Distance in the Plane
The absolute value function is defined as
(
x if x ≥ 0; and
|x|
=
fills up the little circle of the second
piece
−x if
x < at
0. the point (0, 0). The result is
that there is no little circle in the graph of the absolute value function
If the number a is positive or zero, then |a| = a. If a is negative, then |a|
is the number you’d get by “erasing” the minus sign in front of it. Thus,
|5| = 5, |17| = 17, |0| = 0, | − 2| = 2, and | − 9| = 9.
Because it’s the job of the absolute value function to erase the minus sign
in front of those numbers that have them, the absolute value of a number
x is the same as the absolute value of its negative, −x. Written with math
symbols
|x| = | − x|
As examples, |4| = 4 = | − 4| and | − 10| = 10 = | − (−10)|.
Whether you square a number or its negative, you’ll get the same result.
That is, x2 = (−x)2 . And since |x| equals either x or −x, depending on
whether x is negative, we know that
|x|2 = x2
As examples, |3|2 = 32 and | − 7|2 = 72 = (−7)2 .
178
inRR
Distancein
Distance
betweenthe
thetwo
numbersa,a,bb∈ERRisis|aa− b|.
Thedistance
distancebetween
twonumbers
bI.
The
—
Examples.
Examples.
Thedistance
distancebetween
betweenthe
thenumbers
numbers55and
and33isis|55− 3|3) == |2|
121 ==2.2.
• •The
—
Thedistance
distancebetween
betweenthe
thenumbers
numbers33and
and55isis|33− 5|5) == | I− 2|2) == 2.2.
• •The
—
—
12) ==| I− 16|
16) ==16.
16.
Thedistance
distancebetween
between−4
—4and
and1212isis| − 44− 12|
• •The
—
—
—
illustrationininthe
firsttwo
examplesabove
abovethat
that
Wesaw
sawan
anillustration
thefirst
twoexamples
We
|ala-b)
− b| ==|blb-al
− a|
betweenaaand
thesame
sameas
as
Thatmakes
makessense.
sense.ItItmeans
meansthat
thatthe
thedistance
distancebetween
andbbisisthe
That
fact using
using
thedistance
distancebetween
betweenb band
anda,a,asasititshould
shouldbe.
be. We
Wecan
cancheck
checkthis
this fact
the
algebra:|aa− b|bl==| −
I (b(b− a)|a)l==|b)b− a|.al.
algebra:
substitutingaa− bbfor
forx,x, we
we have
have
Aswas
wasdiscussed
discussedabove,
above,|x|
.
Thus,substituting
2 = x
2
2
1x1
As
= x2
. Thus,
2 = (a
2
|aa− b|b)
= (a − b)2
—
—
—
—
—
—
—
Lengthsin
inright
right triangles
triangles
Lengths
triangleisisthree
threepoints
pointsininthe
theplane,
plane, with
witheach
eachpair
pairofofpoints
points joined
joined by
by
AAtriangle
thestraight
straightline
linesegment
triangle isis aa figure
figure with
with three
segmentbetween
between them.
them. AA triangle
three
the
sides,orora atrigon.
trigon.We
usuallyplace
placesosomuch
muchemphasis
emphasison
onthe
theangles
anglesofofthree
three
Weusually
sides,
sidedfigures
figuresthat
thatwe
usuallycall
callthem
themtriangles
trianglesinstead
insteadofoftrigons.
trigons.
weusually
sided
righttriangle
triangleisisa atriangle
triangleone
oneofofwhose
whoseangles
anglesisisaaright
rightangle.
angle. We’ll
We’llhave
have
AAright
moretotosay
sayabout
aboutright
rightangles
anglesand
andangles
anglesiningeneral
generalsoon.
soon. Most
Mostpeople
peopleare
are
more
comfortablewith
withwhat
angleis,is,although
althoughthere
thereare
aredifferent
different names
names
whata aright
rightangle
comfortable
163
179
for them. The three most common are an angle of 900, an angle of or one
intersecting
perpendicular
of four
resultingare
from
for them.
The equal
three angles
most common
an the
angle
of 90◦ , an of
angle
of π2 , or onelines.
of four equal angles resulting from the intersecting of perpendicular lines.
,
rit
r
L J
The side lengths of right triangles satisfy a well known equation. It’s famous
enough
andofuseful
enough that
instead
calling
it a “claim”
we call it a
The side
lengths
right triangles
satisfy
a well of
known
equation.
It’s famous
enough “theorem”.
and useful enough that instead of calling it a “claim” we call it a
“theorem”.
The Pythagorean Theorem (1). If a, b, and c are the three lengths of
sides of a right
triangle,
andIf if
theclength
the side
opposite
the
The the
Pythagorean
Theorem
(1).
a, cb,isand
are theofthree
lengths
of
right
the triangle,
then
the sides
of aangle
rightintriangle,
and if
c is the length of the side opposite the
right angle in the triangle, then
a+b=c2
a2 + b2 = c2
9
C.
9
b
0
Proof: We can draw the same triangle four times to create a giant square,
each
whose
sides
length
picture
is drawn
the next page.
a + b.four
This
on square,
Proof:
Weofcan
draw
thehave
same
triangle
times
to create
a big
164
each of whose sides have length a + b. This picture is drawn on the next page.
180
6
a.
6
b
b
Notice that the area of the big square is
Notice that the area of the giant
is
(a +square
b)2
2
The big square can be dissected into(a+b)
4 triangles and a smaller square. Each
of theThe
4 triangles
has acan
basebeofdissected
a and a into
height
of b, so they
have square.
area
giant square
4 triangles
and each
a smaller
1
Theof
smaller
in has
the amiddle
of athe
hasofsides
the 4 square
triangles
base of
andpicture
a height
b, sowhose
they lengths
each have
2 ab.Each
area
in the
middle
of the
picture
has issides
whose
equal
c. -ab.
Thus,The
the smaller
area of square
the square
in the
middle
of the
picture
c2 . The
Thus, is
thethe
area
of of
thethe
square
picture
arealengths
of the equal
entirec.picture
sum
areasinofthe
themiddle
4 triangles
the is
of theand
. The
2
c
areainofthe
themiddle
entire picture is the sum of the areas of the 4 triangles
smaller
square
and the smaller square in the middle
1
4( ab) + c2
24(ab) c
+2
We’ve calculated the area of the entire picture in two different ways, and
We’vebecalculated
the area of the entire picture in two different ways, and
they must
equal
1
they must be equal
(a + b)2 = 4( ab) + c2
2
=4(ab)+c2
2
(a+b)
We can multiply out the left side of the equation and simplify the right side:
We can multiply out the
side+of
equation
a2 left
+ 2ab
b2 the
= 2ab
+ c2 and simplify the right side:
2 + 2ab + b
a
2 = 2ab + c
2
Now subtract 2ab from both sides:
Now subtract 2ab from both asides:
2
+ b2 = c2
2+b
a
2
181
165
=
Distance in R2
The Pythagorean Theorem allows us to determine the distance between
any pair of points in the plane.
Proposition (2). The distance between two points (x1 , y1 ) and (x2 , y2 ) is
p
(x1 − x2 )2 + (y1 − y2 )2
Proof: The distance between (x1 , y1 ) and (x2 , y2 ) is the length, c, of a side
of a right triangle. Notice that c is a length, so c ≥ 0.
(x)
C
a
I x-x1
The other two sides of the triangle have length |x1 − x2 | and |y1 − y2 | so
the Pythagorean theorem tells us that the distance, c, between (x1 , y1 ) and
(x2 , y2 ) satisfies the equation
c2 = |x1 − x2 |2 + |y1 − y2 |2
Therefore, either
p
c = |x1 − x2 |2 + |y1 − y2 |2
or
p
c = − |x1 − x2 |2 + |y1 − y2 |2
Recall that c ≥ 0, so the only solution for c is
p
c = |x1 − x2 |2 + |y1 − y2 |2
We discussed earlier in this chapter that |a − b|2 = (a − b)2 for any numbers
a and b, so we can write that c, the distance between (x1 , y1 ) and (x2 , y2 ),
equals
p
(x1 − x2 )2 + (y1 − y2 )2
182
Example.
• The distance between the points (2, −3) and (−5, 8) is
p
p
√
√
(2 − (−5))2 + (−3 − 8)2 = (7)2 + (−11)2 = 49 + 121 = 170
Norms of vectors
The norm of a vector (a, b) is a number, usually written as ||(a, b)||, that
is the distance
between the point (a, b) and the point (0, 0). Thus, ||(a, b)||
p
equals (a − 0)2 + (b − 0)2 which simplifies as
(-1,3)
p
3
||(a, b)|| = a2 + b2
If you think of a vector as an arrow, then its norm is the length of the arrow.
Example.
• The norm of the vector (−1, 3) is
p
√
√
2
2
||(−1, 3)|| = (−1) + 3 = 1 + 9 = 10
(-1,3)
3
b
0.
Norms via dot products
Recall from the exercises in the chapter “The Plane of Vectors” that
a
(a, b)
= a2 + b2
b
b
is sometimes called the “dot product” of the vectors (a, b) and (a, b). Thus,
the norm of (a, b) is the square root of the dot product of the vectors (a, b)
and (a, b). That is how some prefer to think about norms.
183
0.
Exercises
For #1-14, provide the value asked for.
1.) |4|
2.) |0|
3.) | − 4|
4.) |3 − 4|
5.) |4 − 7|
6.) |10 − (−4)|
7.) the distance between 7 and −4
8.) the distance between −5 and 9
9.) the distance between (8, −2) and (2, 5)
10.) the distance between (3, 4) and (1, 7)
11.) the distance between (10, 4) and (−2, 4)
12.) the distance between (−3, −7) and (−1, −5)
13.) the distance between (−9, 3) and (0, 0)
14.) | − (−6)|
For #15-18, give the norms of the vectors.
15.) ||(1, 3)||
16.) ||(−4, 6)||
17.) ||(−1, −5)||
18.) ||(7, −3)||
For #19-21, find the solutions of the equations in one variable.
19.) (2x2 − 3x − 2)2 = 1
20.) (2 − ex )2 = 9
184
21.)
p
loge (x) + 27 = −2
y)
y2)
of a right triangle.
y2)
Yi)
Distance in R
2
between
distance
The
Pythagorean
allows
us to determine
thethe
distance
to #22-27,
determine
orem allows usFor
usethe
theTheorem
Pythagorean
Theorem
to determine
value ofbetween
x, the
any
pair
of
points
in
plane.
the
plane.
length of the specified side of the right triangle.
is
, y2)between
2
(x
Proposition
Theand
distance
two points 1
(x Y2) is
,
two points (2).
, yj)
1
(x
e distance between
(x Yi) and 2
,
____________________
_____________________
) +25.)
2
x
) +22.)
2
x
1
(x
(Yi Y2)
2
(Yi Y2)
2
—
—
—
—
a side
distance
between
,
1
c, of (x
the length,
, y2) is the length, c, of a side
2
, Yl) Proof:
1
and (x
, The
2
etween (x
Yi) and (x
y2) is
15
of a right triangle.
9
3
Distance in R
2
distance between
orem allows us The
to determine
the Theorem
Pythagorean
allows us to determine the distance between
plane.
any pair of points in the plane.
is
e distance between
, y’)
1
(x
, Y2)between
2
(x
two points (xi, yl) and (12, y2) is
two points (2).
Proposition
Theand
distance
triangle
length x
The
other
two
sides
of
the
have
1 x
23.)
26.)
the triangle have length lxi x
and lYi Y2l °
2 and ly’ Y21 °
1
2
12)2
)
2
1
(y distance,
+the
Y2)
2
Y2)
(yl
Pythagorean
tells
us that
c, between (x
, y) and
1
m tells us x
thatthe
distance,
c, theorem
between (x
and +the
, Yi)
1
(x
satisfies
the
,
2
tion
Y2)(x
isdistance
length,
c, of (xi,
a side
etween
, Yi)(x
1
(x
and
,
2
the equation
Proof:
The
between
Y2) I0
Yi) and (12, Y2) isxthe length, c, of a side
triangle.
2 = jx
C
1 x
j + Ii Y21
2
= jx
1 x
2
lof +a right
2
IYi Y21
Therefore,
C = VIxi
2 + ly’
X2
C = lxi
2
Y2
1 + lYi Y21
2
x
2
7
We jadiscussed
earlier
that a
= (a
2 for any numbers
b)
his chapter that
2 = (a
bj
2inforthis
b)
anychapter
numbers
write
that (x
, y2),
2
and b, between
so we can(xi,
c, the
, yi) and (x
1
e that c, the adistance
,
2
y2),distance between (x
Yi) and
equals
) +27.)
2
x
) have
2
x
+The
2 lxi
Y2)
(Y1 length
2 two
Y2)
(Yi length
other
of the
1 sides
x
f the triangle
1 and lYi Y21 °
2
x
1 and
2
x
Y2l °have
li triangle
24.)
us that
and the distance, c, between (x
, Yi) and
1
, Yl)
1
(x
distance, c, theorem
between tells
thePythagorean
m tells us thatthe
166
i66
, Y2) satisfies the equation
2
(x
ation
—
—
—
—
—
—
—
—
—
—
—
—
—
—
—
—
—
—
—
—
—
—
—
—
—
—
-
I0
= lxi
1 + lYi Y21
2
x
Y2l
x
x
Therefore,
C = \,/JXi
C = lxi
1 + Yi Y21
2
x
1 + li Y21
2
x
2
2
We discussed
this
that a b1
2 for any numbers
anychapter
numbers
(a12 b)
2infor
his chapter that
2 = (a9 b)
a bl
2 = earlier
and b, so
we can(Xi,
write
c, the
, y2),
2
and (x
,
2
between
e that c, the adistance
y) that
y2),distance between (xi, Yi) and (x
equals
) + (Yi Y2)
2
x
) + (Yi Y2)
2
x
2
2
f the triangle have
1 sides
x
and
The length
other two
of
the
triangle
1
2
x
Ii Y2i °have length x 121 and lyi Y21 °
m tells us that
distance, c, theorem
between tells
, y)
1
(x
and
thethePythagorean
us that
166 the distance, c, between (Xj, Yl) and
i66
ation
(12, y2) satisfies the equation
185
2 + IYi
xj
2 =
C
= lxi
121 + yi Y21
Y21
=
l
2
lxi x
—
+
lYi
—
—
—
—
—
—
—
—
—
—
—
—
—
—
—
—
—
—
—
—
—
—
Therefore,
1212 + Iyl Y2l
C =
C = Ixi
1 + lYi
2
x
2
Y21
numbers
his chapter that
2 = earlier
bj
(a b)
2infor
anychapter
We Ia
discussed
this
that Ia bI
2 = (a
—
—
—
—
—
—
—
2 for any numbers
b)