[email protected] mth55_lec-67_sec_10-1_dist-n-mid_formulas.ppt 1 bruce mayer, pe chabot...

[email protected] • MTH55_Lec-67_sec_10-1_Dist-n-Mid_Formulas.ppt1

Bruce Mayer, PE Chabot College Mathematics

Bruce Mayer, PELicensed Electrical & Mechanical Engineer

[email protected]

Chabot Mathematics

§10.1 Distance§10.1 DistanceMIdPoint EqnsMIdPoint Eqns



Review §Review §

Any QUESTIONS About• §9.6 → Exponential Decay & Growth

Any QUESTIONS About HomeWork• §9.6 → HW-48

9.6 MTH 55



The Distance FormulaThe 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|.



Pythagorean DistancePythagorean 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|.



Pythagorean DistancePythagorean 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



Distance Formula FormallyDistance Formula Formally

The distance d between any two points (x1, y1) and (x2, y2) is given by



Example Example Find Distance 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(5 3) ( 6 1)d

2 2(2) ( 7)

53

7.280.

Substituting

This is exact.

Approximation



Example Example Verify Rt TriAngle 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.




Soln a.SketchTriAngle




Soln b. Find the length of each side of the triangle → Use Distance Formula

d A, B 4 1 2 3 4 2 9 1 10

d B,C 1 2 2 4 5 2

9 81 90

d B,C 4 2 2 3 5

2

36 64 100 10




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 length 10 units.

d A, B 2 d B,C

210 90

100 10 2 d A,C 2.

It follows from the converse of the Pythagorean Theorem that the triangle ABC IS a right triangle.



Example Example BaseBall Distance 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.



Example Example BaseBall Distance 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




Find from the DiagramThe coordinates of home plate (O), first base (A) second base (C) and third base (B)




Find the distance between points A & B

d A, B 90 0 2 0 90 2

90 2 90 2

2 90 2

90 2

127.28 feet12

7.3

ft



The MidPoint FormulaThe 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.



The MidPoint FormulaThe MidPoint Formula

If the endpoints of a segment are (x1, y1) and (x2, y2), then the coordinates of the midpoint are

1 2 1 2, .2 2

x x y y (x1, y1)

(x2, y2)

x

y

That is, to locate the midpoint, average the x-coordinates and average the y-coordinates



Example Example MidPoint Formula 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

Midpoint x1 x2

2,y1 y2

2

31

2,6 4

2

1, 5



CIRCLE DefinedCIRCLE 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.



CIRCLE GraphedCIRCLE Graphed

The graph of a circle with center (h, k) and radius r.



CIRCLE - EquationCIRCLE - Equation

The equation of a circle with center (h, k) and radius r is

This equation is also called the standard form of an equation of a circle with radius r and center (h, k).



Example Example Find Circle Eqn Find Circle Eqn

Find the center-radius form of the equation of the circle with center (−3, 4) and radius 7.

Solution:

x h 2 y k 2 r2

x 3 2 y 4 2 72

x 3 2 y 4 2 49



Example Example Graph Circle Graph Circle

Graph each equation

b. x 2 2 y 3 2 25a. x2 y2 1

Solution:a. x2 y2 1

x 0 2 y 0 2 12

• Center: (0, 0)

• Radius: 1– Called the

unit circle



Example Example Graph Circle Graph Circle

Solution: b. x 2 2 y 3 2 25

• Center: (−2, 3)

• Radius: 5

b. x 2 2 y 3 2 25

x 2 2 y 3 2 52



Equation ↔ CircleEquation ↔ Circle

Note that stating that the equation:

x 3 2 y 4 2 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



Circle Eqn → General FormCircle Eqn → General Form

The general form of the equation of a circle is



Example Example General Form 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

x2 6x y2 8y 10

x2 6x 9 y2 8y 16 10 9 16

x 3 2 y 4 215

Center: (3, – 4) Radius: 15 3.9




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

(x + 1)2 + (y – 3)2 = 4

x2 + 2x + y2 – 6y = –6

x2 + 2x + 1 + y2 – 6y + 9 = –6 + 1 + 9

(x – (–1))2 + (y – 3)2 = 2 2




Solution: Graph• Center: (–1, 3)

• Radius: 2

SketchGraph (–1, 3)

x

y

(x – (–1))2 + (y – 3)2 = 2 2



WhiteBoard WorkWhiteBoard Work

Problems From §10.1 Exercise Set• 16, 26, 38, 48, 54, 56

CircleEqns 111 22 yx

912 22 yx



All Done for TodayAll Done for Today

Circle asConic

Section



Bruce Mayer, PELicensed Electrical & Mechanical Engineer

[email protected]

Chabot Mathematics

AppendiAppendixx

–

srsrsr 22



ReCall Logarithmic LawsReCall Logarithmic Laws

Solving Logarithmic Equations Often Requires the Use of the Properties of Logarithms

[email protected] mth55_lec-67_sec_10-1_dist-n-mid_formulas.ppt 1 bruce mayer, pe chabot...

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