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ivann1987 [24]
3 years ago
14

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
Mathematics
1 answer:
Dafna1 [17]3 years ago
8 0

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:

*<em> Lets explain how to solve the problem</em>

- 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

 (x_{1},y_{1})  and (x_{2},y_{2}), then

 x=\frac{x_{1}+x_{2}}{2} and y=\frac{y_{1}+y_{2}}{2}

* <em>Lets solve the problem</em>

∵ Point A has coordinates (p , q)

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

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

∵ x=\frac{p+p+2q}{2}

∴ x=\frac{2p+2q}{2}

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

∴ x=\frac{2(p+q)}{2}

- Divide up and down by 2

∴ x = p + q

∵ y=\frac{q+q+2p}{2}

∴ y=\frac{2q+2p}{2}

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

∴ y=\frac{2(q+p)}{2}

- Divide up and down by 2

∴ y = q + p

∴ <em>The mid point of AB is (p + q , q + p)</em>

- 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

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