Ask your own question, for FREE!
Mathematics 8 Online
OpenStudy (icalibear):

How do you find the circumcenter of a triangle?

OpenStudy (icalibear):

Find the circumcenter of \[\Delta EFG \] with \[E(2,6), F(2,6), G(6,4)\]

OpenStudy (anonymous):

do you know what the circumcenter of a triangle is?

OpenStudy (icalibear):

isnt it where all of the medians intersect

OpenStudy (anonymous):

no, that's the centroid...

OpenStudy (icalibear):

hey i was writing that down

OpenStudy (anonymous):

the point where all the perpendicular bisectors of the sides meet is called the circumcenter...

OpenStudy (anonymous):

sorry... i wanted to get the terms right....

OpenStudy (icalibear):

how do i find it do i graph the triangle or..? and thats ok

OpenStudy (anonymous):

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?

OpenStudy (icalibear):

yes how do we set that up o:

OpenStudy (anonymous):

graphing MIGHT be easier though...

OpenStudy (icalibear):

i should probably know the algebra way for tests when theres no graph paper..

OpenStudy (anonymous):

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?

OpenStudy (icalibear):

Yep evidently

OpenStudy (icalibear):

wait no

OpenStudy (icalibear):

E(2,6) and F(2, 4)

OpenStudy (icalibear):

i typoed

OpenStudy (anonymous):

ok find the midpoint of E(2,6) and F(2,4) using the midpoint formula... do you remember the midpoint formula?

OpenStudy (icalibear):

what was it again

OpenStudy (icalibear):

it was like x2/x1 or something

OpenStudy (anonymous):

midpoint = (average of the x's, average of the y's)

OpenStudy (anonymous):

mipoint = \(\large (\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}) \)

OpenStudy (icalibear):

\[(\frac{ 2 + 2 }{ 2 }\ ), (\frac{ 6 + 4 }{ 2 })\]

OpenStudy (anonymous):

ok... simplified you have the midpoint of EF as (2, 5) correct?

OpenStudy (icalibear):

umm how did u get 5

OpenStudy (anonymous):

6+4=10 10/2=5

OpenStudy (icalibear):

oh -_- sorry ok so what do we do next

OpenStudy (icalibear):

5?

OpenStudy (icalibear):

no 2

OpenStudy (anonymous):

what is the slope that is PERPENDICULAR to side EF?

OpenStudy (icalibear):

wait wht hold on

OpenStudy (icalibear):

2?

OpenStudy (anonymous):

slope = \(\large \frac {y_2-y_1}{x_2-x_1} \)

OpenStudy (icalibear):

\[\frac{ 4-6}{ 2-2 } =\frac{ -2 }{ 0 }\]

OpenStudy (icalibear):

slope is -2

OpenStudy (anonymous):

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?

OpenStudy (icalibear):

right

OpenStudy (anonymous):

so the perpendicular bisector of EF has slope 0, so the equation of the line through (2,5) is y=5... agreed?

OpenStudy (shubhamsrg):

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..

OpenStudy (sirm3d):

if you plot the points, you'll be looking at the vertices of a right triangle.

OpenStudy (icalibear):

:'s

OpenStudy (shubhamsrg):

which is obviously evident from any figure..

OpenStudy (anonymous):

this is a right triangle?

OpenStudy (icalibear):

well i wanted to know the algebraic way so that i can do it when i dont have graph paper

OpenStudy (icalibear):

It is?

OpenStudy (sirm3d):

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.

OpenStudy (anonymous):

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.

OpenStudy (icalibear):

well how do we do that

OpenStudy (shubhamsrg):

exactly!

OpenStudy (anonymous):

uh.... like you said... maybe we can graph it and see which is the hypotenuse.... (don't kill me!!!)

OpenStudy (icalibear):

:'s..

OpenStudy (icalibear):

i ripped my notes paper erasing -__- which are we actually gonna do lol

OpenStudy (anonymous):

don't do that.. the procedure we went through works on ALL triangles....

OpenStudy (icalibear):

..but we only did 2 steps

OpenStudy (anonymous):

that's ok.... because the other steps is doing the same thing to another side...

OpenStudy (shubhamsrg):

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..

OpenStudy (icalibear):

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

OpenStudy (anonymous):

@shubhamsrg , be my guest....???? pls... i need to be quiet now.... sorry @icalibear .... :(

OpenStudy (icalibear):

D':

OpenStudy (sirm3d):

EF is vertical, FG is horizontal. so EG is the hypotenuse.

OpenStudy (icalibear):

ok got it

OpenStudy (icalibear):

wats next 0.0

OpenStudy (shubhamsrg):

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..

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!