Parallel lines, perpendicular lines and meeting points

Geometry, grade 10 (5 units) Lesson 3 of 26Learning

Parallel lines, perpendicular lines and meeting points

Parallel lines

Two lines that have slopes are parallel exactly when their slopes are equal:

\displaystyle m_1

\displaystyle =

\displaystyle m_2

The same slope and a different point on the $\displaystyle y$ -axis – two parallel lines: $\displaystyle {y}={2x+1}$ and $\displaystyle {y}={2x-3}$ .
The same slope and the same point on the $\displaystyle y$ -axis – one line, written twice: $\displaystyle {y}={2x+1}$ and $\displaystyle {4x-2y+2}={0}$ .
Different slopes – the lines meet at one point.

Perpendicular lines

Two lines that have slopes are perpendicular exactly when the product of their slopes is $\displaystyle -1$ :

\displaystyle m_1\cdot m_2

\displaystyle =

\displaystyle -1

\displaystyle y=2x

\displaystyle y=-\frac{1}{2}x

\displaystyle A=(1,2)

\displaystyle B=(-2,1)

Why, for the lines through the origin $\displaystyle {y}={2x}$ and $\displaystyle {y}={-\frac{1}{2}x}$ :

$\displaystyle A(1,2)$ is on the first line: from $\displaystyle O$ , $\displaystyle 1$ to the right and $\displaystyle 2$ up.
Turn the triangle of that step a quarter turn about $\displaystyle O$ : $\displaystyle 1$ to the right becomes $\displaystyle 1$ up, and $\displaystyle 2$ up becomes $\displaystyle 2$ to the left – the point $\displaystyle B(-2,1)$ , and $\displaystyle OB\perp OA$ .
The slope of $\displaystyle OB$ : $\displaystyle {\frac{1}{-2}}={-\frac{1}{2}}$ , and $\displaystyle {2\cdot\left(-\frac{1}{2}\right)}={-1}$ .

For any slope $\displaystyle m$ the step $\displaystyle (1,m)$ turns into $\displaystyle (-m,1)$ , of slope $\displaystyle -\frac{1}{m}$ . Lines that do not pass through the origin are moved there without changing their slopes. A line parallel to the $\displaystyle x$ -axis is perpendicular to a line parallel to the $\displaystyle y$ -axis, which has no slope.

Exercise 1

How do the two lines lie: the same line, parallel, perpendicular, or meeting without being perpendicular?

$\displaystyle {y}={-3x+2}$ and $\displaystyle {6x+2y-4}={0}$
$\displaystyle {y}={\frac{2}{5}x-1}$ and $\displaystyle {5x+2y}={7}$
$\displaystyle {y}={4x+1}$ and $\displaystyle {y}={4x-1}$
$\displaystyle {x-2y}={0}$ and $\displaystyle {y}={x}$

Exercise 2

Write the equation of the line through the point $\displaystyle P(1,1)$ parallel to the line $\displaystyle {2x-y+5}={0}$ .

\displaystyle 2x-y+5=0

\displaystyle P=(1,1)

Exercise 3

Write the equation of the line through the point $\displaystyle P(2,5)$ perpendicular to the line $\displaystyle {3x+4y}={0}$ .

\displaystyle 3x+4y=0

\displaystyle P=(2,5)

The meeting point of two lines

The meeting point is on both lines, so its coordinates satisfy both equations: it is the solution of their system of equations.

One solution – the lines meet at one point.
No solution – the lines are parallel: the system leads to something false, like $\displaystyle {0}={4}$ .
Infinitely many solutions – it is one line: the system leads to something always true, like $\displaystyle {0}={0}$ .

Example: $\displaystyle {y}={2x-1}$ and $\displaystyle {x+y}={8}$ .

$\displaystyle {x+(2x-1)}={8}$ , so $\displaystyle {3x}={9}$ and $\displaystyle {x}={3}$ ;
$\displaystyle {y}={2\cdot3-1}={5}$ ;
the lines meet at $\displaystyle (3,5)$ .

Exercise 4

Find the meeting point of the lines $\displaystyle {y}={3x-5}$ and $\displaystyle {2x+y}={10}$ .

The distance from a point to a line

The distance from a point to a line is the length of the perpendicular from the point to the line. Three steps find it:

the line through the point perpendicular to the given line;
the foot of the perpendicular – where the two lines meet;
the distance between the point and the foot.

Any other segment from the point to the line is longer: it is the hypotenuse of a right-angled triangle whose leg is the perpendicular.

Exercise 5

The line $\displaystyle \ell$ : $\displaystyle {y}={x+1}$ , and the point $\displaystyle P(5,0)$ .

Write the equation of the line through $\displaystyle P$ perpendicular to $\displaystyle \ell$ .

\displaystyle y=x+1

\displaystyle P=(5,0)

Find the foot of the perpendicular.

Find the distance from

\displaystyle P

\displaystyle \ell

, in its simplest form.

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – check and a new line · * – times · ^ – power, ↓ – end of the power

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

Exercise 6

The lines: $\displaystyle {kx+2y}={6}$ and $\displaystyle {3x-y}={4}$ .

For which

\displaystyle k

are the lines parallel? (A hint: the slope of the first is

\displaystyle -\frac{k}{2}

For which

\displaystyle k

are the lines perpendicular?

How many solutions has the system of the two equations for the $\displaystyle k$ of part 1?
And for every other $\displaystyle k$ ?
Explain.

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 $\displaystyle P$
$\displaystyle PT^{2}$ $\displaystyle =$ $\displaystyle PA\cdot PB$