The endpoints of AB are A(2, 3) and B(8, 1). The perpendicular bisector of AB is CD, and point C lies on AB. The length of CD is (root of 10) units.
The coordinates of point C are (-6, 2) (5, 2) (6, -2) (10, 4) . The slope of is -3 -1/3 1/3 3 . The possible coordinates of point D are (4, 5) (5, 5) (6, 5) (8, 3) and (2, 1) (4, -1) (5, -1) (6, -1) .
Please pick 1 answer for all the questions

Comment
If C lies on AB Then it must be the midpoint of AB because CD is the Perpendicular Bisector of AB

Midpoint
A(2,3) B(8,1)
Midpoint formula = (x1 + x2)/2 + (y1 + y2)/2
Midpoint = (2 + 8)/2 + (3 + 1) / 2
Midpoint = 10/2 + 4/2
Midpoint = (5,2) <<<<< answer

Slope AB
slope = (y2 - y1)/(x2 - x1)
slope = (3 - 1) / (2 - 8)
slope = 2/-6 = -1/3

Slope CD
CD_slope * AB_slope = - 1
CD_slope * -1/3 = -1 multiply both sides by - 3
CD_slope * 1 = - 1 * -3
CD_slope = 3 <<<<< Slope CD Answer

Equation CD
Given the midpoint of AB which is C( and the slope of CD
y = 3x + b
2 = 3*5 + b
b = - 13
Equation y = 3x - 13

Using Distance to get D
d^2 = (x1 - x2)^2 + (y2 - y1)'^2
d^2 = (x2 - 5)^2 + (y2 - 2)^2
y2 = 3*x2 - 13
10 = x2^2 - 10x2 + 25 + (3x2 - 13 - 2)^2
10 = x2^2 - 10x2 + 25 + (3x2 - 15)^2
10 = x1^2 - 10x2 + 25 + 9x^2 - 90x2 + 225
10 = 10x2^2 - 100x2 +250
0 = 10x2^2 - 100x2 + 240 Divide through by 10
0 = x2^2 - 10x^2 + 24
0 = (x2 - 6)(x2 - 4)

x2 = 6 or
x2 = 4

Using y2 = 3x - 13
for x2 = 4
then y2 = 3(4) - 13
y2 = - 1

for x2 = 6
y2 = 3*6 - 13
y2 = 5

Possible Answers for D
D = (4,-1)
D = (6,5)

I've included a graph to show a graphical solution.

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mkenya570 mkenya570 · Answer 2 · 2021-02-18T18:21:15+01:00

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)