Loci

Geometry, grade 10 (5 units) Lesson 9 of 26Learning

Loci

A locus

A locus is the set of all the points that have a given property – and only they. To prove that a set of points is the locus of a property, prove two directions:

every point of the set has the property;
every point that has the property belongs to the set.

The first direction alone is not enough: a few points of a circle are at distance $\displaystyle R$ from its centre, but the circle is the locus because every such point is on it.

The first example is the definition of the circle: the locus of the points at distance $\displaystyle R$ from the point $\displaystyle O$ .

The perpendicular bisector

The perpendicular bisector of a segment is the line perpendicular to it through its midpoint.

Theorem: the perpendicular bisector of a segment is the locus of the points at equal distances from its ends.

Given: $\displaystyle M$ is the midpoint of $\displaystyle AB$ , and $\displaystyle P$ is on the perpendicular to $\displaystyle AB$ at $\displaystyle M$ .
To prove: $\displaystyle {PA}={PB}$ .

The proof below shows the first direction by congruent triangles (SAS):

	Claim	Reason
1	$\displaystyle {AM}={MB}$	Given ( $\displaystyle M$ is the midpoint of $\displaystyle AB$ )
2	$\displaystyle {\angle PMA}={\angle PMB}$	$\displaystyle PM\perp AB$ : both angles are right
3	$\displaystyle {PM}={PM}$	A common side
4	$\displaystyle \triangle PMA\cong\triangle PMB$	(1), (2), (3); SAS
5	$\displaystyle {PA}={PB}$	(4); corresponding sides of congruent triangles

The second direction

Given: $\displaystyle {PA}={PB}$ ; $\displaystyle M$ is the midpoint of $\displaystyle AB$ .
To prove: $\displaystyle PM\perp AB$ – so $\displaystyle P$ is on the perpendicular bisector.

The proof below shows that the triangles $\displaystyle PMA$ and $\displaystyle PMB$ are congruent (SSS), so the two angles at $\displaystyle M$ are equal – and they add up to $\displaystyle 180^{\circ}$ :

	Claim	Reason
1	$\displaystyle {PA}={PB}$	Given
2	$\displaystyle {AM}={MB}$	Given ( $\displaystyle M$ is the midpoint of $\displaystyle AB$ )
3	$\displaystyle {PM}={PM}$	A common side
4	$\displaystyle \triangle PMA\cong\triangle PMB$	(1), (2), (3); SSS
5	$\displaystyle {\angle PMA}={\angle PMB}$	(4); corresponding angles of congruent triangles
6	$\displaystyle PM\perp AB$	(5); two equal adjacent angles add up to $\displaystyle 180^{\circ}$ , so each is $\displaystyle 90^{\circ}$

Exercise 1

$\displaystyle ABC$ and $\displaystyle DBC$ are isosceles triangles with the common base $\displaystyle BC$ : $\displaystyle {AB}={AC}$ and $\displaystyle {DB}={DC}$ .
$\displaystyle A$ and $\displaystyle D$ are on opposite sides of $\displaystyle BC$ .
Prove: $\displaystyle AD\perp BC$ .
(A hint: where are $\displaystyle A$ and $\displaystyle D$ , relative to the segment $\displaystyle BC$ ?)

The perpendicular bisector in coordinates

The perpendicular bisector of $\displaystyle AB$ is the line through the midpoint of $\displaystyle AB$ whose slope makes $\displaystyle -1$ with the slope of $\displaystyle AB$ .

Example: $\displaystyle A(0,1)$ and $\displaystyle B(4,3)$ .

The midpoint: $\displaystyle M(2,2)$ ;
the slope of $\displaystyle AB$ : $\displaystyle {\frac{3-1}{4-0}}={\frac{1}{2}}$ , so the perpendicular bisector’s slope is $\displaystyle -2$ ;
the perpendicular bisector: $\displaystyle {y-2}={-2(x-2)}$ , or $\displaystyle {y}={-2x+6}$ .

Exercise 2

$\displaystyle A(-2,1)$ and $\displaystyle B(4,5)$ are two points.

Write the equation of the perpendicular bisector of $\displaystyle AB$ .

\displaystyle A=(-2,1)

\displaystyle B=(4,5)

The point $\displaystyle P$ is on the line $\displaystyle {y}={x-3}$ , at equal distances from $\displaystyle A$ and from $\displaystyle B$ .
Find $\displaystyle P$ .

The angle bisector

The distance from a point to a line is the length of the perpendicular from the point to the line.

Theorem: the bisector of an angle is the locus of the points inside the angle at equal distances from its arms.

Given: $\displaystyle AP$ bisects the angle $\displaystyle BAC$ ; $\displaystyle PK\perp AB$ and $\displaystyle PL\perp AC$ .
To prove: $\displaystyle {PK}={PL}$ .

The proof below shows the first direction by congruent triangles (ASA); the second direction is Exercise 3.

	Claim	Reason
1	$\displaystyle {\angle KAP}={\angle LAP}$	Given ( $\displaystyle AP$ bisects the angle)
2	$\displaystyle {\angle AKP}={\angle ALP}$	$\displaystyle PK\perp AB$ , $\displaystyle PL\perp AC$ : both angles are right
3	$\displaystyle {\angle APK}={\angle APL}$	(1), (2); the angles of a triangle add up to $\displaystyle 180^{\circ}$
4	$\displaystyle {AP}={AP}$	A common side
5	$\displaystyle \triangle AKP\cong\triangle ALP$	(1), (3), (4); ASA
6	$\displaystyle {PK}={PL}$	(5); corresponding sides of congruent triangles

Exercise 3

The point $\displaystyle P$ is inside the angle $\displaystyle BAC$ .
$\displaystyle PK\perp AB$ , $\displaystyle PL\perp AC$ , and $\displaystyle {PK}={PL}$ .
Prove: $\displaystyle {\angle KAP}={\angle LAP}$ – that is, $\displaystyle P$ is on the bisector of the angle.
(A hint: two sides and the angle opposite the larger of them.)

Lines parallel to a line

Two more loci are made of parallel lines:

the locus of the points at distance $\displaystyle d$ from a line $\displaystyle \ell$ is two lines parallel to $\displaystyle \ell$ , one on each side of it, at distance $\displaystyle d$ ;
the locus of the points at equal distances from two parallel lines is the line parallel to both, midway between them.

Exercise 4

Describe each locus in words, and explain.

The centres of all the circles that pass through two given points $\displaystyle A$ and $\displaystyle B$ .
The points at distance $\displaystyle 3$ from the line $\displaystyle {y}={1}$ – write their equations.
The points inside a triangle at equal distances from two of its sides, $\displaystyle AB$ and $\displaystyle AC$ .

Formulas: geometry, grade 10 (5 units)

Open as a page

Introduction to analytic geometry

The distance between the points $\displaystyle (x_1,y_1)$ and $\displaystyle (x_2,y_2)$
$\displaystyle d$ $\displaystyle =$ $\displaystyle \sqrt{(x_1-x_2)^{2}+(y_1-y_2)^{2}}$
The midpoint of the segment
$\displaystyle M\left(\frac{x_1+x_2}{2},\ \frac{y_1+y_2}{2}\right)$
The slope of a line through two points
$\displaystyle m$ $\displaystyle =$ $\displaystyle \frac{y_2-y_1}{x_2-x_1}$ $\displaystyle x_1\neq x_2$
The line of slope $\displaystyle m$ through the point $\displaystyle (x_1,y_1)$
$\displaystyle y-y_1$ $\displaystyle =$ $\displaystyle m(x-x_1)$
The explicit equation of a line – $\displaystyle m$ the slope, $\displaystyle b$ where it crosses the $\displaystyle y$ -axis
$\displaystyle y=mx+b$
The general equation of a line
$\displaystyle Ax+By+C=0$
Parallel lines
$\displaystyle m_1$ $\displaystyle =$ $\displaystyle m_2$
Perpendicular lines
$\displaystyle m_1\cdot m_2$ $\displaystyle =$ $\displaystyle -1$
A canonical circle – its centre at the origin, $\displaystyle R$ its radius
$\displaystyle x^{2}+y^{2}=R^{2}$
The circle with centre $\displaystyle (a,b)$ and radius $\displaystyle R$
$\displaystyle (x-a)^{2}+(y-b)^{2}=R^{2}$

Midsegments and the meeting point of the medians

A triangle’s midsegment – $\displaystyle DE$ is parallel to $\displaystyle BC$
$\displaystyle DE$ $\displaystyle =$ $\displaystyle \frac{1}{2}BC$
A trapezoid’s midsegment – $\displaystyle a$ , $\displaystyle b$ the bases
$\displaystyle EF$ $\displaystyle =$ $\displaystyle \frac{a+b}{2}$
The meeting point of the medians $\displaystyle G$ divides the median $\displaystyle AD$ in the ratio $\displaystyle 2:1$
$\displaystyle AG$ $\displaystyle =$ $\displaystyle 2GD$
Pythagoras’ theorem and its converse: the triangle is right-angled, the right angle opposite $\displaystyle c$
$\displaystyle a^{2}+b^{2}=c^{2}$

Proportion and similarity

A property of a proportion
$\displaystyle \frac{a}{b}=\frac{c}{d}\ \Rightarrow\ \frac{a+b}{b}=\frac{c+d}{d}$
Thales’ theorem – $\displaystyle DE\parallel BC$
$\displaystyle \frac{AD}{DB}$ $\displaystyle =$ $\displaystyle \frac{AE}{EC}$
The extended Thales theorem – $\displaystyle DE\parallel BC$
$\displaystyle \frac{AD}{AB}=\frac{AE}{AC}=\frac{DE}{BC}$
The angle bisector theorem – $\displaystyle AD$ bisects the angle $\displaystyle A$
$\displaystyle \frac{BD}{DC}$ $\displaystyle =$ $\displaystyle \frac{AB}{AC}$
The similarity ratio – corresponding sides
$\displaystyle \frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}=k$
Heights, medians, perimeters, the radii of the circumscribed and of the inscribed circles
$\displaystyle \frac{h_1}{h_2}=\frac{m_1}{m_2}=\frac{P_1}{P_2}=\frac{R_1}{R_2}=\frac{r_1}{r_2}=k$
The ratio of the areas
$\displaystyle \frac{S_1}{S_2}$ $\displaystyle =$ $\displaystyle k^{2}$

The circle

An inscribed angle is half the central angle on the same arc
$\displaystyle \angle ACB$ $\displaystyle =$ $\displaystyle \frac{1}{2}\angle AOB$
An inscribed angle on the diameter $\displaystyle AB$
$\displaystyle \angle ACB=90^{\circ}$
A quadrilateral $\displaystyle ABCD$ inscribed in a circle – opposite angles
$\displaystyle \angle A+\angle C$ $\displaystyle =$ $\displaystyle 180^{\circ}$
The angle between the tangent $\displaystyle AT$ and the chord $\displaystyle AB$ equals an inscribed angle on $\displaystyle AB$ from its other side
$\displaystyle \angle BAT$ $\displaystyle =$ $\displaystyle \angle ACB$
Two tangents from the point $\displaystyle P$
$\displaystyle PA=PB$
A quadrilateral $\displaystyle ABCD$ circumscribing a circle
$\displaystyle AB+CD$ $\displaystyle =$ $\displaystyle BC+AD$
The chords $\displaystyle AB$ and $\displaystyle CD$ meet at the point $\displaystyle E$
$\displaystyle AE\cdot EB$ $\displaystyle =$ $\displaystyle CE\cdot ED$
Two secants from the point $\displaystyle P$
$\displaystyle PA\cdot PB$ $\displaystyle =$ $\displaystyle PC\cdot PD$
A tangent $\displaystyle PT$ and a secant $\displaystyle PAB$ from the point $\displaystyle P$
$\displaystyle PT^{2}$ $\displaystyle =$ $\displaystyle PA\cdot PB$