Transcript Geometry - San Ramon Valley High School
Geometry
13.1 The Distance Formula
A(-5, 4) . .
B(-1, 4)
Example 1
Find the distance between the two points.
a.
A
( –5, 4) and | -5 – (-1) |
B
( –1, 4) b.
C
(2, –5) and
D
(2, 7) | 7 – (-5) | 4 units 12 units As you can see, if two points share an x coordinate or a y coordinate, the distance between the two points can be found by finding the distance (the absolute value of the difference) between the other coordinates.
What if two points do not lie on a horizontal or vertical line?
When two points do not lie on a horizontal or vertical line, you can find the distance between points by using the distance formula which is derived from the Pythagorean Theorem.
The distance d between points
( , 1 1 )
and
( , 2 2 )
is:
Why? Let’s try an example to find out!
d
(
x
2
x
1 ) 2 (
y
2
y
1 ) 2
Example 2
Find the distance between ( –3, 4) and (1, –4).
3 1 4 ( 4 ) 2 16 64 80 (-3, 4) 4 √5 .
4 Pythagorean Theorem!
.
8 (1, -4) 4 5
Use the distance formula to find the distance between the two points.
4. (4, 2) and (1,-1) Answer: 3 √2
Example 3
Given points
A
(1, 1),
B
( –3, –11), and
C
(4, 10), find
AB
,
AC
, and
BC
. Are
A
,
B
, and
C
collinear? If so, which point lies between the other two?
AB = √(1+3) 2 + (1+11) 2 = √16 + 144 = √160 = 4√10 BC = √(-3 – 4) 2 + (-11 - 10) 2 = √49 + 441 = √490 = 7√10 AC = √(1 – 4) 2 + (1 – 10) 2 = √9 + 81 = √90 = 3√10 Are A, B, and C collinear?
.
.
If yes, the two small distances add to the large distance.
4 √10 + 3√10 = 7√10 The points are collinear!!
.
Since BC is the longest distance, B and C are the endpoints.
Thus, point A must be in the middle.
Pick one, #1 or #2!!
12. Show that the triangle with vertices P(5, 0), E( –1, –2), and T(3, 6) is isosceles.
PE = √(5+1) 2 + (0+2) 2 ET = √(-1 – 3) 2 + (-2 – 6) 2 PT = √(5 – 3) 2 + (0 – 6) 2 = √36 + 4 = √40 = 2√10 = √16 + 64 = √80 = 4√5 = √4 + 36 = √40 = 2√10 The triangle is isosceles because two of its sides are the same length!
13. Quadrilateral GEMA has vertices G(3, 8), E(8, –3), M(–2, –5), and A(–5, 2). Show that its diagonals are congruent.
GM = √(3+2) 2 + (8+5) 2 EA = √(8+5) 2 + (-3 – 2) 2 = √25 + 169 = √194 No need to simpilfy… = √169 + 25 = √194 The diagonals are congruent!
An equation of the circle with center (a, b) and radius r is:
(
x
a
) 2 (
y
b
) 2
r
2 This is the circle formula to remember!!
Why is the circle formula in the section about the distance formula?
If you rearrange the formula above it reads…
r
(
x
a
) 2 (
y
b
) 2 Now that we know the distance formula, this represents all points that are “r distance” away from a point (a, b).
This is precisely a circle!!!
An equation of the circle with center (a, b) and radius r is:
(
x
a
) 2 (
y
b
) 2
r
2 How could this be a circle?
Let’s analyze (x – 0) 2 + (y – 0) 2 = 81 to see if it really is a circle!!
Example 4
Find the center and radius of the circle with the equation: (
x
3) 2 (
y
5) 2 4 Center: (-3, 5) Radius = 2
Write an equation of the circle that has the given center and radius.
14.
C
(0, 0);
r
= 9 15.
C
(-3, -8);
r
= 6 16.
C
(1, -2);
r
= 3 x 2 + y 2 = 81 (x + 3) 2 + (y + 8) 2 = 36 (x – 1) 2 + (y + 2) 2 = 3
Find the center and radius of each circle. Sketch the graph.
17. (
x
2) 2 (
y
4) 2 9 18.
(
x
7) 2 (
y
3) 2 25 .
Center: (2, -4) Radius = 3 .
Center: (-7, -3) Radius = 5 19.
x
2 (
y
1) 2 49 20.
(
x
1) 2 (
y
2) 2 9 16 Center: (1, 2) Radius = 3/4 .
Radius = 7
• • • • • • • • • • • •
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