The coordinates of point a are (p, q) and coordinates of point b are (p+2q, q+2p). provide your complete solutions and proofs in your paper homework and respond to questions or statements online. show that the midpoint of ab is the same distance from the x-axis and y-axis

The mid-point (p + q , q + p) of AB is the same distance from the x-axis and the y-axis

Step-by-step explanation:

* Lets explain how to solve the problem

- Any point will be equidistant from the x-axis and the y-axis must have

 equal coordinates

- Ex: point (4 , 4) is the same distance from the x-axis and the y-axis

 because the distance from the x-axis to the point is 4 (y-coordinate)

 and the distance from the y-axis and the point is 4 (x-coordinate)

- If (x , y) is the mid-point of a segment its endpoints are

   and , then

  and

* Lets solve the problem

∵ Point A has coordinates (p , q)

∵ Point B has coordinates (p + 2q , q + 2p)

- The mid-point of AB is (x , y)

∵

∴

- Take 2 as a common factor from the terms of the numerator

∴

- Divide up and down by 2

∴ x = p + q

∵

∴

- Take 2 as a common factor from the terms of the numerator

∴

- Divide up and down by 2

∴ y = q + p

∴ The mid point of AB is (p + q , q + p)

- p + q is the same with q + p

∵ The x-coordinate of the mid point of AB is p + q

∵ The y-coordinate of the mid point of AB is q + p

∵ p + q = q + p

∴ The coordinates of the mid-point of AB are equal

- According the explanation above

∴ The mid-point (p + q , q + p) of AB is the same distance from the

   x-axis and the y-axis

you need to include the information

the height of the building is 60.08 feet