Mathematics Competition Training Class Notes Coordinates ...ylmass.edu.hk/~mathsclub/MCTCN/cge.pdf · Mathematics Competition Training Class Notes Coordinates Geometry 109 the vertical

Mathematics Competition Training Class Notes Coordinates Geometry

108

CCoooorrddiinnaatteess GGeeoommeettrryy Precaution: This chapter requires some knowledge in Elementary Geometry.

Coordinates System

Locating a point: Introduction the coordinates system

In mathematics and daily life we often need to describe the position of an object. But

mostly the description is imprecise, for example:

How would you describe the location of the gray circle? It is above the red line and on

the right of the blue line. But there is a problem: this description occupies a large area,

but the circle is only a small object inside it. That means this method is not suitable to

find out the actual position of an object. One solution is to tell exactly how far the

object is from the referencing ones. For example, the gray circle is 0.8 cm above the

red line and 1.3 cm on the right of the blue line. This will determine a unique point on

the plane and thus the object is accurately positioned. This is an example of

coordinates system. In a coordinates system defined on a plane, a point can be

defined precisely by a pair of numbers. This pair of numbers is called the coordinates

of the point. In our example above the system is called a rectangular coordinates

system. This system is constructed by followings:

1. Create a horizontal real line. Identify the zero on it, and make the positive side

points right.

2. Create another vertical real line that its zero matches the horizontal one. Make

the positive side points up.

The horizontal real line is called the x-axis and is marked with a letter “x”. Similarly,

y

0

1

2

1 2 3 4 5 -1

x

(2, 1)


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the vertical real line is called the y-axis. The intersection of the two axes is called the

origin, which is marker with a letter “O”. In this system if we place an object, then we

can move this horizontally until reaching the y-axis. The point it coincide with the

y-axis is called the ordinate or y-coordinate of the object. Likewise, we can move the

object vertically and arrive at the x-axis. The point is called the abscissa or

x-coordinate. For the figure on the last page, the x-coordinate of black circle is 2 and

the y-coordinate is 1. If the x- and y-coordinate of a point is ζ and η respectively, then

the coordinates of this point are (ζ, η).

Conversely, if we consider the coordinates system is the set of all points on a plane,

then this set is defined as {(x, y) | x, y ∈ R}. However, this is also the definition of R2.

Therefore, the set R2 and the coordinates system is actually the same thing. From now

on, when we say R2 it means the coordinates system unless otherwise specified.

The coordinates system can be divided into four parts by the two axes. Each part is

called a quadrant. As shown below, the part {(x, y) | x, y ∈ R+} is called quadrant I,

{(x, y) | x ∈ R–, y ∈ R+} is called quadrant II, etc. The axes are not included in any

quadrant.

Application of the coordinates system*

The coordinates system is useful in solving geometrical problems and algebraic

equations. The first use is obvious. We can plug the geometric figures into the system

directly. And for algebraic equations, we can draw curves about the equations and find

the corresponding intersections in the system (to be discussed in the next section).

I II

III IV


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Line Segment

Curves on the coordinates system

Curves are one-dimensional objects that is “bent” from a line or line segment. Circles,

lines are examples of curves.

Example of curves

In coordinates system, the shape of a curve can be described by collection of points.

For example, the red points below are some points that can make up a circle, and the

green curve is the “circle” created by infinite points like that.

x

y

0 0.5 1

0.5

1

We can say a curve in the coordinates systems is a set of points that satisfies some

predefined conditions. That means a curve C is a subset of R2 that is equals to

( ) ( ){ }2, , 0x y f x y∈ =ℝ for some function f: R2 → R. Since only f is important in

how the curve is shaped, this curve is usually just denoted “f(x, y) = 0”. The following

show two curves. The purple one is y = x (that is the same as x – y = 0) and the green

one is y = x2.


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x

y

-2 0 2

2

4

Line segments

Line segments are a special kind of curve that is not “curved” at all, and the length is

finite.

Examples of line segments

It is guaranteed that the shortest path that an object travels from one point to another

is a line segment and there is only one unique line segment for this. That means two

points on the coordinates system are enough to fix a line segment. Problems involving

line segments are usually “how long is that segment?” or “how steep is that segment?”

or “what is the area of a polygon enclosed by n segments?” or something like these.

In this section we will go in depth answering these questions.

Length of segment

(12.1)

In the figure there is a line segment ended with two points (x1, y1) and (x2, y2). The

length of the segment is easy to find. As figured we assume the segment is the

hypotenuse of a right triangle as shown. Then by Pythagoras’ theorem we have:

( ) ( )2 2

1 2 1 2Length of segment x x y y= − + − (12.2)

(x2, y2)

(x1, y1)

x1 – x2

y1 – y2


112

Example 12a: Show that the points (2, 1), (3, 8) and (5, 2) forms a right triangle.

Solution 12a:

Let A = (2, 1), B = (3, 8) and C = (5, 2). By (11.2), we have got:

50

40

10

AB

BC

CA

=

=

=

(12.3)

Therefore, by the converse of Pythagoras’ theorem, ∆ABC is a right triangle. Q.E.D.

Slope of segment

A measure of steepness is to find out the rise in one horizontal step for a segment. We

define this measure to be the slope of the segment. In the following, the red segment

has slope 0.5, the blue has slope 1 and green has slope 3.

However, the horizontal distance of the two end points may not be 1. To solve this, we

can shrink or enlarge the segment to have horizontal distance to be 1. If the original

horizontal distance is X, then the rise will be divided by X after the dilation. Referring

to the notation in (11.1), the slope of a line segment is defined as:

1 2

1 2

y y

x x

−

− (12.4)

But the equation may give negative value. By definition, a negative slope means the

segment is “falling”:

If the slope of a segment is 0, then it is “flat”, i.e., horizontal. The slope for a vertical

segment is not defined, because there can never be any “horizontal step”.

1

1

2

3

Positive / Rise

Negative / Fall

Slope > 0

Slope < 0


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Point of division

Given a line segment we can always cut it into two different segments. The place of

the cut is called the point of division. A common question is, “how to determine the

point of division if I want to divide the segment into ratio r : s?”. This can be easily

done with help of coordinate geometry.

Example 12b: Find out the formula for determining the point of division.

Solution 12b:

First of all, observe that the ratio does not change if we project the segment onto

either axis. Thus, we may find out the point of division of the projection on the two

axes to determine the coordinates of the point of division of the segment.

Now assume the end points of the green segment are (x1, y1) and (x2, y2), and the point

of division is (x0, y0). So, on the blue segment, the point x0 divides the interval [x1, x2]

into ratio r : s. This means 0 1

2 1

x x r

x x r s

−=

− + and 2 0

2 1

x x r

x x r s

−=

− +. Solving both

equations give 2 1

0

rx sxx

r s

+=

+. Similarly, 2 1

0

ry syy

r s

+=

+. Therefore:

The point 2 1 2 1,rx sx ry sy

r s r s

+ + + +

divides the segment

Connecting (x1, y1) and (x2, y2) into ratio r : s. (12.5)

The formula (12.5) is known as the section formula. In particular, if the point divides

the segment into ratio 1:1, i.e., the point is the midpoint of the segment, then its

coordinates are:

1 2 1 2,2 2

x x y y+ +

(12.6)

This is called the midpoint formula.

External division

Till now we can only cut the segment inside it into ratio r : s. But we can also “cut”

the segment outside, and also result in ratio r : s. Consider the segment connecting the

r : s

r : s

r : s


114

origin and A = (2, 0). For a point P that 1

3

AP

OP= , the coordinates of P can be

3,0

4

by section formula. But P can also be (3, 0) and the ratio is still 1 : 3. Such a point that

“cut” the segment outside it is called an external point of division. Likewise, the one

that cut inside it is an internal point of division. To classify the external and internal

point of division, we define the external one divides a negative ratio. In our example,

the P divides the segment into ratio 1 : -3. With this definition, the section formula

holds for external point of division as well.

Example 12c: Given two points A (5, 6) and B (-2, 5). If a point P (x, -1) divides the

segment AB into ratio r : s with r < s, find:

� r : s

� x

Solution 12c:

Applying the section formula we have 5 6

1r s

r s

+− =

+. Solving this gives r : s = -7 : 6.

Apply section formula again gives x = -44.

Example 12d: Show that the point A (1, 1.5), B (-4, 2) and C (3, 1) are not collinear.

Solution 12d:

Assume A, B and C are collinear. Then C must be a point of division of AB, no matter

internal or external. So let the ratio be r : s. By section formula, for y-coordinates,

21.5

r s

r s

+=

+, thus r : s = 1 : 1. However, for x-coordinates,

3 41

r s

r s

−=

+, which

implies r : s = 5 : 2, a contradiction. Q.E.D.

Area of polygon enclosed by segments

Area of triangle

A triangle is form by three joined segments, which in turn is defined by three distinct

points. Thus we can find the area of a triangle using the coordinates of the points.

Example 12e: Given three points A (x1, y1), B (x2, y2) and C (x3, y3). Find the area of

∆ABC.


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Solution 12e:

Enclose a rectangle ARQP around ∆ABC, and without loss of generality assume A is

the one of the vertices.

From the figure, we have:

3 1 2 1

2 1 3 1

3 2 2 3

AP x x AR y y

RB x x PC y y

BQ x x CQ y y

= − = −

= − = −

= − = −

(12.7)

Therefore, the area of( )( ) ( )( )3 1 3 1 3 2 2 3

∆ , ∆2 2

x x y y x x y yACP BQC

− − − −= = ,

( ) ( ) ( )( )2 1 2 1

3 1 2 1∆ and2

x x y yARB ARQP x x y y

− −= = − − . But area of ∆ABC =

ARQP – ∆ACP – ∆BQC – ∆ARB. Therefore, after simplifying, the area of ∆ABC is:

1 2 2 1 2 3 3 2 3 1 1 3

2

x y x y x y x y x y x y− + − + − (12.8)

(12.8) is not easy to remember. So we define:

1 1

2 2 1 1

3 3 1 1

1 11 1

n nk k

k k k k

k kk k

n n

a b

a ba b

a b a b a ba b

a b

− −

+ += =+ +

= = −∑ ∑⋮ ⋮

(12.9)

So the area of triangle is:

1 1

2 2

3 3

1 1

1

2

x y

x y

x y

x y

(12.10)

A (x1, y1)

B (x2, y2)

C (x3, y3)

P (x3, y1)

Q (x3, y2) R (x1, y2)


116

Example 12f: Redo example 12d.

Solution 12f:

If A, B and C are collinear then the area of ∆ABC should be zero. But area of ∆ABC =

3 1

1 1.510.75 0

4 22

3 1

= ≠−

, so A, B, C are not collinear. Q.E.D.

Note that when using (12.10) please ensure the points are arranged counterclockwise.

Otherwise a negative value will be obtained. If you don’t know whether the points are

counterclockwise or not, take an absolute value.

Area of polygon

A polygon can be dissected into triangles. So we can just find out the area of each

triangle and sum up them to give the area of the polygon. In fact, if the vertices of the

polygon is (x1, y1), (x2, y2), …, (xn, yn) in counterclockwise order, then its area is:

1 1

2 2

1 1

1

2n n

x y

x y

x y

x y

⋮ ⋮ (12.11)

Example 12g: Let X be 1 1,

2 2

−

, Y be 2 1,

3 3

, O be the origin, C be the curve

y = 3 – x2 and P be a point on C in quadrant I. Find the maximum area of

quadrilateral OYPX.

Solution 12g:

Let P be (x, y). So 0 3x< < . Then:

0 0

2 3 1 31

Area of 2

1 2 1 2

0 0

1 2

2 3 2

1 7

2 6

OYPX x y

y x x y

x y

=

−

− += +

+=

(12.12)


117

2

2

7

2

2

21 7

12

7 3

12

7 1 589

12 14 196

7 589 1

12 196 14

7 589

12 196

589

336

x

x x

x

x

x

+ −=

− −=

= − −

= − −

≤

=

Equality occurs when x = 1⁄14. Thus, the maximum area is

589⁄336.


118

Lines

Lines in coordinates system

Lines are curves that are straight and infinite in length. There are no end points for

lines. (A “line” with two end points is a segment, and one is a ray.) A line can be

thought as an extension of a line segment.

A line and a segment

As two points can fix a segment, it is also true for lines. There exists a unique line that

passes through two given distinct points. Since line is a type of curve, we are

interested in finding out the equation of a line.

Equation of a line

Two-point form

It is given two points A and B and a line passes through them. We may assume there is

an arbitrary point P lies on the line. Then P must satisfy the followings:

� The slope of segment AP is the same as that of AB.

� P is a point of division of segment AB.

We will find the equation of a line using the first condition. Readers may try the

second condition.

Example 13a: Given two points A (x1, y1) and B (x2, y2). Find the equation of the line

passing though them.

Solution 13a:

Let P (x, y) be an arbitrary point lies on the line. Then slope of AP is equal to slope of

AB. Thus:

1 2 1

1 2 1

y y y y

x x x x

− −=

− − (13.1)

This equation holds for all P on the line. Thus (13.1) is the equation for the line

passing through A and B.

(13.1) is known as the two-point form of a line.

Example 13b: Find the equation of the line joining (5, -1) and (2, 7).


119

Solution 13b:

The equation of the line is:

( ) ( )

7 1 7

2 5 2

3 7 8 2

8 3 37 0

y

x

y x

x y

− − −=

− −− = − −

+ − =

(13.2)

x

y

0 5

5

Point-slope form

Besides determining a line by two points, one can also find it out by its slope and a

point on it. Using a similar approach in example 13a, if the point is (x1, y1) and the

slope is m, then the equation is:

1

1

y ym

x x

−=

− (13.3)

Or:

y – y1 = m (x – x1) (13.4)

(13.4) is known as the point-slope form of a line.

Slope-intercept form

An intercept of a curve is a point that cut the axis. If the axis is an x-axis, the

intercept is called an x-intercept; if the axis is y-axis, it is called a y-intercept. For

example, the x-intercept of the line in example 13b is 37⁄8 and the y-intercept is

37⁄3.

If the y-intercept and slope of a line is given, we can determine that line. It is because

the y-intercept is just a point and we can use the point-slope form to find out the

equation of the line. If the y-intercept is c and the slope is m, then the equation is:

8x + 3y – 37 = 0

(2, 7)

(5, -1)


120

y = mx + c (13.5)

This is the slope-intercept form of a line.

Intercept form

If we are given the two intercepts of the line, we can also tell the equation the line

using two-point form. If the x-intercept is b and y-intercept is c, then:

1x y

b c+ = (13.6)

This is known as the intercept form of a line.

General form

Recall that we define a curve as set of points satisfying f(x, y) = 0. If we can arrange

the equation for a curve in the form “= 0”, then we say that is the general form of the

curve. For a line, its general form is:

Ax + By + C = 0 (13.7)

For real numbers A, B and C. By rearranging (13.7) into slope-intercept and intercept

form, we have:

-intercept

-intercept

Slope

Cx

A

Cy

B

A

B

= −

= −

= −

(13.8)

Distance between point and line*

The distance between a point (x0, y0) and a line L: Ax + By + C = 0 can be given by:

0 0

2 2

Ax By Cd

A B

+ += ±

+ (13.9)

The sign is taken opposite to that of C. If the point and the origin are lying on the

same side of L, the distance is negative. Otherwise, it is positive. Usually we just use

the absolute value of (13.9) in practical use.


121

Intersecting Lines

Intersection of two lines

When there are two lines in a coordinates system, we may ask, “where will they

intersect?” This will be introduced here.

Assume we have two lines L1: ax + by + c = 0 and L2: dx + ey + f = 0. If P (x0, y0) is

the intersection of L1 and L2 then this should satisfy both equations L1 and L2, i.e.,

0 0

0 0

0

0

ax by c

dx ey f

+ + =

+ + = (14.1)

This is just a set of ordinary simultaneous linear equations in 2 unknowns. Using the

notation in set theory we may write P = L1 ∩ L2.

Example 14a: Find Area of red triangle

Area of rectangle in the following figure, if AM : MB =

CN : DN = 1 : 2 and BP = CP.

Solution 14a:

Without loss of generality assume the figures lies on a coordinates system with origin

at D and A = (0, 2) and C = (3, 0). It is because ratio is invariant in scaling. Therefore

B = (3, 2), P = (3, 1), M = (1, 2) and N = (2, 0). Thus:

: 3

: 2

: 2

DP x y

DM y x

AN x y

=

=

+ =

(14.2)

Thus I = DM ∩ AN = 2 4,

3 3

and J = DP ∩ AN = 3 1,

2 2

. So the area of the red

triangle is

0 0

3 2 1 21 5

2 3 4 32 6

0 0

= . But the area of the rectangle is 6. Thus the answer is5

36.

A B

C D

M

N

P I

J


122

Angle between two lines

We are also interested in finding the angle formed between two lines. But we need a

lemma:

Another representation of slope

Using the definition of slope and the above figure, we know that the slopePR

XR= . But

from definition of trigonometric function we know tanPR

θXR

= . Hence we have:

Slope = tan θ (14.3)

The value θ is known as the inclination of the line, which is the angle between the

line and the x-axis.

Angle between two lines

Refer to the following figure,

It is easy to show that β2 – β1 = θ. If the slope of L1 is m1 and L2 is m2, then:

θ X

P

R

x

β2

θ

β1

L2 L1

X2 X1


123

( )1 2

1 2

1 2

1 2

1 2

tan tan

tan tan

1 tan tan

1

θ β β

β β

β β

m m

mm

= −

−=

+

−=

+

(14.4)

However, the last value may be negative. If we are only concentrated to the acute

angle between the two lines, we can takes its absolute value, i.e.,

1 2

1 2

tan1

m mθ

mm

−=

+ (14.5)

When two lines parallel, the angle between is 0°. Thus from (14.5),

(L1 // L2) ⇔ (m1 = m2) (14.6)

When two lines are perpendicular to each other, i.e., the angle between them is 90°,

the denominator of its tangent is zero. Thus:

(L1 ⊥ L2) ⇔ (m1m2 = –1) (14.6)

Example 14b: (HKCEE 1997/A) L is the line y = 2x + 3. A line with slope m makes

an angle of 45° with L. Find the value(s) of m.

Solution 14b:

The slope of L is 2. From (14.4),

( )

2tan 45

1 2

1 1 2 2

1 2 2 or 1 2 2

13 or

3

m

m

m m

m m m m

m m

−° =

+

+ = −

+ = − + = −

= − =

(14.7)


124

Locus and Parametric Equation

Locus

In mathematics, locus is a curve traced out by a moving point. This point usually

satisfies some conditions we want. To find out the equation of a locus, we usually let

the point be (x, y) and join the other conditions to make up an equation in terms of x

and y.

Example 15a: Find the locus of a point P such that the distant between P and (2, 5) is

always 3.

Solution 15a:

Let P = (x, y). Hence:

( ) ( )2 2

2 2

2 2

2 3 3

4 4 6 9 9

4 6 4 0

x y

x x y y

x y x y

− + − =

− + + − + =

+ − − + =

(15.1)

Is the required locus.

Example 15b: Given two curves H: x2 – y

2 = 1 and C: x

2 + y

2 = 4. X is an intersection

of C and H in the first quadrant. Let P be a point on C such that the line XP intersects

H at Q. Find the locus of the midpoint of PQ.

Solution 15b:

x

y

-4 -2 0 2 4

-2

2

H

C

XP

QM

Locus of M

This figure shows a brief look of the graph. Here we let M to be the midpoint of PQ.


125

Since we are going to find the locus of M, we let M = (x, y). For P, Q and X, we let

them to be (a, b), (c, d) and (m, n). Since X is an intersection of H and C, we have:

2 2

2 2

1

4

, 0

m n

m n

m n

− =

+ = >

(15.2)

Thus 5 3and

2 2m n= = . If we let the equation of XP to be y – n = k (x – m) for

some constant k denoting the slope, then:

( )

2 2

0

4

ka b n km

a b

− + − =

+ = (15.3)

And:

( )

2 2

0

1

kc d n km

c d

− + − =

− = (15.4)

Solving (15.3) gives( )

( )( )( )

2 2

2 2

5 1 2 3 3 1 2 5,

2 1 2 1

k k ka b

k k

− − − − −= =

+ +. Solving (15.4)

gives ( )

( )( )

( )

2 2

2 2

5 1 2 3 3 1 2 5,

2 1 2 1

k k k kc d

k k

+ − − + += =

− −. Since ,

2 2

a c b dx y

+ += = ,

we have:

( )( )

( )( )

4 3

4

4

4

5 1 2 3

2 1

3 1 2 5

2 1

k kx

k

k ky

k

+ −=

−

− + +=

−

(15.5)

This is a parametric equation, which will be introduced in this section:

Parametric equation

Parametric equations are equations describing a curve in the form ( )( )

x f t

y g t

=

= or

(f(t), g(t)). Here f and g are real-valued functions and t is a real variable, called the

parameter of the curve. As t changes, the point (f(t), g(t)) moves. If we find the locus

of this point the result will be a curve this parametric equation describing.


126

Example 15c: Sketch the graph with x = t + 5, y = 5t – 1 for t ∈ [-3, 3].

Solution 15c:

We can plot a few points with respect to t then join them up to get a rough image of

the graph.

x

y

0 2 4 6 8

-10

10

t = -3

t = -2

t = -1

t = 0

t = 1/2

t = 1

t = 2

t = 3

The curve is a segment.

Converting general equation into parametric equation*

If the curve is of the form y = f(x), then we can just set x = t and y = f(t) to form a

parametric equation. If the curve is f(x, y) = 0, then we may (I) convert it into the form

y = g(t) or (II) find two functions h1(t) and h2(t) such that f(h1(t), h2(t)) = 0. Then

x = h1(t) and y = h2(t).

Example 15d: Convert the curve x2 + y

2 = 1 into parametric form.

Solution 15d: Since sin2 t + cos

2 t = 1 for all t, the parametric form is (sin t, cos t).

Converting parametric equation into general equation

To convert a parametric equation into its general form, the only way is to eliminate

the parameter and connect x and y in one equation. To eliminate the parameter one can

(I) express t in terms of x or y then substitutes this into the other equation, or (II) find

a function such that f(x(t), y(t)) = constant.

Example 15e: Convert 2,2

tt

t

+ − into general form.


127

Solution 15e:

Let x = t + 2 and 2

ty

t=

−.

Method (I):

Hence t = x – 2. Substitute this into y gives:

2

4

xy

x

−=

− (15.6)

That means xy – x – 4y + 2 = 0 is the general form.

Method (II):

First of all, notice that 2 4x t

y t

−= , which is simpler in expressing. If we subtract

2 2 4t

y t

−= from it, the result will be

22 22

x t tt

y t

− −= = − . Subtract x again from it

gives 2

4x

xy

−− = − . Rearranging this also result in xy – x – 4y + 2 = 0.


128

Circles

What are circles?

In coordinates geometry, circles are loci of a point P that is equidistant from another

point O. The point O is called the center of the circle and the length OP is the radius.

If the center of the circle is (x0, y0) and the radius is r, then the equation of the circle

is:

(x – x0)2 + (y – y0)

2 = r

2 (16.1)

This is called the standard form of the circle. We can expand (16.1) into:

x2 + y

2 – 2x0 x – 2y0 y + (x0

2 + y0

2 – r

2) = 0 (16.2)

To get its general form. If it is given the general form of a circle to be:

x2 + y

2 + ax + by + c = 0 (16.3)

Then:

2 2

Center ,2 2

4Radius

2

a b

a b c

= − −

+ − =

(16.4)

So, when a2 + b

2 – 4c = 0, the “circle” reduced into a point. This is called a point

circle. When a2 + b

2 – 4c < 0, the radius will be unreal and no graph can be drawn.

Then that equation is said to be representing an imaginary circle.

Example 16a: Given a triangle ∆ABC such that the vertices A, B, C are (2, 5), (1, 7)

and (-4, 2) respectively. Find:

� The equation of the circumcircle Γ of ∆ABC, and

� The intersection (other than A) of Γ and the altitude of ∆ABC from A.

Solution 16a:

The first question means to find a circle Γ such that A, B, C all lies on it. We let Γ to

be x2 + y

2 + ax + by + c = 0. So:

( )

2 2

2 2

2 2

2 5 2 5 0

1 7 7 0

5 2 5 2 0

a b c

a b c

a b c

+ + + + =

+ + + + =− + − + + =

(16.5)

Solving this gives Γ: x2 + y

2 + 3x – 9y + 10 = 0.


129

For the second equation, let the altitude be L. Therefore, L ⊥ BC. Since the slope of

BC is 1, the slope of L is –1. By point-slope form, the equation of L is x + y – 7 = 0.

Thus the intersection satisfies:

2 2

7 0

3 9 10 0

x y

x y x y

+ − =

+ + − + = (16.6)

Hence the intersection is (-1, 8). (The other solution (2, 5) should be rejected because

it is just the point A.)

8

7

6

5

4

3

2

1

-6 -4 -2 2 4

L

Γ

C

B

A

Tangent of circle

If a curve C intersects a curve C’ at exactly one point somewhere and C’ does not

appear on both sides separated by C there, C and C’ are tangent to each other. This is

especially important if C is a line; this will be called the tangent of C’. In this section

we will show how to find the tangent of a circle.

Tangent of circle passing through a point on circle

Let C: x2 + y

2 + ax + by + c be a circle with center O and T (x0, y0) be

a point on the C such that a tangent L passes through it. Thus OT ⊥ L.

Since slope of OT is 0 0

0 0

2 2

2 2

y b y b

x a x a

+ +=

+ +, the slope of L is

0

0

2

2

x a

y b

+−

+. Therefore, by point-slope form, the equation of L is:

CL

O

T


130

( )

( )

0

0 0

0

2 2

0 0 0 0 0 0

2

2

2 2 2 0

x ay y x x

y b

ax by x x y y ax by x y

+− = − − +

+ + + − − − + =

(16.7)

But because T is on C, we have:

2 2

0 0 0 0x y ax by c+ = − − − (16.8)

Substitute this back to (16.7) and simplifying gives:

0 0

0 0: 02 2

x x y yL x x y y a b c

+ + + + + + =

(16.9)

Example 16b: Using the figure in example 16a, let the LA, LB and LC be the three

tangents of Γ passing through A, B and C respectively. Find the length of the segment

connecting LA ∩ LB and LA ∩ LC.

Solution 16b:

The tangent through A (LA) is:

( ) ( )3 9

2 2: 2 5 2 5 10 0

: 7 19 0

AL x y x y

x y

+ + + − + + =

+ − = (16.10)

Similarly, the LB is x + y – 8 = 0 and LC is x + y + 2 = 0. Thus the two intersections are

7 11 11 37, and ,

2 2 6 6

−

and the distance is 25 2

3.

Lengths of tangent from external point

If a point is outside a circle and tangents passes through it, there will

be exactly two tangents. Nevertheless, their lengths (XT and XU in

the figure) are the same. We wish to find the value of this. If the

circle is (x – ζ)2 + (y – η)

2 = r

2 and X = (x0, y0), then by Pythagoras’

theorem, XT2 + r

2 = OX

2. But OX

2 = (x0 – ζ)

2 + (y0 – η)

2. So:

( ) ( )2 2 2

0 0XT x ζ y η r= − + − − (16.11)

But this is just the value of the equation of the circle when (x, y) = (x0, y0). Hence, if

the circle in general form is x2 + y

2 + ax + by + c = 0, the length of tangent is:

2 2

0 0 0 0x y ax by c+ + + + (16.12)

Many circles

If we are given two circles on a coordinates system, we can tell their relationship by

the distance of their centers and their radii. If the radii of the two circles are r1 and r2

U

T

O X


131

respectively with r1 > r2, then:

Distance of centers in… Implication

(r1 + r2, ∞) Two circles separates

{r1 + r2} They are tangent to

each other externally

(r1 – r2, r1 + r2) They intersect at 2 points

{r1 – r2} They are tangent to

each other internally

[0, r1 – r2) One circle is inside

the other

If the two circles intersect, there will be a common chord which passes through the

two intersections. If they are C1: x2 + y

2 + a1x + b1y + c1 = 0 and

C2: x2 + y

2 + a2x + b2y + c2 = 0, simple calculation suggests that the equation of the

common chord is:

L: (a1 – a2) x + (b1 – b2) y + (c1 – c2) (16.13)

Or an easy-to-remember representation:

L: C1 – C2 (16.14)

If C1 and C2 are tangent to each other, then L reduced to one of the common tangents

of them.

Example 16c: Prove that the common chords of any three intersecting circles are

always concurrent.

Solution 16c:

Let the three circles be C1: x2 + y

2 + a1x + b1y + c1 = 0, C2: x

2 + y

2 + a2x + b2y + c2 = 0

and C3: x2 + y

2 + a3x + b3y + c3 = 0. Thus the three chords are

L3: (a1 – a2) x + (b1 – b2) y + (c1 – c2) = 0, L1: (a2 – a3) x + (b2 – b3) y + (c2 – c3) = 0

and L2: (a3 – a1) x + (b3 – b1) y + (c3 – c1) = 0. Let the intersection of L3 and L1 be X.

Since X satisfy both L1 and L3, it would also satisfy:

L1 + L3: (a1 – a3) x + (b1 – b3) y + (c1 – c3) = 0 (16.15)

But this is just L2. Therefore, X is on L1, L2 and L3. That means the three chords are

concurrent. Q.E.D.


132

Parabola

The quadratic polynomial in coordinates geometry

If we graph the curve y = x2 on a coordinates system, the result would be something

like this:

This curve is called a parabola. In physics, the (upside-down) parabola resembles the

path of an object thrown at an angle, and it is why this curve named like this (it means

“to throw across” in Greek).

In mathematics, a parabola is the curve y = ax2 + bx + c for a ≠ 0. The following

shows some example of parabolas:

x

y

-4 -2 0 2 4

-2

2

4


133

Some properties associated to parabola

x-Intercepts

In Elementary Algebra, we learnt that the discriminant ∆ = b2 – 4ac determines how

many real roots a does a quadratic equation has. In coordinates geometry, the root of a

curve is same as its x-intercept. Therefore, ∆ decides how the parabola will cross the

x-axis. The x-intercepts would be the roots of the quadratic polynomial.

x

y

-2 0 2 4

2

4

Three parabolas with different values of ∆.

The purple one: ∆ < 0. The yellow one: ∆ = 0. The blue one: ∆ > 0.

y-Intercept

A function will cross the y-axis when x = 0. By substituting this, we know that the

y-intercept of a parabola is the constant term of its equation, i.e., c.

x

y

0 5 10

-10

-5

The parabola y = ½ x2 – 5x + 1. The y-intercept is 1 according to the graph.


134

Extremum and direction of opening

An extremum means the maximum or minimum point of a function. By completing

the square, a quadratic polynomial can be expressed as:

a (x – h)2 + k (17.1)

Where ∆

,2 4

bh k

a a= − = − . Because n

2 ≥ 0 for any real number n, the maximum /

minimum of (17.1) will be k depending on the sign of a. The value k can be achieved

when x = h. Thus, (h, k) is the extremum of a parabola.

y

0 5 10

-10

-5

(5, -11.5)

(0.5, -2.75)

Two parabolas: y = -x2 + x – 3 and y = ½ x2 – 5x + 1.

The extremum will be a minimum when a > 0 (Why?). It will be a maximum when

a < 0. If the extremum of a parabola is a minimum, then no points can lower than this

point. Thus the parabola opens upwards. Similarly, if the extremum is a maximum, the

parabola opens downwards. That means when a is positive, the parabola opens

upwards. When a is negative, the parabola opens downwards.

In some books the point “extremum” is called the “vertex” of a parabola.

Example 17a: (HKCEE 1990/2) The graph of y = ax2 + bx + c is given as shown.

Which of the following is/are negative: a, b or c?

y

x O


135

Solution 17a:

From the graph, the y-intercept is negative. Thus c is negative. Also, the parabola

opens downward, thus a is also negative. The sum of roots is positive, thus 0b

a− > .

This means b is positive. Therefore, a and c are negative.


136

Coordinates Geometric Transformation

What is transformation?

In reality, if you have an object, you can move it, rotate it, squeeze it, pull it, destroy it,

etc. These are transformations of the object. Transformation of something means to

change the shape and/or position of something in a specific way. In coordinates

geometry, a transformation can be represented by a function τ: R2 → R

2. This function

maps a point to a different (or same) coordinates. Since a curve is composed of points,

if we transform every point on the original curve, the result will be another curve

transformed by τ. Usually if we transform a curve C by τ, we simply write τ(C), which

actually means the set {τ(x) | x ∈ C}.

Translation

Translation is a transformation. This transformation will move every point by a fixed

distance and direction. If the origin is mapped to (a, b), then the transformation is

given by:

T(x, y) = (x + a, y + b) (18.1)

If the original curve it C: f(x, y) = 0, the translated curve will be f(x – a, y – b) = 0. We

say the curve is translated by (a, b).

x

y

-2 0 2 4 6

-2

2

The curve x2 + y2 = 4 is translated by (5, 1).

Scaling

Scaling, also called homothety, is a transformation that makes an object bigger or

smaller. To demonstrate the mechanism, imagine you want to take a picture of

something far away using a digital camera. This thing looks too small in preview. To

have a clear look of this thing, you would zoom in. During zooming, the objects

around the thing are getting farther away (in the picture!), and the result would be the


137

whole view getting larger and larger.

This is exactly how scaling works. One point is fixed during scaling, and the other

will move away from / towards that point. The points that are farther apart from the

fixed point are moving “faster”. If the fixed point is (a, b), and a line segment of

length 1 is scaled up / down to length r, than the transformation function is:

H(x, y) = (r (x – a) + a, r (x – b) + b) (18.2)

Usually the fixed point is taken to be the origin. Hence:

H0(x, y) = (rx, ry) (18.3)

If a curve C: f(x, y) = 0 is transformed by H, the new curve will be

, 0x a y b

f a br r

+ + − − =

. The curve C is said to be scaled by a factor of r around

(a, b).

x

y

-2 0 2 4

-4

-2

2

The curve x2 + y2 = 4 is scaled by a factor of 2 around (-1, 1).

If the factor is greater than 1, the figure is enlarged. If the factor is in between 0 and 1,

the figure is shrunk. If the factor is negative, the figure is inverted.

Reflection

If you look at a mirror, you would find yourself behind the mirror with the direction

inverted. This is a kind of reflection. In geometry, reflection is a transformation that

maps every point P to another point P’ with respect to a line L such that the distance

between P’ and L is same as that of P and L and PP’ is perpendicular to L.


138

Example 18a: Find the function of reflection if L is Ax + By + C = 0.

Solution 18a:

Let P = (x, y), P’ = (x’, y’) and the function be S: R2 → R

2. Therefore:

S(x, y) = (x’, y’) (18.4)

Since PP’ is perpendicular to L, the slope of PP’ is B

A. That means:

y y B

x x A

′−=

′−. (18.5)

Now because the distance from P and P’ to L are the same, and P and P’ are on

opposite sides of L (otherwise P = P’, and this is not a reflection), we have:

Ax + By + C = –Ax’ – By’ – C (18.6)

Using (13.9). Solving (18.5) and (18.6) for x’ and y’ gives:

( )( ) ( ) ( ) ( )2 2 2 2

2 2 2 2

2 2, ,

B A x A C By A B y B C AxS x y

A B A B

− − + − − + = + +

(18.7)

If a curve f(x, y) = 0 is transformed by S, the new curve is

( ) ( ) ( ) ( )2 2 2 2

2 2 2 2

2 2, 0

B A x A C By A B y B C Axf

A B A B

− + + − + + = + +

. We say the curve is

reflected along L.

x

y

-10 -5 0

-5

5

The curve y = x2 is reflected along x + 2y + 3 = 0.

The resulting curve is 9x2 – 24xy + 16y2 – 16x + 63y + 96 = 0.


139

O P

P’

P’

The formula (18.7) is extremely hard to memorized, so we recommend readers do not

use this if possible.

Rotation*

Rotation is a transformation that moves a point P to another point P’ around a fixed

point O such that ∠POP’ is constant. If this angle is α, we say that P is

rotated by an angle of α about O. However, there may be two

choices of P’, as shown on the right. In mathematics, we

usually measure angle in counterclockwise direction.

Therefore, if α is positive, we take the blue P’. If we want the

red one, make α negative.

In coordinates geometry, rotation is evaluated as:

R(x, y) = (cos α (x – a) – sin α (y – b) + a, sin α (x – a) + cos α (y – b) + b) (18.8)

Where (a, b) is the center of rotation (the point O) and α is the angle. A curve is

transformed as:

( ) ( ) ( ) ( )( )cos sin , sin cos 0f α x a α y b a α x a α y b b− + − + − − + − + = (18.9)

x

y

-4 -2 0 2

2

4

6

The curve y = x2 is rotated about (0, -1) by 45°.

Invariants in transformations*

In the transformations listed above, some quantities are not changed. These are the

invariants of that transformation. The table below shows some invariants:


140

Invariants Translation Scaling Reflection Rotation

Length and area � � � �

Ratio � � � �

Angle between two lines � � � �

Slope � � � �

Type / number of intersections � � � �

Coordinates � � Points on L �


141

Summary � Coordinates System

� In a coordinates system, a point can be defined by a pair of numbers called

coordinates.

� The rectangular coordinates system is created by two perpendicular real

lines, called the x-axis and y-axis respectively, that are crossing at their

zeros.

� The coordinates of a point are given by the readings on the axes. They are

denoted as (x, y).

� Coordinates system = {(x, y) | x, y ∈ R} = R2.

� The axes divide the coordinates system into four quadrants.

� Line Segment

� Curves are defined as ( ) ( ){ }2, , 0C x y f x y= ∈ =ℝ , where f: R2 → R.

Usually C is just written as “f(x, y) = 0”.

� Line segments are of curve that is not “curved” at all, and length is finite.

� If a segment connects (x1, y1) and (x2, y2), then:

� ( ) ( )2 2

1 2 1 2Length x x y y= − + −

� 1 2

1 2

Slopey y

x x

−=

−

� The point 2 1 2 1,rx sx ry sy

r s r s

+ + + +

divides the segment into ratio r : s

internally.

� A negative ratio of division means the point divides the segment

externally.

� Area of triangle

1 1

2 2

3 3

1 1

1

2

x y

x y

x y

x y

= . Area of polygon

1 1

2 2

1 1

1

2n n

x y

x y

x y

x y

= ⋮ ⋮ .

I II

III IV


142

� Lines

� Lines are curves that are straight and infinite in length.

� There exists a unique line that passes through any two distinct points.

� Two-point form of a line: 1 2 1

1 2 1

y y y y

x x x x

− −=

− −

� Point-slope form: y – y1 = m (x – x1).

� Slope-intercept form: y = mx + c

� Intercept form: 1x y

b c+ = .

� General form of a line: Ax + By + C = 0, where

-intercept

-intercept

Slope

Cx

A

Cy

B

A

B

= −

= −

= −

.

� Distance between a point and a line is 0 0

2 2

Ax By Cd

A B

+ += ±

+, where the sign

is taken opposite to that of C.

� Intersecting Lines

� Finding intersection P of two lines L1, L2 is same as finding solution to a set

of simultaneous linear equations in 2 unknowns. We write P = L1 ∩ L2.

� Slope = tan θ, where θ is the inclination of a line, which is the angle

between it and the x-axis.

� Angle between two lines: 1 2

1 2

tan1

m mθ

m m

−=

+.

� (L1 // L2) ⇔ (m1 = m2)

� (L1 ⊥ L2) ⇔ (m1m2 = –1)

� Locus and Parametric Equation

� Locus is a curve traced out by a moving point.

� We let the moving point be (x, y) to make up the equation of locus.

� Parametric equations are equations describing a curve in the form (f(t), g(t)).

� Circles

� Circles are loci of a point that is equidistant from another point.

*


143

� If the center is (x0, y0) and the radius is r, then the circle is

(x – x0)2 + (y – y0)

2 = r

2. This is the standard form of a circle.

� General form of circle: x2 + y

2 + ax + by + c = 0, where

2 2

Center ,2 2

4Radius

2

a b

a b c

= − −

+ − =

.

� If a curve C intersects a curve C’ at exactly one point somewhere and C’

does not appear on both sides separated by C there, C and C’ are tangent to

each other.

� The tangent line of x2 + y

2 + ax + by + c = 0 at (x0, y0) is

0 0

0 0 02 2

x x y yx x y y a b c

+ + + + + + =

.

� Length of the tangent from an external point (x0, y0) is

2 2

0 0 0 0x y ax by c+ + + + .

� Relationship of two circles:

Distance of centers in… Implication

(r1 + r2, ∞)

{r1 + r2}

(r1 – r2, r1 + r2)

{r1 – r2}

[0, r1 – r2)

� Common chord/tangent of two circles: L: C1 – C2.

� Parabola

� Parabola is the curve y = ax2 + bx + c for a ≠ 0.

� x-intercepts of parabola are roots of that quadratic polynomial.

� ∆ decides how the parabola will cross the x-axis.

� y-intercept = c.

� Extremum / Vertex = ∆

,2 4

b

a a

− −

.


144

� (a > 0) ⇔ (parabola opens upwards).

(a < 0) ⇔ (parabola opens downwards).

� Coordinates Geometric Transformation

� Transformation means to change shape and/or position of something.

� Translation: T(x, y) = (x + a, y + b)’

� Scaling: H(x, y) = (r (x – a) + a, r (x – b) + b)

� Reflection:

( )( ) ( ) ( ) ( )2 2 2 2

2 2 2 2

2 2, ,

B A x A C By A B y B C AxS x y

A B A B

− − + − − + = + +

.

� Rotation:

R(x, y) = (cos α (x – a) – sin α (y – b) + a, sin α (x – a) + cos α (y – b) + b).

� Invariants in transformation:

Invariants Translation Scaling Reflection Rotation

Length and area � � � �

Ratio � � � �

Angle between two lines � � � �

Slope � � � �

Type / number of intersections � � � �

Coordinates � � Points on L �

*

*


145

Self Assessment Exercises

Level 1

1) Solve the followings:

a) Find and express the equation of the line passing through (cos t, sin t) and

(-sin t, cos t) in terms of t.

b) Show that this line is tangent to the circle 2x2 + 2y

2 – 1 = 0.

[Hint: Consider the distance from the origin to the line]

2) (HKCEE 1994/2) In the figure, the line y = mx + k cuts the curve y = x2 + bx + c at

x = α and x = β. Find the value of αβ in terms of b, c, m and/or k.

3) Prove the length is invariant in translation.

4) (HKMO 1996/HG Q1) In the figure, the quadratic curve y = f(x) cuts the x-axis at

two points (1, 0) and (5, 0) and the y-axis at point (0, -10). Find the value of p.

5) (IMSC 2000/FG1 Q1) Let u, v be integers such that 0 < v < u. Let A = (u, v), B be

the reflection of A along the line y = x, C be reflection of B along the y-axis, D be

the reflection of C along the x-axis and E be the reflection of D along the y-axis.

The area of the pentagon ABCDE is 451. Find u + v.

Level 2

6) (JSMQ 1999/FG Q2) A belt fits around three circles x2 + y

2 = 1, (x – 14)

2 + y

2 = 1

and (x – 9)2 + (y – 12)

2 = 1. Find the length of the belt.

7) (Apollonian Circle) Let A be (-1, 0) and B be (1, 0). Find the locus of P such that

APr

PB= and express it in terms of r. What kind of curve is it?

y

x

y = x2 + bx + c

y = mx + k

α β

y

x 1 5

-10

(4, p)


146

8) (PSMSIMC 2004/S Q14c) In the figure, the radius of the circle is 1 and AP is a

diameter of the circle. B is a point on the circumference and C is a point on BP

such that PC : CD = 5 : 3. If M is the mid-point of AC and BM is extended to meet

AP at Q, find the maximum possible area of ∆BPQ.

[Hint: Find the locus of C as B moves and show that Q is a fixed point]

Q

M C

A P

B

9) (IMOPHK 2004 Q13) Find the area enclosed by the graph x2 + y

2 = |x| + |y| on the

xy-plane.

10) (ISMC 2003/I Q7) A piece of graph is folded once so that (0, 2) is matched with

(4, 0) and (7, 3) is matched with (m, n). Find m + n.

11) In the figure, P is a point on the line y = x – 7. Find the minimum value of

AP + BP and the corresponding coordinates of P.

12) Solve the followings:

a) Find the locus P of point X such that the distance from X to the line

L: y = -¼ and the point F (0, ¼) is the same.

b) The circle C: x2 + y

2 = 1 intersects P and L at four points. Find the area of

the quadrilateral enclosed by these four points.

c) It is given that if X (x0, y0) is a point on P, the slope of the tangent to P

passing through X is 2x0. Find the common tangent(s) of C and P.

A (5, -3)

B (3, -6)

P

y = x – 7


147

Answers

1) a) dfdf

4) 6

5) 21

6) 2π + 42

7) ( )2 22 1

1 01

rx y x

r

++ − + =

−. This is a curve with center on the x-axis.

8) 8⁄11

9) π + 2

10) 6.8

11) ( )10113 3

17; ,P = −

12) a) y = x2. b)

( ) ( )2 5 1 15 2 2 5 2

16

− + −. c) 2 2 5 2 5y x= ± + − − .


148

Chinese Translation of Important Terms Abscissa 橫坐標

Altitude 垂線

Center 圓心

Circle 圓

Circumcircle 外接圓

Collinear 共線

Common chord 公共弦

Common tangent 公切線

Concurrent 共點

Coordinates 坐標

Coordinates system 坐標系統

Curve 曲線

Equidistant 同距

External point of division 外分點

Extremum 極值

General form 一般式

Homothety 位似變換

Hypotenuse 斜邊

Imaginary circle 虛圓

Inclination 傾角

Intercept 截距

Intercept form 截距式

Internal point of division 內分點

Invariant 不變量

Line segment 線段

Locus (pl. loci) 軌跡

Midpoint formula 中點公式

Ordinate 縱坐標

Origin 原點

Parabola 抛物線

Parameter 參數

Parametric equation 參數方程

Point circle 點圓

Point of division 分點

Point-slope form 點斜式


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Polygon 多邊形

Project 投影

Pythagoras’ theorem 畢氏定理

Quadrant 象限

Quadrilateral 四邊形

Radius (pl. radii) 半徑

Ray 射線

Rectangular coordinates system 直角坐標系統

Reflection 反射

Right triangle 直角三角形

Rotation 旋轉

Scaling 縮放

Section formula 截點公式

Slope 斜率

Slope-intercept form 斜截式

Standard form 標準式

Tangent 相切、切線

Transformation 變換

Translation 平移

Two-point form 兩點式

Vertex (pl. vertices) 頂點

x-axis x軸

x-coordinate x坐標

x-intercept x軸截距

y-axis y軸

y-coordinate y坐標

y-intercept y軸截距


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References:

� Wikipedia, the free encyclopedia. [http://en.wikipedia.org/]

� MathWorld – A Wolfram Web Resource. [http://mathworld.wolfram.com/]

� Mathematics Database. [http://www.mathdb.org/]

� MathsWorld-2001 [http://mathsworld.ath.cx/]

Original documents:

� Coordinates System: Chapter 0 “Before You Start” Section 2; Chapter 2

“Linear Equations” Section 1; Chapter 8 “Trigonometry” Section 5.

� Line Segment: Chapter 2 “Linear Equations” Section 1, 5, 7 and 8.

� Lines: Chapter 2 “Linear Equations” Section 2 and 5.

� Intersecting Lines: Chapter 2: “Linear Equations” Section 3 and 6; Chapter 8

“Trigonometry” Section 12.

� Locus and Parametric Equation: Chapter 13: “Parametric Equation and

Locus” Section 1, 2 and 3.

� Circles: Chapter 14: “Quadratic Curve” Section 10.

� Parabola: Chapter 3: “Quadratic Equation and Polynomial Factorization”

Section 8.

� Coordinates Geometric Transformation: Chapter 14: “Quadratic Curve”

Section 9.

Documents

Mathematics Competition Training Class Notes Coordinates ...ylmass.edu.hk/~mathsclub/MCTCN/cge.pdf · Mathematics Competition Training Class Notes Coordinates Geometry 109 the vertical