Question

1. What are the coordinates of the circumcenter of a triangle with vertices A(0,1), B(2, 1) , and C(2, 5) ?

2. What are the coordinates of the centroid of a triangle with vertices A(−6, 0) , B(−4, 4) , and C(0, 2) ?

Answer 1

your answer is a. (-6,0)

Answer 2

Answer:

1.

We know that a circumcenter is the intersection of all three perpendicular bisectors of a triangle, which divides the sides equally. So, basically, we need to find a points which is equidistant from its ends of the side.

So, for points $A(0,1)$ and $B(2,1)$, we have

$(x-0)^{2}+(y-1)^{2} =(x-2)^{2} +(y-1)^{2}$, you can observe that we just replace the coordinates in an expression which states that the distance of the two segments are equal, becuase the points $(x,y)$ divdes them equally. Now, we solve the expression

$(x-0)^{2}+(y-1)^{2} =(x-2)^{2} +(y-1)^{2}\\x^{2} +y^{2} -2y+1=x^{2} -4x+4+y^{2}-2y+2\\ 1=-4x+4+2\\4x=6-1\\x=\frac{5}{4}$

Now, we have $B(2,1)$ and $C(2,5)$, we do the same process

$(x-2)^{2} +(y-1)^{2} =(x-2)^{2} +(y-5)^{2} \\(y-1)^{2} =(y-5)^{2}\\y^{2}-2y+1=y^{2} -10y+25\\ -2y+10y=25-1\\8y=24\\y=3$

Therefore, the circumcenter is at $(\frac{5}{4},3)$

2.

The centroid of a triangle is defined as

$C_{x}=\frac{A_{x}+B_{x} +C_{x} }{3}\\ C_{y}=\frac{A_{y}+B_{y} +C_{y} }{3}$

Where

$A_{x}=-6; A_{y}=0\\B_{x}=-4; B_{y}=4\\C_{x}=0; C_{y}=2$

Replacing all values, we have

$C_{x}=\frac{A_{x}+B_{x} +C_{x} }{3}=\frac{-6-4+0}{3}=-\frac{10}{3} \\ \\C_{y}=\frac{A_{y}+B_{y} +C_{y} }{3}=\frac{0+4+2}{3}=\frac{6}{3}=2$

Therefore, the centroid is at $C(-\frac{10}{3} ;2)$