Q:

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

Accepted Solution

A:
Answer:The mid-point (p + q , q + p) of AB is the same distance from the x-axis and the y-axisStep-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  [tex](x_{1},y_{1})[/tex]  and [tex](x_{2},y_{2})[/tex], then  [tex]x=\frac{x_{1}+x_{2}}{2}[/tex] and [tex]y=\frac{y_{1}+y_{2}}{2}[/tex]* 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)∵ [tex]x=\frac{p+p+2q}{2}[/tex]∴ [tex]x=\frac{2p+2q}{2}[/tex]- Take 2 as a common factor from the terms of the numerator∴ [tex]x=\frac{2(p+q)}{2}[/tex]- Divide up and down by 2∴ x = p + q∵ [tex]y=\frac{q+q+2p}{2}[/tex]∴ [tex]y=\frac{2q+2p}{2}[/tex]- Take 2 as a common factor from the terms of the numerator∴ [tex]y=\frac{2(q+p)}{2}[/tex]- 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