Distance, Circles, And Quadratic Equations Worksheets Page 2
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Appendix H: Distance, Circles, and Quadratic Equations
Example 2
It can be shown that the converse of the Theorem of Pythagoras is true;
+ b
= c
2
2
2
that is, if the sides of a triangle satisfy the relationship a
, then the triangle must
be a right triangle. Use this result to show that the points A(4, 6), B(1, −3), and C(7, 5)
are vertices of a right triangle.
y
Solution.
The points and the triangle are shown in Figure H.2. From (1), the lengths of
A(4, 6)
the sides of the triangles are
√
√
C(7, 5)
d(A, B) =
(1 − 4)
+ (−3 − 6)
=
9 + 81 =
2
2
90
√
√
d(A, C) =
(7 − 4)
+ (5 − 6)
=
9 + 1 =
2
2
10
√
√
x
d(B, C) =
(7 − 1)
+ [5 − (−3)]
=
36 + 64 =
100 = 10
2
2
Since
[d(A, B)]
+ [d(A, C)]
= [d(B, C)]
2
2
2
B(1, −3)
Figure H.2
it follows that ABC is a right triangle with hypotenuse BC.
THE MIDPOINT FORMULA
It is often necessary to find the coordinates of the midpoint of a line segment joining two
points in the plane. To derive the midpoint formula, we will start with two points on a coor-
b − a
dinate line. If we assume that the points have coordinates a and b and that a ≤ b, then,
a
b
as shown in Figure H.3, the distance between a and b is b − a, and the coordinate of the
1
(b − a)
midpoint between a and b is
2
Figure H.3
a +
(b − a) =
a +
b =
(a + b)
1
1
1
1
2
2
2
2
which is the arithmetic average of a and b. Had the points been labeled with b ≤ a, the
y
same formula would have resulted (verify). Therefore, the midpoint of two points on a
y
2
P
(x
, y
)
coordinate line is the arithmetic average of their coordinates, regardless of their relative
2
2
2
positions.
y
If we now let P
) and P
) be any two points in the plane and M(x, y) the
(x
, y
(x
, y
M(x, y)
1
1
1
2
2
2
midpoint of the line segment joining them (Figure H.4), then it can be shown using similar
triangles that x is the midpoint of x
and x
on the x-axis and y is the midpoint of y
and y
1
2
1
2
y
1
P
(x
, y
)
on the y-axis, so
1
1
1
x =
+ x
) and y =
+ y
1
1
x
(x
(y
)
1
2
1
2
2
2
x
x
x
1
2
Thus, we have the following result.
Figure H.4
H.2 theorem
The midpoint of the line segment joining two
(The Midpoint Formula)
points (x
) and (x
) in a coordinate plane is
, y
, y
1
1
2
2
+ x
+ y
1
1
(2)
(x
),
(y
)
1
2
1
2
2
2
y
Find the midpoint of the line segment joining (3, −4) and (7, 2).
Example 3
(x, y)
r
Solution.
From (2) the midpoint is
(3 + 7),
(−4 + 2) = (5, −1)
1
1
(x
, y
)
0
0
2
2
CIRCLES
x
If (x
) is a fixed point in the plane, then the circle of radius r centered at (x
) is the
, y
, y
0
0
0
0
set of all points in the plane whose distance from (x
) is r (Figure H.5). Thus, a point
Figure H.5
, y
0
0
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