How do you find the circumcenter of a triangle?
Find the circumcenter of \[\Delta EFG \] with \[E(2,6), F(2,6), G(6,4)\]
do you know what the circumcenter of a triangle is?
isnt it where all of the medians intersect
no, that's the centroid...
hey i was writing that down
the point where all the perpendicular bisectors of the sides meet is called the circumcenter...
sorry... i wanted to get the terms right....
how do i find it do i graph the triangle or..? and thats ok
no need to graph.... just use algebra... if we can find the equation of two perpendicular bisectors and find where they intersect, then we're done... u wanna do it the algebra way?
yes how do we set that up o:
graphing MIGHT be easier though...
i should probably know the algebra way for tests when theres no graph paper..
ok... first we need to find the midpoints of two sides.... let's pick EF and FG. so find the midpoint of E(2, 6), F(2, 6) wait E and F have the same coordinates?
Yep evidently
wait no
E(2,6) and F(2, 4)
i typoed
ok find the midpoint of E(2,6) and F(2,4) using the midpoint formula... do you remember the midpoint formula?
what was it again
it was like x2/x1 or something
midpoint = (average of the x's, average of the y's)
mipoint = \(\large (\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}) \)
\[(\frac{ 2 + 2 }{ 2 }\ ), (\frac{ 6 + 4 }{ 2 })\]
ok... simplified you have the midpoint of EF as (2, 5) correct?
umm how did u get 5
6+4=10 10/2=5
oh -_- sorry ok so what do we do next
5?
no 2
what is the slope that is PERPENDICULAR to side EF?
wait wht hold on
2?
slope = \(\large \frac {y_2-y_1}{x_2-x_1} \)
\[\frac{ 4-6}{ 2-2 } =\frac{ -2 }{ 0 }\]
slope is -2
ok... you have a denominator of zero, which means the slope of EF is undefined... so the slope that is perpendicular to EF has slope 0, agreed?
right
so the perpendicular bisector of EF has slope 0, so the equation of the line through (2,5) is y=5... agreed?
actually there is an alternate definition of circumcenter which will help us solve the question quicker.. circumcenter is the point which is equidistant from all the 3 vertices of the triangle..
if you plot the points, you'll be looking at the vertices of a right triangle.
:'s
which is obviously evident from any figure..
this is a right triangle?
well i wanted to know the algebraic way so that i can do it when i dont have graph paper
It is?
you may continue with what you've started. but as a precautionary, check first if the triangle is a right triangle because if it's a right triangle, the solution is very simple.
yes it is.. and yes, i shuda realized that too.... @shubhamsrg is correct then, the easier way would be to take the mipoint of the hypotenuse.
well how do we do that
exactly!
uh.... like you said... maybe we can graph it and see which is the hypotenuse.... (don't kill me!!!)
:'s..
i ripped my notes paper erasing -__- which are we actually gonna do lol
don't do that.. the procedure we went through works on ALL triangles....
..but we only did 2 steps
that's ok.... because the other steps is doing the same thing to another side...
anyways, if you dont know what kinda triangle it is,, just use distance formulla twice, you'll get 2 linear eqns which are easier to solve..
well can we hurry its only a few hours till morning here and ive got to get this assignment done.. and i have no idea how to do this
@shubhamsrg , be my guest....???? pls... i need to be quiet now.... sorry @icalibear .... :(
D':
EF is vertical, FG is horizontal. so EG is the hypotenuse.
ok got it
wats next 0.0
i also have to leave now..sorry! :P anyways, i can help a bit.. |dw:1355734954961:dw| let O be circumcenter we know OA = OB =OC you must be knowing distance formula, use that in any 2 equalities and solve for x and y also, you can find out that this is a right triangle which will then be very easy to solve..
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