AG1 / briefcourseinanal-geom-Yefimov

The Slope-intercept Equation of a Line

57

Let

M be a variable point, with x, y as

its (current)

coordi-

nates,

and consider also the point B (0, b)

at which the line cuts

the axis Oy. Let us compute the value of the right member in

formula (2) of Art. 52,

taking the point B as Mi and the point M

as

Afa.

If

the

point M

lies on

the

given

line,

the computation

will yield the slope of the line,

that is,

y-b = *;

(3)

on

the

other

hand,

if

M does

not lie on the given line, this

relation will not be valid. Con-

sequently,

(3)

is

the

equation

of the given straight line

(this

is

also

apparent

from

Fig. 37,

Fig.

37.

when taking into

account

that

k = tana). Clearing of fractions and transposing b to the right

side, we get

54. Thus, every straight line

not

perpendicular

to

the

axis

Ox can be represented by an equation of the form

(4).

Conversely,

every

equation

of

the

form

(4)

represents

a

straight line having slope k and y-intercept b. For,

if y = kx + b

is the given equation, it is always possible to draw the line

with

the given

slope k

and

making the given intercept b on the axis

Oy, no matter what the numbers k and b\

but, according to the foregoing,

the given

equation will then be the equation of the

line drawn. An equation of

the

form

(4)

is called the slope-intercept equation of a

straight line.

Example. Draw the line whose equation is

pig

38'

Solution. Mark off the segment OB =* 2 on the

axis Oy (Fig. 38); through the point B, draw the

segment BAT= 4 in the "right"

direction, parallel

to the axis CU; and lay off

the segment NM = 3 from the point N "upwards"

(in the direction of the axis

Oy). Joining the points B and

M will then

give

the

required

line (which cuts

off

the

intercept b

2 on

Oy and

makes with

Ox

an angle

whose tangent is

3 \

.

-j j

58

Curves of the First Order

55. The function

is called linear.

From the foregoing, we can

say that the graph

of a linear function is a straight line.

When b = 0, we have

y'^kx.

(5)

The variables

x and y, connected by such

a relation, are said

point MI (Xi, j/i). The desired equation is obtained directly from

formula (2) of Art. 52. Let M be a Variable point with

(current)

coordinates x, y. If M lies on the line of slope k passing through

the point M, then, by virtue of formula (2) of Art. 52,

if the point M does not lie on the line, relation (6) is

not valid.

yy\ = k(x xl ).

(7)

Note. In the particular case when the point B (0,

b) is

taken

as Afi

(jci,

j/i), equation (7) assumes the form

(4).

57.

By

using relation (7), the following

problem

can

easily

be solved: to find the equation of the line passing through the two

given points MI (xi, y\) and

M2 (x*, yi)-

Using formula (2) of Art. 52, we

find the slope of the line:

after

which, from

(7),

we obtain

the

required equation

*

-M

x2

X\ V

17

It is

customary to

write this equation in the form

Calculation of the Angle Between Two Lines

59

or 3*

2y

7 = 0.

Two

18. Calculation of the Angle Between

Straight

Lines. Conditions

for

the Parallelism

and

Perpendicularity of Two

Straight

Lines

58.

The

problem

of

determining

the

angle

between

two

straight

lines

is

one

of

the standard

problems of

analytic

ge-

ometry. We shall now derive a formula for calculating the angle

between

straight lines whose slopes are known

(we assume that

neither of the lines

is

perpen-

dicular

to the axis Ox).

~

^

Consider

two

straight

one of them

lines;

(no matter

which)

will

be

referred

to

as

the

first

line,

and

the

other as the second line (see

Fig. 39). Let fe, and k2

de-

note the respective slopes of

these lines, and let <p be the

angle which the second line

makes

with

the

first,

that

is, the angle through which

the first

line has to be turned

Fig.

39.

to

reach

coincidence with

one of the directions of the

first line; if

the axis Ox is then given

an additional turn through

the

angle cp,

it will coincide with one

of the directions of the second

line. Thus, by adding the angle cp

60

Curves of the First Order

Now, tana1 = A1 ,

tana2 = 2 ; therefore,

Example. Given the lines y =

y*1 + !2,

y = 3

Ar-f-3. Find the angle be-

tween them.

Solution. By formula (1),

tan<p = -5

_l\

21+4

7 j

,

--'=-_!~__= 1.

angles

of

inclination are equal, that

is, if

tan cti **

tan 0,2- But

tan cci

= ki,

tan a2 = k2 . Hence, the condition

for the

parallelism

of two straight lines is the equality of their slopes:

The given straight lines will be perpendicular if,

and only

if,

the angle cp between them is equal to

that is, if tancp has

no

-g-,

hand

member in formula (1) will become zero, so that

we shall

have

1 + kik2 = 0. Consequently, the condition for the perpendicu-

larity of two straight lines is expressed by the relation

This last relation is usually written as

*2 = -

(2)

The Straight Line as the First-order Curve

61

Applying the formulas just derived,

we can say

at

once that,

the

lines

2

2

are

for instance,

y ~-^x-}~ 1,

y -g x-{-5

parallel,

y =

O

A

whereas the

lines

-^x~{-2, y = ~^x-{-3

are

mutually

y-9:=*(*-4),

(*)

where k is the slope (unspecified as yet). The required line

must be perpendicu-

lar to AB; consequently, its

slope must

satisfy

the condition

for

perpendicular-

ity with respect to the line AB. The slope of AB being equal

to

'/2, it

follows

from formula (2) that

k =

2. Inserting

this

value of k

in

equation

(*), we

obtain

=

4) or y =

y

9

2 (x

2jc-f 17.

Solving simultaneously the

equations

1

1

y=T*

5-.

y-

2* +17,

we find the coordinates of

the projection

sought:

x = 7,

y = 3.

62

Curves of the First Order

have the same abscissa, equal

to the ^-intercept

of

the line

(that is, to the value of the

segment cut off by the line on the

axis Ox\ see Fig. 40); denoting

this A'-intercept by the

letter

a,

we

thus

obtain the equation of the line

in

the form x = a which, again, is an equation

of

the

first degree.

Hence, every

straight

line is represented by an equation of the first

degree

in

cartesian

coordinates;

the first

part of

the theorem is thus proved.

z

We

now

proceed

to prove

the

converse

statement. Let there be given an equation of

the first

degree,

Fig. 40.

A* + 5y +C = 0,

(1)

y

=

A

x

C

--B

^-

A

C

6, we obtain y = kx + b\ by

Replacing

-g- by k and

-g-

by

_

C

Q

X-j.

Denoting

by

we

get

x = a,

that

the

equation

of a

line

-j

a,

is,

perpendicular

to the

axis

Ox. Thus, every

equation of

the

first

61. An equation

of

the form Ax + By +C =

is

called the

general equation of

a

straight line (inasmuch as

it is

a general