How to find the vertex of a triangle
I was given two coordinates of vertex - A(1,2) and B(5,-4) and also a circumcentre O(6,1). Is it possible to find out the 3rd vertex?
My try is first find the midpoint of AC , which is D[(1+x)/2,(2+y)/2]. However, it is hard to white the equation of AD and OD. Does anyone have any ideas?
$\endgroup$2 Answers
$\begingroup$It is impossible to have the third vertex with your informations.. If you draw the circle of center O containing A and B, every point C of the circle give a triangle ABC for which the circumcentre is O..
$\endgroup$ $\begingroup$Notice, let the third vertex be $(x, y)$.
We know that the distance of all three vertices from the circumscribed center of the circle is equal to the circumradius hence, the distance of each of the vertices $(x, y)$, $(1, 2)$ & $(5, -4)$ from the circumcenter $O(6, 1)$ $$\sqrt{(x-6)^2+(y-1)^2}=\sqrt{(6-1)^2+(1-2)^2}=\sqrt{(6-5)^2+(1-(-4))^2}$$$$\sqrt{(x-6)^2+(y-1)^2}=\sqrt{26}$$ $$(x-6)^2+(y-1)^2=26$$ Above equation represents that the locus of third unknown vertex $(x, y)$ is a circle with center $(6, 1)$ & a radius $\sqrt {26}$ i.e. there are infinitely many points as third vertex.
Hence, it is not possible to find third vertex with the help of given information.
$\endgroup$
