, while y can be anything.

d. Reflection in the line y = b

The line y = b is parallel to the X-axis, b units from it: every point on it has y-coordinate b.

All six reflections together:

The X-axis and Y-axis are the cases b = 0 and a = 0 of the last two.

Examples

Question 1

Question 2

Question 3

Find the images of A(1, 1), B(−3, 0), C(4, 2), D(−5, −3) and E(2, −3) when they are reflected in:

Question 4

A(−2, 0), B(6, 2) and C(5, 3) are the vertices of △ABC. Find the image △A′B′C′ when △ABC is reflected in:

Question 5

P(−1, 3), Q(−3, −1), R(3, −4) and S(2, 1) are the vertices of a quadrilateral PQRS.

Question 6

Find the axis of reflection.

Question 7

Question 8

Question 9

A(2, 5), B(−3, 3) and C(1, −4) are the vertices of △ABC, and the image of A is A′(2, −5).

A rotation turns a shape about a fixed point, the center of rotation, through a given angle. An angle measured anticlockwise is positive (+90∘° is a positive quarter turn), and clockwise is negative (−90∘° is a negative quarter turn), the same rule as for angles in trigonometry. A rotation through θ about a center O is written R[O, θ].

The hands of a clock, a door on its hinges, the wheel of a bicycle, and a photo turned on a phone all rotate. A clock’s hands turn clockwise, so in 15 minutes the minute hand makes a rotation of −90∘°.

a. Rotation through +90° about the origin

In the figure, △ABC is rotated through +90∘° (a positive quarter turn) about the origin O(0, 0) to give △A′B′C′.

Discuss these with your friends:

• a. Has every vertex turned through the same angle in the same direction? Yes: $\angle AOA' = \angle BOB' = \angle COC' = 90^\circ$, all anticlockwise.
• b. Join AA′, BB′ and CC′ and draw their perpendicular bisectors. Do they meet at one point? Yes, at the center, O.
• c. Is △ABC congruent or similar to △A′B′C′? Congruent: its corresponding sides and angles are equal. A rotation only turns the triangle.
• d. How are the coordinates related? A(2, 4) → A′(−4, 2), B(5, 1) → B′(−1, 5), C(−3, 2) → C′(−2, −3). The coordinates swap places and the new x changes sign:

$$R[O, +90^\circ]: \; P(x, y) \to P'(-y, x)$$

The properties of rotation:

1. Every point of the shape turns through the same angle in the same direction.
2. The perpendicular bisector of the segment joining any point to its image passes through the center of rotation.
3. The object and its image are congruent (and, unlike a reflection, not inverted: A → B → C goes round the same way as A′ → B′ → C′).
4. Only the center of rotation is invariant (unless the rotation is a full turn).

b. Rotation through −90° about the origin

In the figure, A(3, 4) is turned through −90∘° (a negative quarter turn, clockwise) about the origin to A′(4, −3). The coordinates swap places and the new y changes sign:
$$R[O, -90^\circ]: \; P(x, y) \to P'(y, -x)$$
For the same reason as before, −90∘° is the same as +270∘°.