Answer:
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
[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
