Transcript lesson 12-7

Distance Formula
d = √ (x1 – x2)2 + (y1 – y2)2,
where d is the distance
between the points (x1, y1)
and (x2, y2).
Example 1
Find the distance between
the points (– 3, 5) and (2, –6).
d = √ (x1 – x2)2 + (y1 – y2)2
= √ (–3 – 2)2 + [5 – (–6)]2
= √ (–5)2 + 112
= √ 146 ≈ 12.1 units
A
(x1, y1)
C
(x1, y2)
B
(x2, y2)
Midpoint Formula
The midpoint of a segment
with endpoints (x1, y1) and
x1 + x2 y1 + y2
(x2, y2) is
,
2
.
2
Example 2
Find the midpoint of the
segment extending from
(–8, –2) to (–1, 5).
–8 + (– 1) , –2 + 5
2
2
–9
3
=
,
= (– 4.5, 1.5)
2
2
Example 3
Parallelogram ABCD has
vertices at A (3, 0), B (7, 1),
C (7, 4), and D (3, 3). Find the
length of each side and the
midpoint of each diagonal.
AD = y1 – y2 = 3 – 0 = 3 units
BC = y1 – y2 = 4 – 1 = 3 units
CD = √ (7 – 3)2 + (4 – 3)2
= √ 16 + 1 = √ 17 ≈ 4.1 units
AB = √ (7 – 3)2 + (1 – 0)2
= √ 16 + 1 = √ 17 ≈ 4.1 units
Midpoint of AC
10 ,
3 + 7, 0 + 4
=
=
2
2
2
= (5, 2)
Midpoint of BD
10 ,
7 + 3, 1 + 3
=
=
2
2
2
= (5, 2)
4
2
4
2
Example
Find the distance between
the points (– 1, 6) and (2, 4).
√ 13
Example
Find the midpoint of the
segment extending from
(4, 7) to (10, 13).
(7, 10)
Example
If (–4, 5) is one endpoint of a
line segment with midpoint
(3, –2), what is the other
endpoint?
(10, –9)
Example
Graph an isosceles triangle
with vertices at (0, 0), (8, 0),
and (4, 6). Find the midpoint
of each side of the triangle.
(4, 0), (2, 3), (6, 3)
Example
Graph an isosceles triangle
with vertices at (0, 0), (8, 0),
and (4, 6). Is the figure
formed by joining the
midpoints similar to the
original triangle?
yes
Exercise
Use the distance formula to
find the lengths of the
diagonals AC and BD of
rectangle ABCD, whose
vertices are A (–1, 4),
B (7, –2), C (4, –6), and
D (–4, 0). How do these two
lengths compare?