Finding mid points after rotating two lines

Question

If $A(a,0)$ and $B(-a,0)$ are two fixed points,$(a>0)$.$P(x,y)$ is any point in the upper half of the $x,y$ plane.The join $AP$ is turned about $A$ clockwise through 90 degree and $BP$ is turned about $B$ in the opposite sense through the same angle so that the new positions of $AP$ and $BP$ are $AP_1$ and $BP_2$ respectively. Prove that for all positions of $P$, the middle point of $P_1P_2$ is the fixed point $(0, a)$.

Answer 1

In a general case (look at the picture!), when you rotate point $P(x,y)$ around point $A(x_A=a, y_A=0)$ clockwise for $90^\circ$ you get point $P_1(x_1,y_1)$ such that:

$$(x_1-x_A)=(y-y_A)$$

$$(y_1-y_A)=-(x-x_A)$$

or:

$$x_1=a+y$$

$$y_1=-(x-a)$$

In a similar way (draw one more picture!), when you rotate point $P(x,y)$ around point $B(x_B=-a, y_B=0)$ counterclockwise for $90^\circ$ you get point $P_2(x_2,y_2)$ such that:

$$(x_2-x_B)=-(y-y_B)$$

$$(y_2-y_B)=(x-x_B)$$

or:

$$x_2=-a-y$$

$$y_2=x+a$$

Midpoint $M(x_M,y_M)$ of $P_1P_2$ has the following coordinates:

$$x_M=\frac{x_1+x_2}2=0$$

$$y_M=\frac{y_1+y_2}2=a$$