Coordinate Geometry

Coordinate geometry describes geometry with numbers. Every point gets an address, an ordered pair (x, y); then a straight line, a circle or any other curve becomes an equation that its points, and only its points, satisfy. A question about a shape (do these lines meet? is this point inside the circle?) becomes a sum.

The French mathematician René Descartes set out this idea, and the Cartesian coordinate system is named after him. Pierre de Fermat found it at the same time on his own. Long before them, the Greek geometer Apollonius of Perga had studied the circle, the ellipse, the parabola and the hyperbola so thoroughly that his book was still the standard work on them two thousand years later. Because it joins algebra and geometry, coordinate geometry is used wherever positions and shapes are computed: in physics, computer graphics, engineering, maps and astronomy.

What you need from Grade 9

This unit builds on these. For two points A(x₁, y₁) and B(x₂, y₂):

Idea	Formula
distance AB	$\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$
midpoint of AB	$\left(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\right)$
slope of AB	$m = \dfrac{y_2 - y_1}{x_2 - x_1}$
slope from the angle θ with the X-axis	m =  tan θ
slope of ax + by + c = 0	$m = -\dfrac{a}{b}$, that is $-\dfrac{\text{coefficient of } x}{\text{coefficient of } y}$
slope–intercept form	y = mx + c (c is where the line cuts the Y-axis)
double-intercept form	$\dfrac{x}{a} + \dfrac{y}{b} = 1$ (cuts the axes at (a, 0) and (0, b))

A line that rises from left to right has a positive slope; one that falls has a negative slope; a level line has slope 0; an upright line (x₁ = x₂) has no slope at all, since the formula would divide by 0.

Two points fix a straight line. If a road on a map runs through two places whose coordinates you know, its equation tells you, for any other place, whether the road passes through it. This section finds that equation, in two steps.

a. Point–slope form

Suppose a line has slope m and passes through a known point Q(x₁, y₁). In slope–intercept form it is

$$y = mx + c \qquad \text{(i)}$$

with some value of c we don’t know yet. Since Q is on the line, its coordinates satisfy (i):

$$y_1 = mx_1 + c \qquad \text{(ii)}$$

Subtract (ii) from (i), and c disappears:

$$\begin{aligned} y - y_1 &= mx + c - (mx_1 + c) \\ &= mx - mx_1 \\ &= m(x - x_1) \end{aligned}$$

So the line with slope m through (x₁, y₁) is

$$y - y_1 = m(x - x_1)$$

This is the point–slope form. Read it as a sentence: for every point (x, y) on the line, the rise from (x₁, y₁) is m times the run.

b. Two-point form

Now suppose we know two points on the line, A(x₁, y₁) and B(x₂, y₂), and let C(x, y) be any other point on it. Three points on one line all have the same slope between them, so

$$\text{slope of } CA = \text{slope of } AB$$

$$\frac{y - y_1}{x - x_1} = \frac{y_2 - y_1}{x_2 - x_1}$$

Multiply both sides by (x − x₁):

$$y - y_1 = \frac{y_2 - y_1}{x_2 - x_1}(x - x_1)$$

This is the two-point form. It is just the point–slope form with the slope worked out from the two points.

Another way: similar triangles

The same formula comes from a picture. A line MN passes through B(x₁, y₁) and C(x₂, y₂), and P(x, y) is any point on it. Drop perpendiculars BG, PR and CD to the X-axis, and draw BE level, meeting PR at F and CD at E.

The sides of the two triangles are differences of coordinates:

$$\begin{aligned} BF &= GR = x - x_1, & PF &= y - y_1 \\ BE &= GD = x_2 - x_1, & CE &= y_2 - y_1 \end{aligned}$$

In similar triangles the sides are in the same ratio, so

$$\frac{PF}{BF} = \frac{CE}{BE}$$

$$\frac{y - y_1}{x - x_1} = \frac{y_2 - y_1}{x_2 - x_1}$$

and again $y - y_1 = \dfrac{y_2 - y_1}{x_2 - x_1}(x - x_1)$. (The ratio “up over across” is exactly the slope, which is why it is the same along the whole line.)

Worked examples

Question 1

Question 2

Question 3

Find the equation of the line in each graph.

Question 4

Find the equation of the straight line through the two points.

Question 5

Prove that the three points are collinear.

Question 6

The points A(b, 0), B(0, c) and C(3, 3) lie on one straight line.

Question 7

The points A(p, 0), B(0, q) and C(5, 5) lie on one straight line.

Question 8

P(1, 0), Q(0, 1) and R(2, 3) are the vertices of △PQR.

Question 9

A(2, 2), B(2, 8) and C(−6, 2) are the vertices of △ABC.

Question 10

X(−1, 3), Y(1, −1) and Z(5, 1) are the vertices of △XYZ.

Question 11

The point A(a, b) lies on the line 6x − y = 1, and B(b, a) on the line 2x − 5y = 5.

Where two roads meet, the angle between them matters: a junction at a small angle is hard to see round, and engineers try to bring roads together nearly at right angles. With coordinates, the angle between two lines comes straight from their slopes.

Take two lines that meet at E: L₁, y = m₁x + c₁, cutting the X-axis at C at an angle θ₁, and L₂, y = m₂x + c₂, cutting it at A at an angle θ₂. Their slopes are the tangents of those angles:

$$m_1 = \tan\theta_1, \qquad m_2 = \tan\theta_2$$

The formula

In △EAC, the angle θ₁ at C is an exterior angle, so it equals the sum of the two interior angles opposite it, θ at E and θ₂ at A:

$$\theta_1 = \theta + \theta_2, \quad\text{so}\quad \theta = \theta_1 - \theta_2$$

Take the tangent of both sides, using the formula for tan (A − B) from Unit 7:

$$\tan\theta = \tan(\theta_1 - \theta_2) = \frac{\tan\theta_1 - \tan\theta_2}{1 + \tan\theta_1\tan\theta_2}$$

$$\tan\theta = \frac{m_1 - m_2}{1 + m_1m_2} \qquad \text{(i)}$$

Two lines that cross make two adjacent angles, θ and $\alpha = 180^\circ - \theta$, side by side on a straight line. Since $\tan(180^\circ - \theta) = -\tan\theta$,

$$\tan\alpha = -\frac{m_1 - m_2}{1 + m_1m_2} \qquad \text{(ii)}$$

Together, (i) and (ii) give the angle between two lines:

$$\tan\theta = \pm\frac{m_1 - m_2}{1 + m_1m_2}, \qquad \theta = \tan^{-1}\left(\pm\frac{m_1 - m_2}{1 + m_1m_2}\right)$$

A positive value of tan θ gives the acute angle; a negative value gives the obtuse one. In practice: work out $\dfrac{m_1 - m_2}{1 + m_1m_2}$, take the acute angle whose tangent is its size, and the obtuse angle is 180∘° minus that.

Parallel lines

Parallel lines make the same angle with the X-axis: θ₁ = θ₂, so tan θ₁ =  tan θ₂:

$$m_1 = m_2$$

The formula agrees: the angle between parallel lines is 0∘°, and $\tan 0^\circ = 0$ makes the top, m₁ − m₂, zero.

Perpendicular lines

If the lines are perpendicular, $\theta_1 = \theta_2 + 90^\circ$. Since $\tan(90^\circ + A) = -\cot A = -\dfrac{1}{\tan A}$,

$$m_1 = \tan(\theta_2 + 90^\circ) = -\frac{1}{\tan\theta_2} = -\frac{1}{m_2}$$

$$m_1m_2 = -1$$

The slopes of perpendicular lines are negative reciprocals of each other: 2 and −1/2, −3/4 and 4/3. In the formula, 1 + m₁m₂ = 0 makes the bottom zero, and tan θ has no value: that happens only at $\theta = 90^\circ$. (This needs both slopes to exist: an upright line and a level line are perpendicular too, but an upright line has no slope.)

Relations between two lines

angle between them	$\theta = \tan^{-1}\left(\pm\dfrac{m_1 - m_2}{1 + m_1m_2}\right)$
parallel	m₁ = m₂
perpendicular	m₁m₂ = −1, that is $m_1 = -\dfrac{1}{m_2}$
parallel to ax + by + c = 0	ax + by + k = 0
perpendicular to ax + by + c = 0	bx − ay + k = 0

Worked examples

Question 1

Question 2

Find the acute angle between the lines.

Question 3

Find the obtuse angle between the lines.

Question 4

Find the angles between the lines.

Question 5

Prove that the lines are parallel.

Question 6

Prove that the lines are perpendicular.

Question 7

The lines are parallel. Find the unknown.

Question 8

The lines are perpendicular. Find the unknown.

Question 9

The lines are px + qy + r = 0 and lx − my + n = 0.

Question 10

Question 11

Question 12

Question 13

Find the perpendicular bisector of the segment joining the points.

Question 14

All four shapes come from one solid: a cone, cut by a flat plane. They are called the conic sections.

8.3.1 Parts of a cone

A right circular cone has a circular base, and its tip, the vertex, is straight above the center of the base.

• The axis is the line from the vertex to the center of the base, at right angles to the base: AB, the cone’s height.
• A generator is a straight line on the cone’s surface from the vertex to the rim: AC, the slant height. Turning AC round the axis sweeps out (generates) the whole surface, hence the name.
• The angle between the axis and a generator, ∠BAC = α, is the semi-vertical angle: half the angle at the vertex.

Double-napped right circular cone

Put two equal cones tip to tip, on one axis, and you get a double-napped right circular cone. Each cone is a nappe: the upper nappe and the lower nappe. Think of the generators as whole straight lines through the vertex, going on in both directions, and they sweep out both nappes at once.

The four sections

Cut a cone with a flat plane that doesn’t pass through the vertex. The cut edge is a curve, the section. Which curve depends only on the angle θ between the plane and the cone’s axis, compared with the semi-vertical angle α.

The plane’s angle with the axis	Section
$\theta = 90^\circ$ (parallel to the base)	circle
$\alpha < \theta < 90^\circ$	ellipse
θ = α (parallel to a generator)	parabola
θ < α (cuts both nappes)	hyperbola

• Circle. A plane at right angles to the axis, that is parallel to the base, cuts the cone in a circle.
• Ellipse. A plane tilted at an angle θ between α and 90∘° cuts right across one nappe in a closed oval, an ellipse.
• Parabola. A plane parallel to a generator (θ = α) cuts one nappe in a curve that never closes: its two arms open out for ever. That is a parabola.
• Hyperbola. A plane steeper than the generators (θ < α) cuts both nappes, so the curve has two separate branches. That is why the double cone is needed: a single cone gives only half of a hyperbola.

Question 1

Question 2

Question 3

Question 4

Define each conic section.

Question 5

Under what condition is each formed? (θ is the angle between the plane and the axis, α the semi-vertical angle.)

A circle can be defined in two ways:

• As a conic section: the curve in which a plane parallel to the base cuts a right circular cone (8.3).
• As a locus: the path of a point that moves so that it is always the same distance from a fixed point. The fixed point O is the center, the fixed distance OP = r the radius.

The second is the one that gives the equation: “a point P(x, y) is on the circle” means “OP = r”, and the distance formula turns that into an equation in x and y.

8.4.1 Equation of a circle

a. Center at the origin, radius r

Let P(x, y) be any point on the circle with center O(0, 0) and radius r. By the distance formula,

$$OP = \sqrt{(x - 0)^2 + (y - 0)^2} = r$$

Square both sides:

$$x^2 + y^2 = r^2$$

b. Center (h, k), radius r

Now let the center be C(h, k), and P(x, y) any point on the circle. CP = r, so $CP^2 = r^2$:

$$(x - h)^2 + (y - k)^2 = r^2$$

This is the center–radius form. With h = k = 0 it is $x^2 + y^2 = r^2$ again.

c. The endpoints of a diameter are (x₁, y₁) and (x₂, y₂)

Let AB be a diameter, with A(x₁, y₁) and B(x₂, y₂), and P(x, y) any other point of the circle. The angle in a semicircle is a right angle, so $\angle APB = 90^\circ$: $AP \perp BP$.

The slopes of AP and BP are $m_1 = \dfrac{y - y_1}{x - x_1}$ and $m_2 = \dfrac{y - y_2}{x - x_2}$. Perpendicular, so m₁m₂ = −1:

$$\begin{aligned} \frac{y - y_1}{x - x_1} \times \frac{y - y_2}{x - x_2} &= -1 \\ (y - y_1)(y - y_2) &= -(x - x_1)(x - x_2) \end{aligned}$$

$$(x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0$$

(When P is A or B itself, the slopes can’t be formed, but A and B plainly satisfy the final equation: one bracket in each product is 0. So the equation holds for every point of the circle.)

d. The general form

Multiply out the center–radius form:

$$\begin{aligned} (x - h)^2 + (y - k)^2 &= r^2 \\ x^2 - 2hx + h^2 + y^2 - 2ky + k^2 &= r^2 \\ x^2 + y^2 + 2(-h)x + 2(-k)y + h^2 + k^2 - r^2 &= 0 \end{aligned}$$

Write g = −h, f = −k and $c = h^2 + k^2 - r^2$:

$$x^2 + y^2 + 2gx + 2fy + c = 0$$

This is the general form of the equation of a circle. Going back, h = −g, k = −f and $r^2 = h^2 + k^2 - c$:

$$\text{center } (-g, -f), \qquad \text{radius } r = \sqrt{g^2 + f^2 - c}$$

So an equation is a circle’s when:

• it is of degree 2 in x and y;
• the coefficients of $x^2$ and $y^2$ are equal (divide by them first if they aren’t 1);
• it has no xy term;
• $g^2 + f^2 - c$ is positive, so that there is a radius.

To read off the center, halve the coefficients of x and y and change their signs: $x^2 + y^2 - 6x + 4y - 3 = 0$ has 2g = −6, 2f = 4, so the center is (3, −2) and $r = \sqrt{9 + 4 + 3} = 4$.

Circles in special places

Where the circle is	Equation
center at the origin	$x^2 + y^2 = r^2$
center (h, k)	$(x - h)^2 + (y - k)^2 = r^2$
diameter from (x₁, y₁) to (x₂, y₂)	(x − x₁)(x − x₂) + (y − y₁)(y − y₂) = 0
touches the X-axis ($r = \lvert k \rvert$)	$(x - h)^2 + (y - k)^2 = k^2$
touches the Y-axis ($r = \lvert h \rvert$)	$(x - h)^2 + (y - k)^2 = h^2$
touches both axes ($r = \lvert h \rvert = \lvert k \rvert$)	$(x - h)^2 + (y - k)^2 = r^2$, h = ±r, k = ±r

A circle that touches the X-axis meets it at one point, straight below (or above) the center, so the radius there is the center’s height above the axis, |k|.

Worked examples

Back to the earthquake

Now the opening question can be answered. Three stations record an earthquake (the numbers are made up; 1 unit = 10 km):

Station	Position	Distance to the start
A	(0, 0)	5
B	(10, 11)	10
C	(−8, −2)	13

Question 1

Question 2

Find the equation of the circle with the given center and radius.

Question 3

Find the center and radius of each circle.

Question 4

Find the circle whose diameter has the given ends.

Question 5

The equation of a circle is $x^2 + y^2 - 4x + 6y - 12 = 0$.

Question 6

The equation of a circle is $x^2 + y^2 - 6x - 8y - 11 = 0$.

Question 7

The equation of a circle is $3x^2 + 3y^2 + 6x - 3y + 3 = 0$.

Question 8

Find the circle.

Question 9

The circle passes through the origin and cuts the given lengths from the positive axes. Find it.

Question 10

Find the circle.

Question 11

Find the circle through the three points.

Question 12

Question 13

Question 14

A round flower bed is made in the middle of the school playground. On a grid in meters, its center is (2, −1) and its radius is 7 m.

In the SEE’s Optional Mathematics paper (75 marks, 3 hours), Coordinate Geometry and Transformation (Unit 9) together make up Geometry. CDC’s specification grid gives Geometry 20 marks, in three kinds of question:

Group	Kind of question	Geometry’s marks	Levels
A	multiple choice, 1 mark each	3	knowledge 1, application 1, higher ability 1
B	within the content area: questions on Geometry alone, each of 2 to 4 parts	11	knowledge 2, understanding 3, application 6
C	across content areas: one question joining Geometry with another area	6	understanding 2, application 1, higher ability 3

In CDC’s model paper this unit gives a 1-mark choice (the diameter of a circle $x^2 + y^2 = r^2$); in group B, the acute angle that a line through two points makes with a given line (tan θ, 2 marks), and a circle in general form: the center–radius form (1 mark), its center and radius (2 marks), and the equation of a radius by the two-point form (2 marks); and in group C, which conic section a plane cutting a cone makes at a given angle.

What each kind asks, and where its marks are:

• A line’s equation: through two points, parallel or perpendicular to a line, through the meeting point of two lines, a median or a perpendicular bisector. A mark for the slope, a mark for the equation.
• The angle between two lines: both slopes, then tan θ and the angle; or an unknown from “parallel” or “perpendicular”.
• Conic sections: naming a section from the angle of the plane, or the condition for one.
• The circle: an equation from its center and radius, or a diameter; the center and radius from the general form; a circle through given points, touching the axes, or concentric with another.

Here is a question of the longer kind, written for this book, with how it is marked.