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This is the full problem: The points $A(5,1)$ and $B(-3,6)$ represent one of the equal sides of an isosceles triangle. Determine one of the possible points that would represent the third vertex of the triangle. Provide calculations to support your answer.

This is easy to figure out by graphing out the triangle, though I have no idea how to do it using calculations. Spent $2$ hours still no luck... Thank you in advance for your help.

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We have two cases.

i) Think of the circle with center in $A$ and radius $AB$. Any point will do.

or

ii) Get the circle with center in $B$ and radius $AB$. Pick any point.

Remember that the equation of a circle is given by $$(x - x_c)^2 + (y - y_c)^2 = r²$$ where $(x_c, y_c)$ is the center, and $r$ is the radius. Remember also, that $$d((x,y),(a,b)) = \sqrt{(x-a)^2 + (y-b)^2}$$

Can you do it?

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  • $\begingroup$ I'm not sure what you mean, can you please give me a step by step explanation? I know that I have to find the radius of the line, but then what do I do with that information? Peter said I can reflect the line, but you can't use graphing so that's out of question. $\endgroup$ – Simon Jul 12 '14 at 19:30
  • $\begingroup$ You can draw stuff to see what is happening, but everything else is just analytic geometry. You can focus on one of the cases, to get the idea. Once you've got the equation of the circle, choose a point $P = (x_p, y_p)$ there, and check that $d(A,P) = d(B,P)$, for example. $\endgroup$ – Ivo Terek Jul 12 '14 at 19:35
  • $\begingroup$ A question, what does (a,b) represent? The unknown points? I'm still confused. So the steps are: 1. Find the radius. 2. Then what's next? I'm bad when it comes to math :( $\endgroup$ – Simon Jul 12 '14 at 19:48
  • $\begingroup$ Don't worry about that $(a,b)$, it's just a point I put there to remind you of the formula. When you need to give names to points, and you've already used $(x,y)$, use your creativity to give names, $(a,b), (u,v), (c,d)$, etc. $\endgroup$ – Ivo Terek Jul 12 '14 at 19:50
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$|AB|=\sqrt{(-3-5)^2+(6-1)^2}=\sqrt{89}$ so the set of all points which work lie on circles centered at $A$ and $B$ each with radius $\sqrt{89}$. Still if you are just looking for a particular point, it is probably easiest just to reflect one of your points over a line through the other point.

For example, we can reflect $B$ through the horizontal line $y=1$ which passes through $A$. This gives a new point $B'=(-3,-4)$.

$$\\$$ Just for the sake of completeness, we can also note that you do get a couple degenerate solutions when you reflect through lines perpendicular to the line segment $AB$.

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  • $\begingroup$ You can't reflect the line since you aren't allowed to graph. There has to be a mathematical method to doing this. $\endgroup$ – Simon Jul 12 '14 at 19:28
  • $\begingroup$ @Simon You don't need to graph to reflect over a line - especially if you are reflecting over one that is parallel to the $x$ or $y$ axis. $\endgroup$ – Peter Woolfitt Jul 12 '14 at 19:34
  • $\begingroup$ @Simon in order to be isosceles, you just need to show that two sides have the same length - this can be done through explicit calculation. $\endgroup$ – Peter Woolfitt Jul 12 '14 at 19:37

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