Distance, Circles, And Quadratic Equations Worksheets

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H
DISTANCE, CIRCLES, AND
QUADRATIC EQUATIONS
DISTANCE BETWEEN TWO POINTS IN THE PLANE
Suppose that we are interested in finding the distance d between two points P
) and
(x
, y
1
1
1
) in the xy-plane. If, as in Figure H.1, we form a right triangle with P
and P
as
P
(x
, y
2
2
2
1
2
vertices, then it follows from Theorem F.4 in Appendix F that the sides of that triangle have
lengths |x
− x
| and |y
− y
|. Thus, it follows from the Theorem of Pythagoras that
2
1
2
1
d =
|x
− x
|
+ |y
− y
|
=
− x
+ (y
− y
2
2
2
2
(x
)
)
2
1
2
1
2
1
2
1
and hence we have the following result.
H.1 theorem
The distance d between two points P
) and P
) in a
(x
, y
(x
, y
1
1
1
2
2
2
coordinate plane is given by
d =
− x
+ (y
− y
2
2
(1)
(x
)
)
2
1
2
1
y
P
(x
, y
)
1
1
1
y
1
d
|y
− y
|
2
1
P
(x
, y
)
2
2
2
y
2
x
x
x
1
2
|x
− x
|
Figure H.1
2
1
To apply Formula (1) the scales on the coordinate axes must be the same; otherwise, we would not
REMARK
have been able to use the Theorem of Pythagoras in the derivation. Moreover, when using Formula
(1) it does not matter which point is labeled
and which one is labeled
, since reversing the points
P
P
1
2
− x
− y
changes the signs of
and
; this has no effect on the value of
because these quantities
x
y
d
2
1
2
1
are squared in the formula. When it is important to emphasize the points, the distance between
P
1
and
is denoted by
or
.
P
d(P
, P
)
d(P
, P
)
2
1
2
2
1
Example 1
Find the distance between the points (−2, 3) and (1, 7).
Solution.
If we let (x
) be (−2, 3) and let (x
) be (1, 7), then (1) yields
, y
, y
1
1
2
2
d =
[1 − (−2)]
+ [7 − 3]
=
+ 4
=
25 = 5
2
2
2
2
3
H1

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