Question

Find the coordinates of the circumcenter of the triangle with the vertices A(4,12) B(14,6) and C(-6,2) This is problem number 4. I tried to graph it to the best of my ability. Image is down below.

Answer 1

The circumcenter can be found as the intersection of the perpendicular bisectors.

The middle points of each one of the segments are

`\begin{gathered} AB=((4+14)/(2),(12+6)/(2))=(9,9) \\ BC=((14-6)/(2),(6+2)/(2))=(4,4) \\ CA=((-6+4)/(2),(2+12)/(2))=(-1,7) \\ \text{Where we used the formula} \\ M=((x_1+x_2)/(2),(y_1+y_2)/(2)) \end{gathered}`

Furthermore, the slopes of each one of the three segments are

`\begin{gathered} m_(AB)=(6-12)/(14-4)=-(6)/(10)=-(3)/(5) \\ m_(BC)=(6-2)/(14+6)=(4)/(20)=(1)/(5) \\ m_(CA)=(2-12)/(-6-4)=-(10)/(-10)=1 \\ \text{where we used the formula} \\ m=(y_2-y_1)/(x_2-x_1) \end{gathered}`

In general, two lines are perpendicular if the product of their slopes is equal to -1. Then, the slopes of a perpendicular line to each one of the segments are

`\begin{gathered} m^(\prime)_(AB)=(5)/(3) \\ m^(\prime)_(BC)=-5 \\ m^(\prime)_(CA)=-1 \end{gathered}`

Given a point on a line and its slope, we can calculate the equation of any line. In our case,

`\begin{gathered} bisectorofAB\colon y-9=(5)/(3)(x-9) \\ \Rightarrow bisectorofAB\colon y=(5)/(3)x-6 \\ bisectorofBC\colon y-4=-5(x-4)_{} \\ \Rightarrow bisectorofBC\colon y=-5x+24 \end{gathered}`

Finally, calculate the intersection point of lines AB and BC as shown below

`\begin{gathered} y=(5)/(3)x-6,y=-5x+24 \\ \Rightarrow(5)/(3)x-6=-5x+24 \\ \Rightarrow(5)/(3)x+5x=30 \\ \Rightarrow(20)/(3)x=30 \\ \Rightarrow x=(9)/(2) \\ \Rightarrow y=-5((9)/(2))+24=(3)/(2) \\ \Rightarrow y=(3)/(2) \\ \Rightarrow((9)/(2),(3)/(2))\to\text{circumcenter} \end{gathered}`

The answer is (9/2,3/2)

`\begin{cases}y=(5x)/(3)-6 \\ y=-5x+24\end{cases}`

Remember that (properties of equality)

`\begin{gathered} a=b\Rightarrow a+c=b+c \\ \text{and} \\ a=b\Rightarrow a\cdot c=b\cdot c \end{gathered}`

From the system of equations,

`\begin{gathered} y=y \\ \Rightarrow(5x)/(3)-6=-5x+24 \end{gathered}`

Then, add +6 to both sides of the equation,

`\begin{gathered} (5x)/(3)-6+6=-5x+24+6 \\ \Rightarrow(5x)/(3)=-5x+30 \end{gathered}`

Add +5x to both sides of the equation,

`\begin{gathered} (5x)/(3)+5x=-5x+30+5x \\ \Rightarrow(20x)/(3)=30 \end{gathered}`

Multiply both sides of the equation by 3

`\begin{gathered} (20x)/(3)=30 \\ \Rightarrow20x=90 \end{gathered}`

Finally, multiply both sides of the equation by 1/20

`\begin{gathered} 20x=90 \\ \Rightarrow(1)/(20)\cdot20x=(1)/(20)\cdot90 \\ \Rightarrow x=(90)/(20) \end{gathered}`

There was no need to use negative reciprocals.