Rectangular Coordinate System

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Rectangular Coordinate SystemEach point in the plane may be addressed by two numbers (x, y) called an ordered pair.

Rectangular Coordinate SystemEach point in the plane may be addressed by two numbers (x, y) called an ordered pair.

(4, -3)an ordered pair

Rectangular Coordinate SystemEach point in the plane may be addressed by two numbers (x, y) called an ordered pair. To locate (x, y), start from the origin (0, 0),

(4, -3)an ordered pair

Rectangular Coordinate SystemEach point in the plane may be addressed by two numbers (x, y) called an ordered pair. To locate (x, y), start from the origin (0, 0),x = the amount to move right (+) or left (–), (4, -3)

an ordered pair

Rectangular Coordinate SystemEach point in the plane may be addressed by two numbers (x, y) called an ordered pair. To locate (x, y), start from the origin (0, 0),x = the amount to move right (+) or left (–), y = the amount to moveup (+) or down (–).

(4, -3)an ordered pair

For example, the point corresponding to (4, -3) is4 right, and 3 down from the origin.

Rectangular Coordinate SystemEach point in the plane may be addressed by two numbers (x, y) called an ordered pair. To locate (x, y), start from the origin (0, 0),x = the amount to move right (+) or left (–), y = the amount to moveup (+) or down (–).

(4, -3)

4 right

3 down

an ordered pair

For example, the point corresponding to (4, -3) is4 right, and 3 down from the origin.

Rectangular Coordinate SystemEach point in the plane may be addressed by two numbers (x, y) called an ordered pair. To locate (x, y), start from the origin (0, 0),x = the amount to move right (+) or left (–), y = the amount to moveup (+) or down (–).

Points on the x-axis have the form (#, 0)

(#, 0)

(4, -3)

4 right

3 down

an ordered pair

For example, the point corresponding to (4, -3) is4 right, and 3 down from the origin.

Rectangular Coordinate SystemEach point in the plane may be addressed by two numbers (x, y) called an ordered pair. To locate (x, y), start from the origin (0, 0),x = the amount to move right (+) or left (–), y = the amount to moveup (+) or down (–).

Points on the x-axis have the form (#, 0) and points on the y-axis have the form (0, #).

(#, 0)

(0, #)

(4, -3)

4 right

3 down

an ordered pair

The axes divide the plane into four quadrants,numbered 1, 2, 3, and 4 counter-clockwise.

Rectangular Coordinate System

The axes divide the plane into four quadrants,numbered 1, 2, 3, and 4 counter-clockwise.

Q1Q2

Q3 Q4

Rectangular Coordinate System

The axes divide the plane into four quadrants,numbered 1, 2, 3, and 4 counter-clockwise. Respectively, the signs of the coordinates of each quadrant are shown.

Q1Q2

Q3 Q4

(+,+)(–,+)

(–,–) (+,–)

Rectangular Coordinate System

The axes divide the plane into four quadrants,numbered 1, 2, 3, and 4 counter-clockwise. Respectively, the signs of the coordinates of each quadrant are shown.

Rectangular Coordinate System

The points (x, y) and (–x , y) are reflections of each other across the y-axis.

Q1Q2

Q3 Q4

(+,+)(–,+)

(–,–) (+,–)

The axes divide the plane into four quadrants,numbered 1, 2, 3, and 4 counter-clockwise. Respectively, the signs of the coordinates of each quadrant are shown.

Rectangular Coordinate System

The points (x, y) and (–x , y) are reflections of each other across the y-axis.

(5,4)(–5,4)

The axes divide the plane into four quadrants,numbered 1, 2, 3, and 4 counter-clockwise. Respectively, the signs of the coordinates of each quadrant are shown.

Rectangular Coordinate System

The points (x, y) and (–x , y) are reflections of each other across the y-axis.The points (x, y) and (x , –y) are reflections of each other across the x-axis.

(5,4)(–5,4)

The axes divide the plane into four quadrants,numbered 1, 2, 3, and 4 counter-clockwise. Respectively, the signs of the coordinates of each quadrant are shown.

Rectangular Coordinate System

The points (x, y) and (–x , y) are reflections of each other across the y-axis.The points (x, y) and (x , –y) are reflections of each other across the x-axis.

(5,4)(–5,4)

(5, –4)

The axes divide the plane into four quadrants,numbered 1, 2, 3, and 4 counter-clockwise. Respectively, the signs of the coordinates of each quadrant are shown.

Rectangular Coordinate System

The points (x, y) and (–x , y) are reflections of each other across the y-axis.The points (x, y) and (x , –y) are reflections of each other across the x-axis.The points (x, y) and (–x , –y) are reflections of each other across the origin.

(5,4)(–5,4)

(5, –4)

The axes divide the plane into four quadrants,numbered 1, 2, 3, and 4 counter-clockwise. Respectively, the signs of the coordinates of each quadrant are shown.

Rectangular Coordinate System

The points (x, y) and (–x , y) are reflections of each other across the y-axis.The points (x, y) and (x , –y) are reflections of each other across the x-axis.The points (x, y) and (–x , –y) are reflections of each other across the origin.

(5,4)(–5,4)

(5, –4)(–5, –4)

Let A be the point (2, 3). Rectangular Coordinate System

A(2, 3)

Let A be the point (2, 3). Suppose it’s x–coordinate is increased by 4 to (2 + 4, 3) = (6, 3)

Rectangular Coordinate System

A(2, 3)

Let A be the point (2, 3). Suppose it’s x–coordinate is increased by 4 to (2 + 4, 3) = (6, 3) – to the point B, this corresponds to moving A to the right by 4.

Rectangular Coordinate System

A B

x–coord. increased by 4

(2, 3) (6, 3)

Let A be the point (2, 3). Suppose it’s x–coordinate is increased by 4 to (2 + 4, 3) = (6, 3) – to the point B, this corresponds to moving A to the right by 4.

Rectangular Coordinate System

A B

Similarly if the x–coordinate is decreased by 4 to (2 – 4, 3) = (–2, 3) – to the point C,

C

x–coord. increased by 4

x–coord. decreased by 4

(2, 3) (6, 3)(–2, 3)

this corresponds to moving A to the left by 4.

Let A be the point (2, 3). Suppose it’s x–coordinate is increased by 4 to (2 + 4, 3) = (6, 3) – to the point B, this corresponds to moving A to the right by 4.

Rectangular Coordinate System

A B

Similarly if the x–coordinate is decreased by 4 to (2 – 4, 3) = (–2, 3) – to the point C,

Hence we conclude that changes in the x–coordinates correspond to moving the point right and left.

C

x–coord. increased by 4

x–coord. decreased by 4

(2, 3) (6, 3)(–2, 3)

this corresponds to moving A to the left by 4.

Let A be the point (2, 3). Suppose it’s x–coordinate is increased by 4 to (2 + 4, 3) = (6, 3) – to the point B, this corresponds to moving A to the right by 4.

Rectangular Coordinate System

A B

Similarly if the x–coordinate is decreased by 4 to (2 – 4, 3) = (–2, 3) – to the point C,

Hence we conclude that changes in the x–coordinates correspond to moving the point right and left. If the x–change is +, the point moves to the right. If the x–change is – , the point moves to the left.

C

x–coord. increased by 4

x–coord. decreased by 4

(2, 3) (6, 3)(–2, 3)

this corresponds to moving A to the left by 4.

Again let A be the point (2, 3).Rectangular Coordinate System

A(2, 3)

Again let A be the point (2, 3). If the y–coordinate is increased by 4 to (2, 3 + 4) = (2, 7) – to the point D, this corresponds to moving A up by 4.

Rectangular Coordinate System

A

Dy–coord. increased by 4

(2, 3)

(2, 7)

Again let A be the point (2, 3). If the y–coordinate is increased by 4 to (2, 3 + 4) = (2, 7) – to the point D, this corresponds to moving A up by 4.

Rectangular Coordinate System

A

D

If the y–coordinate is decreased by 4 to (2, 3 – 4) = (2, –1) – to the point E,

E

y–coord. increased by 4

y–coord. decreased by 4

(2, 3)

(2, 7)

(2, –1)

this corresponds to moving A down by 4.

Again let A be the point (2, 3). If the y–coordinate is increased by 4 to (2, 3 + 4) = (2, 7) – to the point D, this corresponds to moving A up by 4.

Rectangular Coordinate System

A

D

If the y–coordinate is decreased by 4 to (2, 3 – 4) = (2, –1) – to the point E,

Hence we conclude that changes in the y–coordinates correspond to moving the point up and down.

E

y–coord. increased by 4

y–coord. decreased by 4

(2, 3)

(2, 7)

(2, –1)

this corresponds to moving A down by 4.

Again let A be the point (2, 3). If the y–coordinate is increased by 4 to (2, 3 + 4) = (2, 7) – to the point D, this corresponds to moving A up by 4.

Rectangular Coordinate System

A

D

If the y–coordinate is decreased by 4 to (2, 3 – 4) = (2, –1) – to the point E,

Hence we conclude that changes in the y–coordinates correspond to moving the point up and down.

If the y–change is +, the point moves up.If the y–change is – , the point moves down.

E

y–coord. increased by 4

y–coord. decreased by 4

(2, 3)

(2, 7)

(2, –1)

this corresponds to moving A down by 4.

Let (x1, y1) and (x2, y2) be two points andD = the distance between them,

The Distance Formula

Let (x1, y1) and (x2, y2) be two points andD = the distance between them,

DD

The Distance Formula

(2, –4)

(–1, 3)

Let (x1, y1) and (x2, y2) be two points andD = the distance between them,

Example A. Find the distancebetween (–1, 3) and (2, –4).

DD

The Distance Formula

(2, –4)

(–1, 3)

Let (x1, y1) and (x2, y2) be two points andD = the distance between them, then D2 = Δx2 + Δy2, where Δx = difference in the x's = x2 – x1, Δy = difference in the y's = y2 – y1,

Example A. Find the distancebetween (–1, 3) and (2, –4).

DD

The Distance Formula

(2, –4)

(–1, 3)

Let (x1, y1) and (x2, y2) be two points andD = the distance between them, then D2 = Δx2 + Δy2, where Δx = difference in the x's = x2 – x1, Δy = difference in the y's = y2 – y1,

Example A. Find the distancebetween (–1, 3) and (2, –4). (–1, 3) – ( 2, –4)

DD

The Distance Formula

(2, –4)

(–1, 3)

Let (x1, y1) and (x2, y2) be two points andD = the distance between them, then D2 = Δx2 + Δy2, where Δx = difference in the x's = x2 – x1, Δy = difference in the y's = y2 – y1,

Example A. Find the distancebetween (–1, 3) and (2, –4). (–1, 3) – ( 2, –4) –3, 7

DD

The Distance Formula

(2, –4)

(–1, 3)

Δx Δy

Let (x1, y1) and (x2, y2) be two points andD = the distance between them, then D2 = Δx2 + Δy2, where Δx = difference in the x's = x2 – x1, Δy = difference in the y's = y2 – y1,

Example A. Find the distancebetween (–1, 3) and (2, –4). (–1, 3) – ( 2, –4) –3, 7

DD

7

-3

The Distance Formula

(2, –4)

(–1, 3)

Δx Δy

Let (x1, y1) and (x2, y2) be two points andD = the distance between them, then D2 = Δx2 + Δy2, where Δx = difference in the x's = x2 – x1, Δy = difference in the y's = y2 – y1,

Example A. Find the distancebetween (–1, 3) and (2, –4). (–1, 3) – ( 2, –4) –3, 7

D = (–3)2 + 72

DD

7

-3

The Distance Formula

(2, –4)

(–1, 3)

Δx Δy

Let (x1, y1) and (x2, y2) be two points andD = the distance between them, then D2 = Δx2 + Δy2, where Δx = difference in the x's = x2 – x1, Δy = difference in the y's = y2 – y1,

Example A. Find the distancebetween (–1, 3) and (2, –4). (–1, 3) – ( 2, –4) –3, 7

D = (–3)2 + 72

= 58 7.62

DD

7

-3

The Distance Formula

(2, –4)

(–1, 3)

Δx Δy

Let (x1, y1) and (x2, y2) be two points andD = the distance between them, then D2 = Δx2 + Δy2, where Δx = difference in the x's = x2 – x1, Δy = difference in the y's = y2 – y1,Hence D = Δx2 + Δy2 orD = (x2 – x1)2+(y2 – y1)2 Example A. Find the distancebetween (–1, 3) and (2, –4). (–1, 3) – ( 2, –4) –3, 7

D = (–3)2 + 72

= 58 7.62

DD

7

-3

The Distance Formula

(2, –4)

(–1, 3)

Δx Δy

Let (h, k) be the center of a circle and r be the radius. Circles

Let (h, k) be the center of a circle and r be the radius.

r(h, k)

Circles

Let (h, k) be the center of a circle and r be the radius. Suppose (x, y) is a point on the circle, then the distance between (x, y) and the center must be r.

r(x, y)

(h, k)

Circles

Let (h, k) be the center of a circle and r be the radius. Suppose (x, y) is a point on the circle, then the distance between (x, y) and the center must be r. Hence,

r(x, y)

(h, k)

r = (x – h)2 + (y – k)2

Circles

Let (h, k) be the center of a circle and r be the radius. Suppose (x, y) is a point on the circle, then the distance between (x, y) and the center must be r. Hence,

r(x, y)

(h, k)

r = (x – h)2 + (y – k)2

orr2 = (x – h)2 + (y – k)2

Circles

Let (h, k) be the center of a circle and r be the radius. Suppose (x, y) is a point on the circle, then the distance between (x, y) and the center must be r. Hence,

r(x, y)

(h, k)

r = (x – h)2 + (y – k)2

orr2 = (x – h)2 + (y – k)2

This is the standard equationof circles.

Circles

Let (h, k) be the center of a circle and r be the radius. Suppose (x, y) is a point on the circle, then the distance between (x, y) and the center must be r. Hence,

r(x, y)

(h, k)

r = (x – h)2 + (y – k)2

orr2 = (x – h)2 + (y – k)2

This is the standard equationof circles.

Circles

Therefore, the specific equation for the circle with the radius = 5 and centered at (–1, 3) is

Let (h, k) be the center of a circle and r be the radius. Suppose (x, y) is a point on the circle, then the distance between (x, y) and the center must be r. Hence,

r(x, y)

(h, k)

r = (x – h)2 + (y – k)2

orr2 = (x – h)2 + (y – k)2

This is the standard equationof circles.

Circles

Therefore, the specific equation for the circle with the radius = 5 and centered at (–1, 3) is

52 = (x – (–1))2 + (y – 3)2

Let (h, k) be the center of a circle and r be the radius. Suppose (x, y) is a point on the circle, then the distance between (x, y) and the center must be r. Hence,

r(x, y)

(h, k)

r = (x – h)2 + (y – k)2

orr2 = (x – h)2 + (y – k)2

This is the standard equationof circles.

Circles

Therefore, the specific equation for the circle with the radius = 5 and centered at (–1, 3) is

52 = (x – (–1))2 + (y – 3)2 or 25 = (x + 1)2 + (y – 3 )2

Completing the SquareAdding a number to an expression so their sum is a perfect square is called completing the square.

Completing the SquareAdding a number to an expression so their sum is a perfect square is called completing the square. This procedure is the main technique in dealing with 2nd degree equations.

Completing the SquareAdding a number to an expression so their sum is a perfect square is called completing the square. This procedure is the main technique in dealing with 2nd degree equations.

Completing the Square:If we are given x2 + bx,

Completing the SquareAdding a number to an expression so their sum is a perfect square is called completing the square. This procedure is the main technique in dealing with 2nd degree equations.

Completing the Square:If we are given x2 + bx, then adding (b/2)2 to the expression makes the expression a perfect square, i.e. x2 + bx + (b/2)2 is a perfect square.

Example B. Fill in the blank to complete the square:a. x2 – 6x + ( )2 =

b. y2 + 12y + ( )2 =

Completing the SquareAdding a number to an expression so their sum is a perfect square is called completing the square. This procedure is the main technique in dealing with 2nd degree equations.

Completing the Square:If we are given x2 + bx, then adding (b/2)2 to the expression makes the expression a perfect square, i.e. x2 + bx + (b/2)2 is a perfect square.

Example B. Fill in the blank to complete the square:a. x2 – 6x + ( )2 =

b. y2 + 12y + ( )2 =

Completing the SquareAdding a number to an expression so their sum is a perfect square is called completing the square. This procedure is the main technique in dealing with 2nd degree equations.

Completing the Square:If we are given x2 + bx, then adding (b/2)2 to the expression makes the expression a perfect square, i.e. x2 + bx + (b/2)2 is a perfect square.

Example B. Fill in the blank to complete the square:a. x2 – 6x + (-6/2)2 =

b. y2 + 12y + ( )2 =

Completing the SquareAdding a number to an expression so their sum is a perfect square is called completing the square. This procedure is the main technique in dealing with 2nd degree equations.

Completing the Square:If we are given x2 + bx, then adding (b/2)2 to the expression makes the expression a perfect square, i.e. x2 + bx + (b/2)2 is a perfect square.

Example B. Fill in the blank to complete the square:a. x2 – 6x + (-6/2)2 = x2 – 6x + 9

b. y2 + 12y + ( )2 =

Completing the SquareAdding a number to an expression so their sum is a perfect square is called completing the square. This procedure is the main technique in dealing with 2nd degree equations.

Completing the Square:If we are given x2 + bx, then adding (b/2)2 to the expression makes the expression a perfect square, i.e. x2 + bx + (b/2)2 is a perfect square.

Example B. Fill in the blank to complete the square:a. x2 – 6x + (-6/2)2 = x2 – 6x + 9 = (x – 3 )2

b. y2 + 12y + ( )2 =

Completing the SquareAdding a number to an expression so their sum is a perfect square is called completing the square. This procedure is the main technique in dealing with 2nd degree equations.

Completing the Square:If we are given x2 + bx, then adding (b/2)2 to the expression makes the expression a perfect square, i.e. x2 + bx + (b/2)2 is a perfect square.

Example B. Fill in the blank to complete the square:a. x2 – 6x + (-6/2)2 = x2 – 6x + 9 = (x – 3 )2

b. y2 + 12y + (12/2)2 =

Completing the SquareAdding a number to an expression so their sum is a perfect square is called completing the square. This procedure is the main technique in dealing with 2nd degree equations.

Completing the Square:If we are given x2 + bx, then adding (b/2)2 to the expression makes the expression a perfect square, i.e. x2 + bx + (b/2)2 is a perfect square.

Example B. Fill in the blank to complete the square:a. x2 – 6x + (-6/2)2 = x2 – 6x + 9 = (x – 3 )2

b. y2 + 12y + (12/2)2 = y2 + 12y + 36 = (y + 6)2

More on Equations of CircleExample C. Complete the square to find the center and radius of x2 – 6x + y2 + 12y = –36. Graph and label the top, bottom, left and right most points.

More on Equations of CircleExample C. Complete the square to find the center and radius of x2 – 6x + y2 + 12y = –36. Graph and label the top, bottom, left and right most points.

Complete the square to put the equation in the standard form:

More on Equations of CircleExample C. Complete the square to find the center and radius of x2 – 6x + y2 + 12y = –36. Graph and label the top, bottom, left and right most points.

Complete the square to put the equation in the standard form:x2 – 6x + + y2 + 12y + = –36

More on Equations of CircleExample C. Complete the square to find the center and radius of x2 – 6x + y2 + 12y = –36. Graph and label the top, bottom, left and right most points.

Complete the square to put the equation in the standard form:x2 – 6x + + y2 + 12y + = –36 complete squares; x2 – 6x + 9 + y2 + 12y + 36 = –36 + 9 + 36

More on Equations of CircleExample C. Complete the square to find the center and radius of x2 – 6x + y2 + 12y = –36. Graph and label the top, bottom, left and right most points.

Complete the square to put the equation in the standard form:x2 – 6x + + y2 + 12y + = –36 complete squares; x2 – 6x + 9 + y2 + 12y + 36 = –36 + 9 + 36 (x – 3 )2 + (y + 6)2 = 9

More on Equations of CircleExample C. Complete the square to find the center and radius of x2 – 6x + y2 + 12y = –36. Graph and label the top, bottom, left and right most points.

Complete the square to put the equation in the standard form:x2 – 6x + + y2 + 12y + = –36 complete squares; x2 – 6x + 9 + y2 + 12y + 36 = –36 + 9 + 36 (x – 3 )2 + (y + 6)2 = 9( x – 3 )2 + (y + 6)2 = 32

More on Equations of CircleExample C. Complete the square to find the center and radius of x2 – 6x + y2 + 12y = –36. Graph and label the top, bottom, left and right most points.

Complete the square to put the equation in the standard form:x2 – 6x + + y2 + 12y + = –36 complete squares; x2 – 6x + 9 + y2 + 12y + 36 = –36 + 9 + 36 (x – 3 )2 + (y + 6)2 = 9( x – 3 )2 + (y + 6)2 = 32

So the center is (3 , –6), and radius is 3

More on Equations of CircleExample C. Complete the square to find the center and radius of x2 – 6x + y2 + 12y = –36. Graph and label the top, bottom, left and right most points.

Complete the square to put the equation in the standard form:x2 – 6x + + y2 + 12y + = –36 complete squares; x2 – 6x + 9 + y2 + 12y + 36 = –36 + 9 + 36 (x – 3 )2 + (y + 6)2 = 9( x – 3 )2 + (y + 6)2 = 32

So the center is (3 , –6), and radius is 3

(3,-6) (6, -6)(0, -6)

(3, -3)

(3, -9)

More on Equations of CircleExample C. Complete the square to find the center and radius of x2 – 6x + y2 + 12y = –36. Graph and label the top, bottom, left and right most points.

Complete the square to put the equation in the standard form:x2 – 6x + + y2 + 12y + = –36 complete squares; x2 – 6x + 9 + y2 + 12y + 36 = –36 + 9 + 36 (x – 3 )2 + (y + 6)2 = 9( x – 3 )2 + (y + 6)2 = 32

So the center is (3 , –6), and radius is 3

(3,-6) (6, -6)(0, -6)

(3, -3)

(3, -9)

The Mid-Point FormulaThe mid-point m between two numbers a and b is the

average of them, that is m = . a + b2

The Mid-Point FormulaThe mid-point m between two numbers a and b is the

average of them, that is m = . a + b2

For example, the mid-point of 2 and 4 is (2 + 4)/2 = 3.

The Mid-Point FormulaThe mid-point m between two numbers a and b is the

average of them, that is m = . a + b2

For example, the mid-point of 2 and 4 is (2 + 4)/2 = 3. In picture:

a b(a+b)/2

mid-pt.

The Mid-Point FormulaThe mid-point m between two numbers a and b is the

average of them, that is m = . a + b2

For example, the mid-point of 2 and 4 is (2 + 4)/2 = 3. In picture:

a b(a+b)/2

mid-pt.

The mid-point formula extends to higher dimensions.

The Mid-Point FormulaThe mid-point m between two numbers a and b is the

average of them, that is m = . a + b2

For example, the mid-point of 2 and 4 is (2 + 4)/2 = 3. In picture:

a b(a+b)/2

mid-pt.

The mid-point formula extends to higher dimensions.

In 2D

(x1, y1)

(x2, y2)

x1

y1

y2

x2

The Mid-Point FormulaThe mid-point m between two numbers a and b is the

average of them, that is m = . a + b2

For example, the mid-point of 2 and 4 is (2 + 4)/2 = 3. In picture:

a b(a+b)/2

mid-pt.

The mid-point formula extends to higher dimensions.In the x&y coordinate the mid-point of (x1, y1) and (x2, y2) is

x1 + x22 ,(

y1 + y22

)

In 2D

(x1, y1)

(x2, y2)

x1

y1

y2

x2

The Mid-Point FormulaThe mid-point m between two numbers a and b is the

average of them, that is m = . a + b2

For example, the mid-point of 2 and 4 is (2 + 4)/2 = 3. In picture:

a b(a+b)/2

mid-pt.

The mid-point formula extends to higher dimensions.In the x&y coordinate the mid-point of (x1, y1) and (x2, y2) is

x1 + x22 ,(

y1 + y22

)

In 2D

(x1, y1)

(x2, y2)

x1

y1

y2

x2(x1 + x2)/2

(y1 + y2)/2

The Mid-Point FormulaExample D. Find the equation of the circle with the line segment from (1, –4) to (5, 2) as a diameter.Draw and label the four cardinal points.

The Mid-Point FormulaExample D. Find the equation of the circle with the line segment from (1, –4) to (5, 2) as a diameter.Draw and label the four cardinal points.We need the center which is the mid-point of the two points:

The Mid-Point FormulaExample D. Find the equation of the circle with the line segment from (1, –4) to (5, 2) as a diameter.Draw and label the four cardinal points.We need the center which is the mid-point of the two points: C = ((1+5)/2, (–4+2)/2) = (3, –1).

The Mid-Point FormulaExample D. Find the equation of the circle with the line segment from (1, –4) to (5, 2) as a diameter.Draw and label the four cardinal points.We need the center which is the mid-point of the two points: C = ((1+5)/2, (–4+2)/2) = (3, –1).We need the radius r which is the distance from the center (3, –1) to one of the points, say (5, 2).

The Mid-Point FormulaExample D. Find the equation of the circle with the line segment from (1, –4) to (5, 2) as a diameter.Draw and label the four cardinal points.We need the center which is the mid-point of the two points: C = ((1+5)/2, (–4+2)/2) = (3, –1).We need the radius r which is the distance from the center (3, –1) to one of the points, say (5, 2).Hence r = (5 – 3)2 + (2 + 1)2 = 13

The Mid-Point FormulaExample D. Find the equation of the circle with the line segment from (1, –4) to (5, 2) as a diameter.Draw and label the four cardinal points.We need the center which is the mid-point of the two points: C = ((1+5)/2, (–4+2)/2) = (3, –1).We need the radius r which is the distance from the center (3, –1) to one of the points, say (5, 2).Hence r = (5 – 3)2 + (2 + 1)2 = 13Therfore the equation is

(x – 3)2 + (y + 1)2 = (13)2

or

(x – 3)2 + (y + 1)2 = 13

The Mid-Point FormulaExample D. Find the equation of the circle with the line segment from (1, –4) to (5, 2) as a diameter.Draw and label the four cardinal points.We need the center which is the mid-point of the two points: C = ((1+5)/2, (–4+2)/2) = (3, –1).We need the radius r which is the distance from the center (3, –1) to one of the points, say (5, 2).Hence r = (5 – 3)2 + (2 + 1)2 = 13Therfore the equation is

(x – 3)2 + (y + 1)2 = (13)2

or

(x – 3)2 + (y + 1)2 = 13

C(3, –1)

The Mid-Point FormulaExample D. Find the equation of the circle with the line segment from (1, –4) to (5, 2) as a diameter.Draw and label the four cardinal points.We need the center which is the mid-point of the two points: C = ((1+5)/2, (–4+2)/2) = (3, –1).We need the radius r which is the distance from the center (3, –1) to one of the points, say (5, 2).Hence r = (5 – 3)2 + (2 + 1)2 = 13Therfore the equation is

(x – 3)2 + (y + 1)2 = (13)2

or

(x – 3)2 + (y + 1)2 = 13

C(3, –1)(3–13, –1) (3+13,–1)

The Mid-Point FormulaExample D. Find the equation of the circle with the line segment from (1, –4) to (5, 2) as a diameter.Draw and label the four cardinal points.We need the center which is the mid-point of the two points: C = ((1+5)/2, (–4+2)/2) = (3, –1).We need the radius r which is the distance from the center (3, –1) to one of the points, say (5, 2).Hence r = (5 – 3)2 + (2 + 1)2 = 13Therfore the equation is

(x – 3)2 + (y + 1)2 = (13)2

or

(x – 3)2 + (y + 1)2 = 13

C(3, –1)

(3, –1+13)

(3, –1–13)

(3–13, –1) (3+13,–1)

The Mid-Point FormulaExample D. Find the equation of the circle with the line segment from (1, –4) to (5, 2) as a diameter.Draw and label the four cardinal points.We need the center which is the mid-point of the two points: C = ((1+5)/2, (–4+2)/2) = (3, –1).We need the radius r which is the distance from the center (3, –1) to one of the points, say (5, 2).Hence r = (5 – 3)2 + (2 + 1)2 = 13Therfore the equation is

(x – 3)2 + (y + 1)2 = (13)2

or

(x – 3)2 + (y + 1)2 = 13

C(3, –1)

(3, –1+13)

(3, –1–13)

(3–13, –1) (3+13,–1)