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The endpoints of are A(2, 3) and B(8, 1). The perpendicular bisector of is , and point C lies on . The length of is units. The coordinates of point C are . The slope of is . The possible coordinates of point D are and .

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Answer:

The Coordinates of Point C are C(5,2)

The slope of CD is 3  

The possible coordinates of point D(a,b) are (6, 5) and (4, -1)  

Step-by-step explanation:

According to the Question,

  • Given, The end points 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 the CD is √10 units.

Now, Let The Coordinates of Point C are C( x , y ) .  

  • Thus, C( x , y ) = ((8 +2) / 2 , ( 1 + 3 )/ 2 ) ⇒ (10/2 , 4/2) ⇒ (5 , 2)

The Coordinates of Point C are C( x , y ) ⇒C(5, 2).  

And, The slope of AB = (1 - 3)/(8 - 2) ⇒ -2/6 ⇔ -1/3 .

  •       Thus, The slope of CD is  -1/(The slope of AB) =  -1/(-1/3) ⇔  3.

Let the coordinate of D be (a, b) then

⇒ √{ (b - 2)² +(a - 5)² } = √10

 on squaring both sides we get,

 ⇒ a² - 10a + 25 + b² - 4b + 4 = 10

 ⇒ a² + b² - 10a - 4b = - 19 ⇔⇔ (Equation 1)  

We Know, the slope of CD = 3

⇒Thus,  (b - 2)/(a - 5) = 3

⇒ b - 2 = 3a - 15

⇒ b = 3a - 13 ⇔⇔ (Equation 2)

Putting value of Equation 2 into Equation 1, We get

⇒a² + (3a - 13)² - 10a - 4(3a - 13) = - 19

⇒a² + 9a² - 78a + 169 - 10a - 12a + 52 = - 19

⇒10a² - 100a + 240 = 0

⇒a² - 10a + 24 = 0

⇒(a - 4)(a - 6) = 0

⇒a = 4 or a = 6

Now,  

When a = 4 , b = 3(4) - 13 ⇒ 12 - 13 ⇒ b = -1

When a = 6, b = 3(6) - 13 ⇒ 18 - 13 ⇒ b = 5

Therefore, the possible coordinates of point D(a,b) are (6, 5) and (4, -1).

Answer:

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Step-by-step explanation:

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