Respuesta :

1) Coordinates of point C.

C is the midpoint of the segment AB.

So, its coordinates are:

x = (2 + 8) / 2 = 10/2 = 5
y = (3 + 1) / 2 = 4 / 2 = 2

So, the point C is (5,2)

2) Slope of the segment CD.

First, calculate the slope of the segment AB, m

m = Δy / Δx = (y2 - y1) / (x2 - x1) = (3 -1) / (2 - 8) = 2 / ( -6) = - 1/3

Second, calculate the slope of CD,m', as the negative inverse of m:

m' = - 1 / m = - 1 / (-1/3) = 3

Slope of segment CD is 3.

3) Possible coordinates of point D.

You have to use this three data:
Point C: (5, 2)
length CD = √10
slope CD = 3

Call the point D (x,y)

Use fhe equation for the slope: [y - 2] / [x - 5] = 3

=> y - 2 = 3( x- 5)
=> y = 3x - 15 + 2
=> y = 3x - 13

Use the equation for the distance: (x - 5)^2 + (y - 2)^2 = (√10)^2

Expand the parenthesis:
x^2 - 10x + 25 + y^2 - 4y + 4 = 10

Replace y with 3x - 13
x^2 - 10x + 25 + (3x - 13)^2 - 4(3x - 13) + 4 = 10
x^2 - 10x + 25 + 9x^2 - 78x + 169 - 12x + 52 + 4 - 10 = 0

Combine like terms:
10x^2 - 100x + 240 = 0

Simplify by 10:
x^2 - 10x + 24 = 0

Factor
(x - 6)(x - 4) = 0 => x = 6 or x = 4

Calculate y for both x-values:
y = 3(6) - 13 = 5 or y = 3(4) - 13 = - 1

Possible coordinates of point D: (6,5) and (4, -1).

Answer and Step-by-step explanation:

Answer:

The coordinates of C are (5 , 2)

The slope of CD is 3

The coordinates of D are (6 , 5) and (4 , -1)

Step-by-step explanation:

* Now lets study the problem

- The ends points of line AB are A = 2 , 3) and B = (8 , 1)

- CD is the perpendicular bisector of AB, and C lies on AB

- That means:

# C is the mid-point of AB

# The slope of AB × the slope of CD = -1 (one of them is a multiplicative

 inverse and additive inverse of the other)

-Ex: the slope of one is a/b, then the slope of the other is -b/a

* The mid-point between two points (x1 , y1) and (x2 , y2) is:

 [(x1 + x2)/2 , (y1 + y2)/2]

∵ C is the mid-point of AB

∴ C = [(2 + 8)/2 , (3 + 1)/2] = [10/2 , 4/2] = (5 , 2)

* The coordinates of C are (5 , 2)

- The slope of a line passing through points (x1 , y1) and (x2 , y2) is:

the slope = (y2 - y1)/(x2 - x1)

∴ The slope of AB = (1 - 3)/(8 -2) = -2/6 = -1/3

∵ CD ⊥ AB

∴ The slope of CD × the slope of AB = -1

∴ The slope of CD = 3

* The slope of CD is 3

- The length of a line passing through points (x1 , y1) and (x2 , y2) is:

the length = √[(x2 - x1)² + (y2 - y1)²]

∵ The length of CD = √10

∵ Point D is (x , y)

∴ (x - 5)² + (y - 2)² = (√10)²

∴ (x - 5)² + (y - 2)² = 10 ⇒ (1)

∵ The slope of CD is (y - 2)/(x - 5) = 3 ⇒ by using cross multiply

∴ (y - 2) = 3(x - 5) ⇒ (2)

- Substitute (2) in (1)

∴ (x - 5)² + [3(x - 5)]² = 10 ⇒ simplify

* [3(x - 5)]² = (3)²(x - 5)² = 9(x - 5)²

∴ (x - 5)² + 9(x - 5)² = 10 ⇒ add the like terms

∴ 10(x - 5)² = 10 ⇒ ÷ 10 both sides

∴ (x - 5)² = 1 ⇒ take √ for both sides

∴ x - 5 = ± 1

∴ x - 5 = 1 ⇒ add 5 to both sides

∴ x = 6

* OR

∴ x - 5 = -1 ⇒ add 5 to both sides

∴ x = 4

- Substitute the values of x in (2)

∴ y - 2 = 3(6 - 5)

∴ y - 2 = 3 ⇒ add 2

∴ y = 5

* OR

∴ y - 2 = 3(4 - 5)

∴ y - 2 = -3 ⇒ add 2

∴ y = -1

* The coordinates of D are (6 , 5) and (4 , -1)