Respuesta :

A(2, 3)  B(8, 1)
distance AB
^2=(8-2)^2+(1-3)^2=40
dAB=2
√10
slope mAB=-2/6=-1/3

The equation of the line through the points
y-y1=(-1/3)((x-x1) 
y-3= (-1/3) (x-2) 
y= -(1/3)x +(2/3)+3 = -(1/3)x+11/3 equation of the line AB
perpendicular line has slope 3 (negative reciprocal)
goes through midpoint which is half the distance between the two points
((2+8)/2,
(3+1)/2)=(5,2)  Point C
Its equation is y-2=3(x-5) 
y=3x-13 
equation of the line CD
calculation of point D    point C (5,2)  point D (x,y)
distance CD=√10
distance CD^2=(x-5)^2+(y-2)^2=10
10=(x-5)^2+(3x-15)^2

solving the quadratic equation
x1=4  y1=-1 
x2=6 y2=5
y=3x-13 equation of the line CD

Point D (6,5)  or (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)