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A is a given point and P is any point on a given straight line. If AQ=AP and AQ makes a constant angle with AP find the locus of Q.

According to me the answer should be two straight lines making angle equal to ±x(angle between AQ and AP) with line on which P moves at P and Q of the only triangle whose Q is on the line on which P moves. These two lines intersect at the image of A with respect to line on which P moves. Just need to confirm it and get a decent looking solution. I did on Geogebra and I get this. enter image description here I used 45 degrees as the given angle here.

  • Welcome to MSE. Please include your question in the body of the question, instead of putting it only in the title. – José Carlos Santos Apr 05 '20 at 09:59
  • Why? Should I change it now? –  Apr 05 '20 at 10:03
  • Because the body of the question is supposed to be understandable without the title. And, yes, change it now. – José Carlos Santos Apr 05 '20 at 10:13
  • Ok I am doing it thanks –  Apr 05 '20 at 10:14
  • Can you clarify this question? As it is written, it makes no sense to me. For a given point $A$ and a given line $l$ and a constant angle $\theta$, there is only 1 value of $P$ that will lead to $ AQ=AP, \angle AQP = \theta$. Can you add an image where you have several loci for $Q$? – Calvin Lin Apr 16 '20 at 13:50

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