Chabot Mathematics
§10.1 Distance
MIdPoint Eqns
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer
BMayer@ChabotCollege.edu
Chabot College Mathematics
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BMayer@ChabotCollege.edu • MTH55_Lec-67_sec_10-1_Dist-n-Mid_Formulas.ppt
Review § 9.6
MTH 55
 Any QUESTIONS About
• §9.6 → Exponential Decay & Growth
 Any QUESTIONS About HomeWork
• §9.6 → HW-48
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BMayer@ChabotCollege.edu • MTH55_Lec-67_sec_10-1_Dist-n-Mid_Formulas.ppt
The Distance Formula
 The distance between the points
(x1, y1) and (x2, y1) on a horizontal
line is |x2 – x1|.
 Similarly, the distance between the
points (x2, y1) and (x2, y2) on a
vertical line is |y2 – y1|.
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BMayer@ChabotCollege.edu • MTH55_Lec-67_sec_10-1_Dist-n-Mid_Formulas.ppt
Pythagorean Distance
 Now consider any
two points (x1, y1)
and (x2, y2).
 These points,
along with (x2, y1),
describe a right
triangle. The lengths
of the legs are |x2 –
x1| and
|y2 – y1|.
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BMayer@ChabotCollege.edu • MTH55_Lec-67_sec_10-1_Dist-n-Mid_Formulas.ppt
Pythagorean Distance
 Find d, the length of the
hypotenuse, by using the
Pythagorean theorem:
d2 = |x2 – x1|2 + |y2 – y1|2
 Since the square of a number is the
same as the square of its opposite, we
can replace the absolute-value signs
with parentheses:
d2 = (x2 – x1)2 + (y2 – y1)2
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BMayer@ChabotCollege.edu • MTH55_Lec-67_sec_10-1_Dist-n-Mid_Formulas.ppt
Distance Formula Formally
 The distance d between any two
points (x1, y1) and (x2, y2) is given by
2
2
d  ( x2  x1)  ( y2  y1) .
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BMayer@ChabotCollege.edu • MTH55_Lec-67_sec_10-1_Dist-n-Mid_Formulas.ppt
Example  Find Distance
 Find the distance between (3, 1) and
(5, −6). Find an exact answer and an
approximation to three decimal places.
 Solution: Substitute into the distance
formula
2
2
d  (5  3)  (6  1)
Substituting
 (2)2  (7)2
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 53
This is exact.
 7.280.
Approximation
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BMayer@ChabotCollege.edu • MTH55_Lec-67_sec_10-1_Dist-n-Mid_Formulas.ppt
Example  Verify Rt TriAngle

Let A(4, 3), B(1, 4) and C(−2, −4) be
three points in the plane. Connect
these Dots to form a Triangle, Then:
a. Sketch the triangle ABC
b. Find the length of each side of the
triangle
c. Show that ABC is a right triangle.
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BMayer@ChabotCollege.edu • MTH55_Lec-67_sec_10-1_Dist-n-Mid_Formulas.ppt
Example  Verify Rt TriAngle
 Soln a.
Sketch
TriAngle
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BMayer@ChabotCollege.edu • MTH55_Lec-67_sec_10-1_Dist-n-Mid_Formulas.ppt
Example  Verify Rt TriAngle
 Soln b. Find the length of each side of
the triangle → Use Distance Formula
d A, B  
4  1  3  4 
2
2
d B,C  
 9  1  10
1  2   4  5 
2
2
 9  81  90
d B,C    4  2    3  5 
2
 36  64  100  10
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BMayer@ChabotCollege.edu • MTH55_Lec-67_sec_10-1_Dist-n-Mid_Formulas.ppt
2
Example  Verify Rt TriAngle
 Soln c.: Show that ABC is a Rt triangle.
 Check that a2 + b2 = c2 holds in this
triangle, where a, b, and c denote the
lengths of its sides. The longest side,
AC, has  d A, B  2   d B,C  2  10  90

 

length
2
2
10 units.  100  10    d A,C  .
 It follows from the converse of the
Pythagorean Theorem that the
triangle ABC IS a right triangle.
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BMayer@ChabotCollege.edu • MTH55_Lec-67_sec_10-1_Dist-n-Mid_Formulas.ppt
Example  BaseBall Distance
 The baseball “diamond” is in fact a
square with a distance of 90 feet
between each of the consecutive bases.
Use an appropriate coordinate system
to calculate the distance the ball will
travel when the third
baseman throws it
from third base to
first base.
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BMayer@ChabotCollege.edu • MTH55_Lec-67_sec_10-1_Dist-n-Mid_Formulas.ppt
Example  BaseBall Distance
 Solution: conveniently choose home plate as
the origin and place the x-axis along the line
from home plate to first base and the y-axis
along the line from home plate to third base
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BMayer@ChabotCollege.edu • MTH55_Lec-67_sec_10-1_Dist-n-Mid_Formulas.ppt
Example  BaseBall Distance
 Find from the DiagramThe coordinates
of home plate (O), first base (A) second
base (C) and third base (B)
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BMayer@ChabotCollege.edu • MTH55_Lec-67_sec_10-1_Dist-n-Mid_Formulas.ppt
Example  BaseBall Distance
 Find the distance between points A & B
d A, B  

90  0   0  90 
2
2
90   90 
2
2
 2 90 
2
 90 2
 127.28 feet
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BMayer@ChabotCollege.edu • MTH55_Lec-67_sec_10-1_Dist-n-Mid_Formulas.ppt
The MidPoint Formula
 Now that we have derived the
Distance formula from the
Pythagorean Theorem we use the
distance formula to develop a
formula for the coordinates of the
MidPoint of a segment connecting
two points.
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BMayer@ChabotCollege.edu • MTH55_Lec-67_sec_10-1_Dist-n-Mid_Formulas.ppt
The MidPoint Formula
 If the endpoints of a segment are (x1, y1)
and (x2, y2), then the coordinates of the
midpoint are
y
 x1  x2 y1  y2 
,

.
2 
 2
(x2, y2)
(x1, y1)
x
 That is, to locate the midpoint, average
the x-coordinates and average the
y-coordinates
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BMayer@ChabotCollege.edu • MTH55_Lec-67_sec_10-1_Dist-n-Mid_Formulas.ppt
Example  MidPoint Formula
 Find the midpoint of the line segment
joining the points P(−3, 6) and Q(1, 4)
 Solution: (x1, y1) = (−3, 6) & (x2, y2) = (1, 4)
x1  3, y1  6, x2  1, y2  4
 x1  x2 y1  y2 
Midpoint  
,

 2
2 
 3  1 6  4 

,

 2
2
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 1, 5 
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BMayer@ChabotCollege.edu • MTH55_Lec-67_sec_10-1_Dist-n-Mid_Formulas.ppt
CIRCLE Defined
 A circle is a set of points in a
Cartesian coordinate plane that are
at a fixed distance r from a
specified point (h, k).
 The fixed distance r is called the
radius of the circle, and
 The specified point (h, k) is called
the center of the circle.
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CIRCLE Graphed
 The graph of a
circle with
center (h, k)
and radius r.
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CIRCLE - Equation
 The equation of a circle with center
(h, k) and radius r is
x  h   y  k 
2
2
r .
2
 This equation is also called the
standard form of an equation of a
circle with radius r and center (h, k).
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BMayer@ChabotCollege.edu • MTH55_Lec-67_sec_10-1_Dist-n-Mid_Formulas.ppt
Example  Find Circle Eqn
 Find the center-radius form of the
equation of the circle with center (−3, 4)
and radius 7.
 Solution:
x  h   y  k   r
2
2
2
 x  3  y  4   7
2
2
x  3  y  4   49
2
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2
2
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BMayer@ChabotCollege.edu • MTH55_Lec-67_sec_10-1_Dist-n-Mid_Formulas.ppt
Example  Graph Circle
 Graph each equation
2
2
2
2
a. x  y  1
b. x  2   y  3  25
 Solution:
2
2
a. x  y  1
x  0   y  0 
2
2
1
2
• Center: (0, 0)
• Radius: 1
– Called the
unit circle
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BMayer@ChabotCollege.edu • MTH55_Lec-67_sec_10-1_Dist-n-Mid_Formulas.ppt
Example  Graph Circle
x  2   y  3
2
2
b. x  2   y  3  25
2
2
2
x

2

y

3

5

   
 Solution: b.
2
2
 25
• Center: (−2, 3)
• Radius: 5
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Equation ↔ Circle
 Note that stating that the equation:
2
2
x  3  y  4   25
represents the circle of radius 5 with
center (–3, 4) means two things:
1. If the values of x and y are a pair of
numbers that satisfy the equation, then
they are the coordinates of a point on the
circle with radius 5 and center (–3, 4).
2. If a point is on the circle, then its
coordinates satisfy the equation
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BMayer@ChabotCollege.edu • MTH55_Lec-67_sec_10-1_Dist-n-Mid_Formulas.ppt
Circle Eqn → General Form
 The general form of the equation
of a circle is
x  y  ax  by  c  0.
2
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2
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BMayer@ChabotCollege.edu • MTH55_Lec-67_sec_10-1_Dist-n-Mid_Formulas.ppt
Example  General Form
 Find the center and radius of the circle
with equation x2 +y2 − 6x + 8y +10 = 0
 Solution: COMPLETE the SQUARE
for both x & y
2
2
x  6x  y  8y  10

x
2
 

 6x  9  y  8y  16  10  9  16
2
x  3
2
  y  4   15
Center: (3, –4)
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2
Radius: 15  3.9
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BMayer@ChabotCollege.edu • MTH55_Lec-67_sec_10-1_Dist-n-Mid_Formulas.ppt
Example  General Form
 Find the center & radius and then graph
the circle x2 + y2 + 2x – 6y + 6 = 0
 Solution: Complete Square for both
x & y to convert to Standard Form
x2 + 2x + y2 – 6y = –6
x2 + 2x + 1 + y2 – 6y + 9 = –6 + 1 + 9
(x + 1)2 + (y – 3)2 = 4
(x – (–1))2 + (y – 3)2 = 2 2
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BMayer@ChabotCollege.edu • MTH55_Lec-67_sec_10-1_Dist-n-Mid_Formulas.ppt
Example  General Form
y
 Solution: Graph
• Center: (–1, 3)
• Radius: 2
 Sketch
Graph
(–1, 3)
x
(x – (–1))2 + (y – 3)2 = 2 2
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BMayer@ChabotCollege.edu • MTH55_Lec-67_sec_10-1_Dist-n-Mid_Formulas.ppt
WhiteBoard Work
 Problems From §10.1 Exercise Set
• 16, 26, 38, 48, 54, 56
 Circle
Eqns
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x  22   y  12  9
x  12   y  12  1
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BMayer@ChabotCollege.edu • MTH55_Lec-67_sec_10-1_Dist-n-Mid_Formulas.ppt
All Done for Today
Circle as
Conic
Section
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Chabot Mathematics
Appendix
r  s  r  s r  s 
2
2
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer
BMayer@ChabotCollege.edu
–
Chabot College Mathematics
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BMayer@ChabotCollege.edu • MTH55_Lec-67_sec_10-1_Dist-n-Mid_Formulas.ppt
ReCall Logarithmic Laws
 Solving Logarithmic Equations Often
Requires the Use of the Properties of
Logarithms
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BMayer@ChabotCollege.edu • MTH55_Lec-67_sec_10-1_Dist-n-Mid_Formulas.ppt