3. The Distance Formula
P = –2 Q = 3For example, if P = –2, and Q = 3
To find distances between two points on a line, we subtract.
4. The Distance Formula
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
To find distances between two points on a line, we subtract.
5. The Distance Formula
The absolute value “| |” is to remind us to disregard the
negative sign if we subtracted (–2) – (3) → –5.
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
To find distances between two points on a line, we subtract.
6. The Distance Formula
The absolute value “| |” is to remind us to disregard the
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|).
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
To find distances between two points on a line, we subtract.
7. The Distance Formula
The absolute value “| |” is to remind us to disregard the
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|). However it’s difficult to
manipulate the | |–notation algebraically, e.g. |x + 3| ≠ |x| + 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
To find distances between two points on a line, we subtract.
8. The Distance Formula
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
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.
To find distances between two points on a line, we subtract.
The absolute value “| |” is to remind us to disregard the
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|). However it’s difficult to
manipulate the | |–notation algebraically, e.g. |x + 3| ≠ |x| + 3.
9. The Distance Formula
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
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.
To find distances between two points on a line, we subtract.
(–2) – 3 = –5
Using the above example, if we calculated the distance as
The absolute value “| |” is to remind us to disregard the
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|). However it’s difficult to
manipulate the | |–notation algebraically, e.g. |x + 3| ≠ |x| + 3.
10. The Distance Formula
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
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.
To find distances between two points on a line, we subtract.
(–2) – 3 = –5
then ( )2
25
Using the above example, if we calculated the distance as
The absolute value “| |” is to remind us to disregard the
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|). However it’s difficult to
manipulate the | |–notation algebraically, e.g. |x + 3| ≠ |x| + 3.
11. The Distance Formula
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
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.
To find distances between two points on a line, we subtract.
(–2) – 3 = –5
then ( )2
25
then √
+5
to get the positive answer.
Using the above example, if we calculated the distance as
The absolute value “| |” is to remind us to disregard the
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|). However it’s difficult to
manipulate the | |–notation algebraically, e.g. |x + 3| ≠ |x| + 3.
12. The Distance Formula
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
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.
To find distances between two points on a line, we subtract.
(–2) – 3 = –5
then ( )2
25
then √
+5
to get the positive answer. We summarize this procedure below.
Using the above example, if we calculated the distance as
The absolute value “| |” is to remind us to disregard the
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|). However it’s difficult to
manipulate the | |–notation algebraically, e.g. |x + 3| ≠ |x| + 3.
14. The Distance Formula
Let the Greek letter Δ (cap. delta) mean “take the difference”. So
given two numbers x1, x2, Δx = x2 – x1.
15. The Distance Formula
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
.
16. The Distance Formula
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
.
Example A.
Let P = –1, find the points that are 2 units away from P using the
distance formula.
17. The Distance Formula
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
.
Example A.
Let P = –1, find the points that are 2 units away from P using the
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
18. The Distance Formula
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
.
Example A.
Let P = –1, find the points that are 2 units away from P using the
distance formula.
Let x be the coordinators of
the points in question.
x x
2
P = –1
2
19. The Distance Formula
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
.
Example A.
Let P = –1, find the points that are 2 units away from P using the
distance formula. x x
2
P = –1
2Let x be the points
in question. Given P = –1 and x, Δx = x – (–1) = x + 1,
20. The Distance Formula
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
.
Example A.
Let P = –1, find the points that are 2 units away from P using the
distance formula.
and “–1 and x are 2 units apart” means √Δx2
is 2
x x
2
P = –1
2Let x be the points
in question. Given P = –1 and x, Δx = x – (–1) = x + 1,
21. The Distance Formula
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
.
Example A.
Let P = –1, find the points that are 2 units away from P using the
distance formula.
and “–1 and x are 2 units apart” means √Δx2
is 2 or that
√(x + 1)2
= 2
x x
2
P = –1
2Let x be the points
in question. Given P = –1 and x, Δx = x – (–1) = x + 1,
22. The Distance Formula
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
.
Example A.
Let P = –1, find the points that are 2 units away from P using the
distance formula.
and “–1 and x are 2 units apart” means √Δx2
is 2 or that
√(x + 1)2
= 2 (x + 1)2
= 4 so
square both sides
x x
2
P = –1
2Let x be the points
in question. Given P = –1 and x, Δx = x – (–1) = x + 1,
23. The Distance Formula
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
.
Example A.
Let P = –1, find the points that are 2 units away from P using the
distance formula.
and “–1 and x are 2 units apart” means √Δx2
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
2Let x be the points
in question. Given P = –1 and x, Δx = x – (–1) = x + 1,
24. The Distance Formula
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
.
Example A.
Let P = –1, find the points that are 2 units away from P using the
distance formula.
and “–1 and x are 2 units apart” means √Δx2
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
2Let x be the points
in question. Given P = –1 and x, Δx = x – (–1) = x + 1,
25. The Distance Formula
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
.
Example A.
Let P = –1, find the points that are 2 units away from P using the
distance formula. x x
2
P = –1
2
and “–1 and x are 2 units apart” means √Δx2
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
in question. Given P = –1 and x, Δx = x – (–1) = x + 1,
26. The Distance Formula
The distance formula for the real line extends nicely to the
distance formula for the plane by the Pythagorean theorem.
27. The Distance Formula
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
28. The Distance Formula
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
29. Δy = the difference between the y's = y2 – y1
The Distance Formula
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
y
P=(x1, y1)
Q = (x2,
y2)
Δy
x
Let P = (x1, y1) and Q = (x2, y2) be two points in the plane and
30. Δy = the difference between the y's = y2 – y1
Δx = the difference between the x's = x2 – x1
The Distance Formula
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
y
P=(x1, y1)
Q = (x2,
y2)
Δx
Δy
x
Let P = (x1, y1) and Q = (x2, y2) be two points in the plane and
31. Example B.
Let P =(–2, 2) and Q = (2, –1). Find Δy, Δx
and the distance r between them.
Δy = the difference between the y's = y2 – y1
Δx = the difference between the x's = x2 – x1
The Distance Formula
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
Δx
Δy
x
y
P=(x1, y1)
Q = (x2,
y2)
x
P =(–2, 2)
Q =(2, –1)
y
Let P = (x1, y1) and Q = (x2, y2) be two points in the plane and
32. Let P = (x1, y1) and Q = (x2, y2) be two points in the plane and
Example B.
Let P =(–2, 2) and Q = (2, –1). Find Δy, Δx
and the distance r between them.
Δy = (–1) – (2) = –3
Δy = –3
Δy = the difference between the y's = y2 – y1
Δx = the difference between the x's = x2 – x1
The Distance Formula
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
Δx
Δy
x
y
P=(x1, y1)
Q = (x2,
y2)
x
P =(–2, 2)
Q =(2, –1)
y
33. Example B.
Let P =(–2, 2) and Q = (2, –1). Find Δy, Δx
and the distance r between them.
Δy = (–1) – (2) = –3
Δx = (2) – (–2) = 4
Δy = –3
Δx = 4
Δy = the difference between the y's = y2 – y1
Δx = the difference between the x's = x2 – x1
The Distance Formula
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
Δx
Δy
x
y
P=(x1, y1)
Q = (x2,
y2)
x
P =(–2, 2)
Q =(2, –1)
y
Let P = (x1, y1) and Q = (x2, y2) be two points in the plane and
34. Example B.
Let P =(–2, 2) and Q = (2, –1). Find Δy, Δx
and the distance r between them.
Δy = (–1) – (2) = –3
Δx = (2) – (–2) = 4
By the Pythagorean theorem
r = √(–3)2
+ 42
Δy = –3
Δx = 4
Δy = the difference between the y's = y2 – y1
Δx = the difference between the x's = x2 – x1
The Distance Formula
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
Δx
Δy
x
y
P=(x1, y1)
x
P =(–2, 2)
Q =(2, –1)
y
r
Q = (x2,
y2)
Let P = (x1, y1) and Q = (x2, y2) be two points in the plane and
35. Example B.
Let P =(–2, 2) and Q = (2, –1). Find Δy, Δx
and the distance r between them.
Δy = (–1) – (2) = –3
Δx = (2) – (–2) = 4
By the Pythagorean theorem
r = √(–3)2
+ 42
= √25 = 5.
Δy = –3
Δx = 4
Δy = the difference between the y's = y2 – y1
Δx = the difference between the x's = x2 – x1
The Distance Formula
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
Δx
Δy
x
y
P=(x1, y1)
x
P =(–2, 2)
Q =(2, –1)
y
r=√(–3)2
+42
= 5
Q = (x2,
y2)
Let P = (x1, y1) and Q = (x2, y2) be two points in the plane and
36. Example B.
Let P =(–2, 2) and Q = (2, –1). Find Δy, Δx
and the distance r between them.
Δy = (–1) – (2) = –3
Δx = (2) – (–2) = 4
By the Pythagorean theorem
r = √(–3)2
+ 42
= √25 = 5.
Δy = –3
Δx = 4
Δy = the difference between the y's = y2 – y1
Δx = the difference between the x's = x2 – x1
The Distance Formula
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
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)
Let P = (x1, y1) and Q = (x2, y2) be two points in the plane and
37. We may also use the √ – method to solve these equations.
The Distance Formula
38. Example C. Find the coordinates of the locations on the y–axis
that are 3 units from the point (2,1).
We may also use the √ – method to solve these equations.
The Distance Formula
39. Example C. Find the coordinates of the locations on the y–axis
that are 3 units from the point (2,1).
We may also use the √ – method to solve these equations.
(2, 1)
y–axis
r = 3
The Distance Formula
40. Example C. Find the coordinates of the locations on the y–axis
that are 3 units from the point (2,1).
We may also use the √ – method to solve these equations.
The coordinate of any point on the y–axis
is of the form (0, y).
(2, 1)
y–axis
r = 3
(0, y)
The Distance Formula
41. Example C. Find the coordinates of the locations on the y–axis
that are 3 units from the point (2,1).
We may also use the √ – method to solve these equations.
The coordinate of any point on the y–axis
is of the form (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)
The Distance Formula
42. Example C. Find the coordinates of the locations on the y–axis
that are 3 units from the point (2,1).
We may also use the √ – method to solve these equations.
where Δy = y – 1, Δx = 0 – 2 = –2.
The coordinate of any point on the y–axis
is of the form (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)
The Distance Formula
43. Example C. Find the coordinates of the locations on the y–axis
that are 3 units from the point (2,1).
We may also use the √ – method to solve these equations.
where Δy = y – 1, Δx = 0 – 2 = –2. Therefore
The coordinate of any point on the y–axis
is of the form (0, y).
For a point (0, y) to be 3 units from (2, 1) means
the distance r = √Δy2
+ Δx2
= 3
√(y – 1)2
+ (–2)2
= 3
(2, 1)
y–axis
r = 3
(0, y)
The Distance Formula
44. Example C. Find the coordinates of the locations on the y–axis
that are 3 units from the point (2,1).
We may also use the √ – method to solve these equations.
where Δy = y – 1, Δx = 0 – 2 = –2. Therefore
The coordinate of any point on the y–axis
is of the form (0, y).
For a point (0, y) to be 3 units from (2, 1) means
the distance r = √Δy2
+ Δx2
= 3
√(y – 1)2
+ 4 = 3
(2, 1)
y–axis
r = 3
(0, y)
The Distance Formula
√(y – 1)2
+ (–2)2
= 3
45. Example C. Find the coordinates of the locations on the y–axis
that are 3 units from the point (2,1).
We may also use the √ – method to solve these equations.
where Δy = y – 1, Δx = 0 – 2 = –2. Therefore
The coordinate of any point on the y–axis
is of the form (0, y).
For a point (0, y) to be 3 units from (2, 1) means
the distance r = √Δy2
+ Δx2
= 3
√(y – 1)2
+ 4 = 3 square both sides
(y – 1)2
+ 4 = 9
(2, 1)
y–axis
r = 3
(0, y)
The Distance Formula
√(y – 1)2
+ (–2)2
= 3
46. Example C. Find the coordinates of the locations on the y–axis
that are 3 units from the point (2,1).
We may also use the √ – method to solve these equations.
where Δy = y – 1, Δx = 0 – 2 = –2. Therefore
The coordinate of any point on the y–axis
is of the form (0, y).
For a point (0, y) to be 3 units from (2, 1) means
the distance r = √Δy2
+ Δx2
= 3
√(y – 1)2
+ 4 = 3 square both sides
(y – 1)2
+ 4 = 9
(y – 1) = ±√5
take the ± square roots
(2, 1)
y–axis
r = 3
(0, y)
The Distance Formula
√(y – 1)2
+ (–2)2
= 3
47. Example C. Find the coordinates of the locations on the y–axis
that are 3 units from the point (2,1).
We may also use the √ – method to solve these equations.
where Δy = y – 1, Δx = 0 – 2 = –2. Therefore
The coordinate of any point on the y–axis
is of the form (0, y).
For a point (0, y) to be 3 units from (2, 1) means
the distance r = √Δy2
+ Δx2
= 3
√(y – 1)2
+ 4 = 3 square both sides
(y – 1)2
+ 4 = 9
(y – 1) = ±√5
take the ± square roots
So y = 1 ±√5
(2, 1)
y–axis
r = 3
(0, y)
The Distance Formula
√(y – 1)2
+ (–2)2
= 3
48. Example C. Find the coordinates of the locations on the y–axis
that are 3 units from the point (2,1).
We may also use the √ – method to solve these equations.
where Δy = y – 1, Δx = 0 – 2 = –2. Therefore
The coordinate of any point on the y–axis
is of the form (0, y).
For a point (0, y) to be 3 units from (2, 1) means
the distance r = √Δy2
+ 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
take the ± square roots
The Distance Formula
√(y – 1)2
+ (–2)2
= 3
49. Example C. Find the coordinates of the locations on the y–axis
that are 3 units from the point (2,1).
We may also use the √ – method to solve these equations.
where Δy = y – 1, Δx = 0 – 2 = –2. Therefore
The coordinate of any point on the y–axis
is of the form (0, y).
For a point (0, y) to be 3 units from (2, 1) means
the distance r = √Δy2
+ 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
take the ± square roots
The Distance Formula
√(y – 1)2
+ (–2)2
= 3
50. On the real line, the average of two points is the half–way point
or the mid–point between them.
The Mid–point Formulas
51. On the real line, the average of two points is the half–way point
or the mid–point between them.
For example, the average of 1 and 5 is (1 + 5)/2 = 3.
m = 3
The Mid–point Formulas
1 5
52. The mid–point m between x1 and x2 is given by
On the real line, the average of two points is the half–way point
or the mid–point between them.
For example, the average of 1 and 5 is (1 + 5)/2 = 3.
m = 3
The Mid–point Formulas
1 5
(x1 + x2)
2
m = .
53. The mid–point m between x1 and x2 is given by
On the real line, the average of two points is the half–way point
or the mid–point between them.
For example, the average of 1 and 5 is (1 + 5)/2 = 3.
m = 3
The Mid–point Formulas
1 5
(x1 + x2)
2
m = .
This mid–point formula extends to points in the x&y–plane.
54. The mid–point m between x1 and x2 is given by
On the real line, the average of two points is the half–way point
or the mid–point between them.
For example, the average of 1 and 5 is (1 + 5)/2 = 3.
m = 3
The Mid–point Formulas
1 5
(x1 + x2)
2
m = .
This mid–point formula extends to points in the x&y–plane.
The mid–point m between (x1, y1) and (x2 , y2) is given by the
mid–point of each coordinate,
55. The mid–point m between x1 and x2 is given by
On the real line, the average of two points is the half–way point
or the mid–point between them.
For example, the average of 1 and 5 is (1 + 5)/2 = 3.
m = 3
The Mid–point Formulas
1 5
(x1 + x2)
2
m = .
This mid–point formula extends to points in the x&y–plane.
The mid–point m between (x1, y1) and (x2 , y2) is given by the
mid–point of each coordinate, (x1, y1)
(x2 , y2)
x1 x2
y1
y2
The mid–point formula in the plane
56. The mid–point m between x1 and x2 is given by
On the real line, the average of two points is the half–way point
or the mid–point between them.
For example, the average of 1 and 5 is (1 + 5)/2 = 3.
m = 3
The Mid–point Formulas
1 5
(x1 + x2)
2
m = .
This mid–point formula extends to points in the x&y–plane.
The mid–point m between (x1, y1) and (x2 , y2) is given by the
mid–point of each coordinate, (x1, y1)
(x2 , y2)
x1 x2
y1
y2
x1 + x2
2
The mid–point formula in the plane
57. The mid–point m between x1 and x2 is given by
On the real line, the average of two points is the half–way point
or the mid–point between them.
For example, the average of 1 and 5 is (1 + 5)/2 = 3.
m = 3
The Mid–point Formulas
1 5
(x1 + x2)
2
m = .
This mid–point formula extends to points in the x&y–plane.
The mid–point m between (x1, y1) and (x2 , y2) is given by the
mid–point of each coordinate, (x1, y1)
(x2 , y2)
x1 x2
y1
y2
x1 + x2
2
y1 + y2
2
x1 + x2
2
( )y1 + y2
2,
The mid–point formula in the plane
58. The mid–point m between x1 and x2 is given by
On the real line, the average of two points is the half–way point
or the mid–point between them.
For example, the average of 1 and 5 is (1 + 5)/2 = 3.
m = 3
The Mid–point Formulas
1 5
(x1 + x2)
2
m = .
This mid–point formula extends to points in the x&y–plane.
The mid–point m between (x1, y1) and (x2 , y2) is given by the
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,
The mid–point formula in the plane
59. The mid–point m between x1 and x2 is given by
On the real line, the average of two points is the half–way point
or the mid–point between them.
For example, the average of 1 and 5 is (1 + 5)/2 = 3.
m = 3
The Mid–point Formulas
1 5
(x1 + x2)
2
m = .
This mid–point formula extends to points in the x&y–plane.
The mid–point m between (x1, y1) and (x2 , y2) is given by the
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,
The mid–point formula in the plane
Hence the mid–point of
(1, 2) and (3, 4) is
1 + 3
2
,m =
2 + 4
2
)(
60. The mid–point m between x1 and x2 is given by
On the real line, the average of two points is the half–way point
or the mid–point between them.
For example, the average of 1 and 5 is (1 + 5)/2 = 3.
m = 3
The Mid–point Formulas
1 5
(x1 + x2)
2
m = .
This mid–point formula extends to points in the x&y–plane.
The mid–point m between (x1, y1) and (x2 , y2) is given by the
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,
The mid–point formula in the plane
Hence the mid–point of
(1, 2) and (3, 4) is
1 + 3
2
,m =
2 + 4
2
)(
= (2, 3)m
61. The Distance Formula
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
the distance formula. Check your answers geometrically.
3. Let P = –2, find the points that are 3 units away from P using
the distance formula. Check your answers geometrically.
4. Let P = –3, find the points that are 2 units away from P using
the distance formula. Check your answers geometrically.
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.
62. The Distance Formula
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).
13. Find the exact and the approximate coordinates of the
locations on the y–axis that are 5 units from the point (–2, 2).
14. Find the exact and the approximate coordinates of the
locations on the y–axis that are 6 units from the point (2, –3).