In a given straight line AB, find a point P such that the difference in the squares on AP and PB is equal to the difference between two given squares.

Question

In a given straight line $AB$, find a point $P$ such that the difference in the squares on $AP$ and $PB$ is equal to the difference between two given squares.

What I did:It is one of the answers(because the question was getting too big). I think my answer is not good so please post a better answer if possible.

Please tell me what I can add before downvoting so that this question can improve instead of just getting negative votes.

Answer 1

Your solution is correct. Here, you can find a more simple way.

Let $A,B$ the area of two given squares. Let $l$ the lenght of the segment $AB$. We have: $$\overline{AP}^2=(l-\overline{PB})^2$$ So, we have: $$\overline{AP}^2-\overline{PB}^2=A-B$$ Substituing, we arrive at: $$l^2-2l\overline{PB}=A-B\leftrightarrow \overline{PB}=\frac{B-A+l^2}{2l}$$

Answer 2

Let the given squares' side equal to $CA$ and $CB'$ ($CB'$ parallel to $AB$) and $\angle CAB' = 90°$ ,then $AB'^2$ will be their squares difference.

Join $B'B$ and construct it's perpendicular bisector. $K$ is the point where this bisector hits $AB$. Join $KB$'. Due to congruency of $BKD$ and $B'KD, BK=B'K$.

$AK^2-KB^2=AK^2-KB'^2=AB'^2$.

Therefore, $K=P$