The coordinates of AB are A(2,3) and B(8,1) The perpendicular bisected of AB is CD and point C lies on AB the length of CD is 10 units

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 ) =  

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

And, The slope of AB = ⇒ ⇔ .

       Thus, The slope of CD is  =    ⇔  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 = -1When 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).

Step-by-step explanation:

We are given a where

As per trigonometric rules,

Comparing the given value of tangent and cosine:

Perpendicular = 3 units

Base = 4 units

Hypotenuse = 5 units

Also, we know that:

Now, it is given that the triangle is dilated by a scale factor of .

i.e. all the sides will become half of its initial values.

New Perpendicular = 1.5 units

New Base = 2 units

New Hypotenuse = 2.5 units

Using the formula for sine: