31 the distance formulas

The Distance Formula
To find distances between two points on a line, we subtract.

P = –2 Q = 3For example, if P = –2, and Q = 3

then the distance between P and Q
is |3 – (–2)| or |–2 – 3| = 5
P = –2 Q = 3
The distance between
P and Q is |P – Q|
For example, if P = –2, and Q = 3

The absolute value “| |” is to remind us to disregard the
negative sign if we subtracted (–2) – (3) → –5.
is |3 – (–2)| or |–2 – 3| = 5
P = –2 Q = 3

negative sign if we subtracted (–2) – (3) → –5. Hence the
distance between two points P = x1 and Q = x2 on the line may
be written as |x2 – x1| (or |x1 – x2|).
is |3 – (–2)| or |–2 – 3| = 5
P = –2 Q = 3

be written as |x2 – x1| (or |x1 – x2|). However it’s difficult to
manipulate the | |–notation algebraically, e.g. |x + 3| ≠ |x| + 3.
is |3 – (–2)| or |–2 – 3| = 5
P = –2 Q = 3

is |3 – (–2)| or |–2 – 3| = 5
P = –2 Q = 3
Hence we use a round about but workable method to discard
the negative sign, i.e. we square the difference then take its
square root.

is |3 – (–2)| or |–2 – 3| = 5
P = –2 Q = 3
square root.
(–2) – 3 = –5
Using the above example, if we calculated the distance as

is |3 – (–2)| or |–2 – 3| = 5
P = –2 Q = 3
square root.
(–2) – 3 = –5
then ( )2
25

is |3 – (–2)| or |–2 – 3| = 5
P = –2 Q = 3
square root.
(–2) – 3 = –5
then ( )2
25
then √
+5
to get the positive answer.

is |3 – (–2)| or |–2 – 3| = 5
P = –2 Q = 3
square root.
(–2) – 3 = –5
then ( )2
25
then √
+5
to get the positive answer. We summarize this procedure below.

Let the Greek letter Δ (cap. delta) mean “take the difference”.

Let the Greek letter Δ (cap. delta) mean “take the difference”. So
given two numbers x1, x2, Δx = x2 – x1.

The Distance Formula on the Real Line
Let P = x1, Q = x2 be two points on the line, then the distance
between P and Q is
√(x2 – x1)2
or √Δx2
.
P = x1 Q = x2
The distance = √Δx2
.

between P and Q is
√(x2 – x1)2
or √Δx2
.
P = x1 Q = x2
.
Example A.
Let P = –1, find the points that are 2 units away from P using the
distance formula.

between P and Q is
√(x2 – x1)2
or √Δx2
.
P = x1 Q = x2
.
Example A.
distance formula.
Note that we may easily draw and
find the solutions geometrically.
The above definition provides an
algebraic method.
x x
2
P = –1
2
x = 1x = –3

between P and Q is
√(x2 – x1)2
or √Δx2
.
P = x1 Q = x2
.
Example A.
distance formula.
Let x be the coordinators of
the points in question.
x x
2
P = –1
2

between P and Q is
√(x2 – x1)2
or √Δx2
.
P = x1 Q = x2
.
Example A.
distance formula. x x
2
P = –1
2Let x be the points
in question. Given P = –1 and x, Δx = x – (–1) = x + 1,

between P and Q is
√(x2 – x1)2
or √Δx2
.
P = x1 Q = x2
.
Example A.
distance formula.
and “–1 and x are 2 units apart” means √Δx2
is 2
x x
2
P = –1

between P and Q is
√(x2 – x1)2
or √Δx2
.
P = x1 Q = x2
.
Example A.
distance formula.
is 2 or that
√(x + 1)2
= 2
x x
2
P = –1

between P and Q is
√(x2 – x1)2
or √Δx2
.
P = x1 Q = x2
.
Example A.
distance formula.
is 2 or that
√(x + 1)2
= 2 (x + 1)2
= 4 so
square both sides
x x
2
P = –1

between P and Q is
√(x2 – x1)2
or √Δx2
.
P = x1 Q = x2
.
Example A.
distance formula.
is 2 or that
√(x + 1)2
= 2 (x + 1)2
= 4 so
x2
+ 2x – 3 = 0
square both sides
x x
2
P = –1

between P and Q is
√(x2 – x1)2
or √Δx2
.
P = x1 Q = x2
.
Example A.
distance formula.
is 2 or that
√(x + 1)2
= 2 (x + 1)2
= 4 so
x2
+ 2x – 3 = 0
(x – 1)(x + 3) = 0 and x = 1, –3.
square both sides
x x
2
P = –1

between P and Q is
√(x2 – x1)2
or √Δx2
.
P = x1 Q = x2
.
Example A.
distance formula. x x
2
P = –1
2
is 2 or that
x2
+ 2x – 3 = 0
(x – 1)(x + 3) = 0 and x = 1, –3. Both answers are good.
√(x + 1)2
= 2 (x + 1)2
= 4 so
square both sides
Let x be the points

The distance formula for the real line extends nicely to the
distance formula for the plane by the Pythagorean theorem.

The Distance Formula in the x&y Plane

x
y
P=(x1, y1)
Q = (x2,
y2)
Let P = (x1, y1) and Q = (x2, y2) be two points in the plane and

Δy = the difference between the y's = y2 – y1
y
P=(x1, y1)
Q = (x2,
y2)
Δy
x

Δx = the difference between the x's = x2 – x1
y
P=(x1, y1)
Q = (x2,
y2)
Δx
Δy
x

Example B.
Let P =(–2, 2) and Q = (2, –1). Find Δy, Δx
and the distance r between them.
Δx
Δy
x
y
P=(x1, y1)
Q = (x2,
y2)
x
P =(–2, 2)
Q =(2, –1)
y

Example B.
Δy = (–1) – (2) = –3
Δy = –3
Δx
Δy
x
y
P=(x1, y1)
Q = (x2,
y2)
x
P =(–2, 2)
Q =(2, –1)
y

Example B.
Δy = (–1) – (2) = –3
Δx = (2) – (–2) = 4
Δy = –3
Δx = 4
Δx
Δy
x
y
P=(x1, y1)
Q = (x2,
y2)
x
P =(–2, 2)
Q =(2, –1)
y

Example B.
Δy = (–1) – (2) = –3
Δx = (2) – (–2) = 4
By the Pythagorean theorem
r = √(–3)2
+ 42
Δy = –3
Δx = 4
Δx
Δy
x
y
P=(x1, y1)
x
P =(–2, 2)
Q =(2, –1)
y
r
Q = (x2,
y2)

Example B.
Δy = (–1) – (2) = –3
Δx = (2) – (–2) = 4
r = √(–3)2
+ 42
= √25 = 5.
Δy = –3
Δx = 4
Δx
Δy
x
y
P=(x1, y1)
x
P =(–2, 2)
Q =(2, –1)
y
r=√(–3)2
+42
= 5
Q = (x2,
y2)

Example B.
Δy = (–1) – (2) = –3
Δx = (2) – (–2) = 4
r = √(–3)2
+ 42
= √25 = 5.
Δy = –3
Δx = 4
Δx
Δy
x
y
P=(x1, y1)
x
The distance r between P and Q is given by
r = √Δy2
+ Δx2
= √(y2 – y1)2
+ (x2 – x1)2
.
P =(–2, 2)
Q =(2, –1)
y
r=√(–3)2
+42
= 5
r = √Δy2
+ Δx2
Q = (x2,
y2)

We may also use the √ – method to solve these equations.

Example C. Find the coordinates of the locations on the y–axis
that are 3 units from the point (2,1).

(2, 1)
y–axis
r = 3

The coordinate of any point on the y–axis
is of the form (0, y).
(2, 1)
y–axis
r = 3
(0, y)

For a point (0, y) to be 3 units from (2, 1) means
the distance r = √Δy2
+ Δx2
= 3 (2, 1)
y–axis
r = 3
(0, y)

where Δy = y – 1, Δx = 0 – 2 = –2.
+ Δx2
= 3 (2, 1)
y–axis
r = 3
(0, y)

where Δy = y – 1, Δx = 0 – 2 = –2. Therefore
+ Δx2
= 3
√(y – 1)2
+ (–2)2
= 3
(2, 1)
y–axis
r = 3
(0, y)

+ Δx2
= 3
√(y – 1)2
+ 4 = 3
(2, 1)
y–axis
r = 3
(0, y)
√(y – 1)2
+ (–2)2
= 3

+ Δx2
= 3
√(y – 1)2
+ 4 = 3 square both sides
(y – 1)2
+ 4 = 9
(2, 1)
y–axis
r = 3
(0, y)
√(y – 1)2
+ (–2)2
= 3

+ Δx2
= 3
√(y – 1)2
(y – 1)2
+ 4 = 9
(y – 1) = ±√5
take the ± square roots
(2, 1)
y–axis
r = 3
(0, y)
√(y – 1)2
+ (–2)2
= 3

+ Δx2
= 3
√(y – 1)2
(y – 1)2
+ 4 = 9
(y – 1) = ±√5
So y = 1 ±√5
(2, 1)
y–axis
r = 3
(0, y)
√(y – 1)2
+ (–2)2
= 3

+ Δx2
= 3
√(y – 1)2
+ 4 = 3
(y – 1)2
+ 4 = 9
(y – 1) = ±√5
(2, 1)
y–axis
r = 3
(0, y)
r = 3
So y = 1 ±√5 and we conclude that there are two points that are
6 units from (2, 1).
square both sides
√(y – 1)2
+ (–2)2
= 3

+ Δx2
= 3
√(y – 1)2
+ 4 = 3
(y – 1)2
+ 4 = 9
(y – 1) = ±√5
(2, 1)
y–axis
r = 3
(0, y)
r = 3
So y = 1 ±√5 and we conclude that there are two points that are
3 units from (2, 1).
They are (0,1 +√5 ≈ 3.24) and (0,1 – √5 ≈ –1.24).
square both sides
√(y – 1)2
+ (–2)2
= 3

On the real line, the average of two points is the half–way point
or the mid–point between them.
The Mid–point Formulas

For example, the average of 1 and 5 is (1 + 5)/2 = 3.
m = 3
1 5

The mid–point m between x1 and x2 is given by
m = 3
1 5
(x1 + x2)
2
m = .

m = 3
1 5
(x1 + x2)
2
m = .
This mid–point formula extends to points in the x&y–plane.

m = 3
1 5
(x1 + x2)
2
m = .
The mid–point m between (x1, y1) and (x2 , y2) is given by the
mid–point of each coordinate,

m = 3
1 5
(x1 + x2)
2
m = .
mid–point of each coordinate, (x1, y1)
(x2 , y2)
x1 x2
y1
y2
The mid–point formula in the plane

m = 3
1 5
(x1 + x2)
2
m = .
(x2 , y2)
x1 x2
y1
y2
x1 + x2
2

m = 3
1 5
(x1 + x2)
2
m = .
(x2 , y2)
x1 x2
y1
y2
x1 + x2
2
y1 + y2
2
x1 + x2
2
( )y1 + y2
2,

m = 3
1 5
(x1 + x2)
2
m = .
mid–point of each coordinate, i.e.
x1 + x2
2 ,m =
y1 + y2
2
)(
(x1, y1)
(x2 , y2)
x1 x2
y1
y2
x1 + x2
2
y1 + y2
2
x1 + x2
2
( )y1 + y2
2,

m = 3
1 5
(x1 + x2)
2
m = .
x1 + x2
2 ,m =
y1 + y2
2
)(
(x1, y1)
(x2 , y2)
x1 x2
y1
y2
x1 + x2
2
y1 + y2
2
x1 + x2
2
( )y1 + y2
2,
Hence the mid–point of
(1, 2) and (3, 4) is
1 + 3
2
,m =
2 + 4
2
)(

m = 3
1 5
(x1 + x2)
2
m = .
x1 + x2
2 ,m =
y1 + y2
2
)(
(x1, y1)
(x2 , y2)
x1 x2
y1
y2
x1 + x2
2
y1 + y2
2
x1 + x2
2
( )y1 + y2
2,
Hence the mid–point of
(1, 2) and (3, 4) is
1 + 3
2
,m =
2 + 4
2
)(
= (2, 3)m

Exercise A.
1. Let P = 1, find the points that are 2 units away from P using
the distance formula. Check your answers geometrically.
2. Let P = 4, find the points that are 2 units away from P using
3. Let P = –2, find the points that are 3 units away from P using
4. Let P = –3, find the points that are 2 units away from P using
Exercise B. Given P and Q, find the distances between them
and their mid–point.
6. If P = (–3, 2), Q = (–2, –1)
7. If P = (5, 3), Q = (2, 0) 8. If P = (3, –4), Q = (–2, 4)
5. I f P = (1, 4), Q = (2, –1)
Exercise C.
9. Find the coordinates of the locations on the y–axis that are
5 units from the point (3,1). Draw.

10. Find the coordinates of the locations on the y–axis
that are 5 units from the point (–1, 4). Draw.
11. Find the coordinates of the locations on the x–axis
that are 10 units from the point (–6, 2). Draw.
12. Find the exact and the approximate coordinates of the
locations on the x–axis that are 4 units from the point (1, 2).
locations on the y–axis that are 5 units from the point (–2, 2).
locations on the y–axis that are 6 units from the point (2, –3).

31 the distance formulas

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31 the distance formulas