Question

Part 1: Centres of a Triangle
Consider the triangle , where . In this part of the assignment, you will use proper mathematical notation and organization to find the following. Remember to show your work and use the rubric to ensure that you have addressed all parts of the evaluation.
 Find the coordinates of the centroid (one type of "centre of a triangle")
 Find the coordinates of the orthocentre (a second type of "centre of a triangle")  Find the coordinates of the circumcentre (a third type of "centre of a triangle")  State any relationship you see between the three key points you found above.

Answer 1

ANSWER on Question #71592 – Math – Analytic Geometry

QUESTION

1) Find the coordinates of the centroid (one type of "centre of a triangle")

2) Find the coordinates of the orthocentre (a second type of "centre of a triangle")

3) Find the coordinates of the circumcentre (a third type of "centre of a triangle")

State any relationship you see between the three key points you found above.

SOLUTION

Suppose we are given a triangle $\Delta ABC$ with coordinates of the vertices:

$$\left\{ \begin{array}{l} A \big (A _ {x}; A _ {y} \big) \\ B \big (B _ {x}; B _ {y} \big) \\ C \big (C _ {x}; C _ {y} \big) \end{array} \right.$$

1) Find the coordinates of the centroid $M\big(M_x;M_y\big)$

1 step: We find the coordinates of $N(N_x; N_y)$ - the middle of the side $BC$.

$$BN = CN \rightarrow \left\{ \begin{array}{l} N _ {x} = \frac {B _ {x} + C _ {x}}{2} \\ N _ {y} = \frac {B _ {y} + C _ {y}}{2} \end{array} \right.$$

(More information: https://en.wikipedia.org/wiki/Midpoint)

2 step: Let us find the coordinates of the $M(M_x; M_y)$ - centroid of the triangle $\Delta ABC$.

The main property of the triangle's centroid: The centroid is twice as close along any median to the side that the median intersects as it is to the vertex it emanates from.

(More information: https://en.wikipedia.org/wiki/Median_(geometry))

Then,

$$A M: M N = 2: 1 \rightarrow \left\{\begin{array}{l}M _ {x} = \frac {2 \cdot N _ {x} + 1 \cdot A _ {x}}{2 + 1}\\ M _ {y} = \frac {2 \cdot N _ {y} + 1 \cdot A _ {y}}{2 + 1}\end{array}\right.\rightarrow \left\{\begin{array}{l}M _ {x} = \frac {2 \cdot \frac {B _ {x} + C _ {x}}{2} + 1 \cdot A _ {x}}{2 + 1}\\ M _ {y} = \frac {2 \cdot \frac {B _ {y} + C _ {y}}{2} + 1 \cdot A _ {y}}{2 + 1}\end{array}\right.\rightarrow$$ $$\left\{ \begin{array}{l} M _ {x} = \frac {A _ {x} + B _ {x} + C _ {x}}{3} \\ M _ {y} = \frac {A _ {y} + B _ {y} + C _ {y}}{3} \end{array} \right.$$

Conclusion,

$$\left\{ \begin{array}{c} A \big (A _ {x}; A _ {y} \big) \\ B \big (B _ {x}; B _ {y} \big) \\ C \big (C _ {x}; C _ {y} \big) \\ M \big (M _ {x}; M _ {y} \big) - c e n t r o i d o f t h e t r i a n g l e \Delta A B C \end{array} \right. \to \left\{ \begin{array}{c} M _ {x} = \frac {A _ {x} + B _ {x} + C _ {x}}{3} \\ M _ {y} = \frac {A _ {y} + B _ {y} + C _ {y}}{3} \end{array} \right.$$

3) Find the coordinates of the circumcenter $U\big(U_x;U_y\big)$

By the definition, the circumcenter $U(U_x;U_y)$ - point is equidistant from all its vertices. We write the equation of the circumscribed of a circle

$$(X - U_x)^2 + (Y - U_y)^2 = R^2$$

(More information: https://en.wikipedia.org/wiki/Circle)

Since the vertex of a triangle is on a given circle, their coordinates satisfy this equation:

$$\left\{ \begin{array}{l} (A_x - U_x)^2 + (A_y - U_y)^2 = R^2 \\ (B_x - U_x)^2 + (B_y - U_y)^2 = R^2 \\ (C_x - U_x)^2 + (C_y - U_y)^2 = R^2 \end{array} \right.$$

We have obtained a system of three equations with respect to three unknowns: $U_x; U_y; R$.

We will try to solve this system and find the coordinates of the center.

$$\left\{ \begin{array}{l} (A_x - U_x)^2 + (A_y - U_y)^2 = R^2 \\ (B_x - U_x)^2 + (B_y - U_y)^2 = R^2 \rightarrow \\ (C_x - U_x)^2 + (C_y - U_y)^2 = R^2 \end{array} \right.$$ $$\left\{ \begin{array}{l} (A_x - U_x)^2 + (A_y - U_y)^2 = (B_x - U_x)^2 + (B_y - U_y)^2 \\ (A_x - U_x)^2 + (A_y - U_y)^2 = (C_x - U_x)^2 + (C_y - U_y)^2 \rightarrow \end{array} \right.$$ $$\left\{ \begin{array}{l} A_x^2 - 2A_xU_x + U_x^2 + A_y^2 - 2A_yU_y + U_y^2 = B_x^2 - 2B_xU_x + U_x^2 + B_y^2 - 2B_yU_y + U_y^2 \\ A_x^2 - 2A_xU_x + U_x^2 + A_y^2 - 2A_yU_y + U_y^2 = C_x^2 - 2C_xU_x + U_x^2 + C_y^2 - 2C_yU_y + U_y^2 \end{array} \right.$$ $$\left\{ \begin{array}{l} -2A_xU_x + 2B_xU_x - 2A_yU_y + 2B_yU_y = B_x^2 - A_x^2 + B_y^2 - A_y^2 \\ -2A_xU_x + 2C_xU_x - 2A_yU_y + 2C_yU_y = C_x^2 - A_x^2 + C_y^2 - A_y^2 \end{array} \right.$$ $$\left\{ \begin{array}{l} 2U_x(B_x - A_x) + 2U_y(B_y - A_y) = B_x^2 - A_x^2 + B_y^2 - A_y^2 \\ 2U_x(C_x - A_x) + 2U_y(C_y - A_y) = C_x^2 - A_x^2 + C_y^2 - A_y^2 \end{array} \right.$$

1 step: From this system we find $U_{x}$

$$\left\{ \begin{array}{l} 2 U _ {x} (B _ {x} - A _ {x}) + 2 U _ {y} \big (B _ {y} - A _ {y} \big) = B _ {x} ^ {2} - A _ {x} ^ {2} + B _ {y} ^ {2} - A _ {y} ^ {2} \big | \times \big (C _ {y} - A _ {y} \big) \\ 2 U _ {x} (C _ {x} - A _ {x}) + 2 U _ {y} \big (C _ {y} - A _ {y} \big) = C _ {x} ^ {2} - A _ {x} ^ {2} + C _ {y} ^ {2} - A _ {y} ^ {2} \big | \times \big (B _ {y} - A _ {y} \big) \end{array} \right.$$ $$\left\{ \begin{array}{l} 2 U _ {x} (B _ {x} - A _ {x}) \big (C _ {y} - A _ {y} \big) + 2 U _ {y} \big (B _ {y} - A _ {y} \big) \big (C _ {y} - A _ {y} \big) = \big (B _ {x} ^ {2} - A _ {x} ^ {2} + B _ {y} ^ {2} - A _ {y} ^ {2} \big) \big (C _ {y} - A _ {y} \big) \\ 2 U _ {x} (C _ {x} - A _ {x}) \big (B _ {y} - A _ {y} \big) + 2 U _ {y} \big (C _ {y} - A _ {y} \big) \big (B _ {y} - A _ {y} \big) = \big (C _ {x} ^ {2} - A _ {x} ^ {2} + C _ {y} ^ {2} - A _ {y} ^ {2} \big) \big (B _ {y} - A _ {y} \big) \end{array} \right.$$ $$\begin{array}{l} 2 U _ {x} \left(B _ {x} - A _ {x}\right) \left(C _ {y} - A _ {y}\right) - 2 U _ {x} \left(C _ {x} - A _ {x}\right) \left(B _ {y} - A _ {y}\right) = \\ = \left(B _ {x} ^ {2} - A _ {x} ^ {2} + B _ {y} ^ {2} - A _ {y} ^ {2}\right) \left(C _ {y} - A _ {y}\right) - \left(C _ {x} ^ {2} - A _ {x} ^ {2} + C _ {y} ^ {2} - A _ {y} ^ {2}\right) \left(B _ {y} - A _ {y}\right)\rightarrow \\ \end{array}$$ $$U _ {x} = \frac {\left(B _ {x} ^ {2} - A _ {x} ^ {2} + B _ {y} ^ {2} - A _ {y} ^ {2}\right) \left(C _ {y} - A _ {y}\right) - \left(C _ {x} ^ {2} - A _ {x} ^ {2} + C _ {y} ^ {2} - A _ {y} ^ {2}\right) \left(B _ {y} - A _ {y}\right)}{2 \left(\left(B _ {x} - A _ {x}\right) \left(C _ {y} - A _ {y}\right) - \left(C _ {x} - A _ {x}\right) \left(B _ {y} - A _ {y}\right)\right)}$$

We transform this expression a little:

$$\begin{array}{l} \left(B _ {x} - A _ {x}\right) \left(C _ {y} - A _ {y}\right) - \left(C _ {x} - A _ {x}\right) \left(B _ {y} - A _ {y}\right) = \\ = B _ {x} C _ {y} - B _ {x} A _ {y} - A _ {x} C _ {y} + A _ {x} A _ {y} - C _ {x} B _ {y} + C _ {x} A _ {y} + A _ {x} B _ {y} - A _ {x} A _ {y} = \\ = \left(A _ {x} B _ {y} - A _ {x} C _ {y}\right) + \left(B _ {x} C _ {y} - B _ {x} A _ {y}\right) + \left(C _ {x} A _ {y} - C _ {x} B _ {y}\right) = \\ = A _ {x} \left(B _ {y} - C _ {y}\right) + B _ {x} \left(C _ {y} - A _ {y}\right) + C _ {x} \left(A _ {y} - B _ {y}\right) \\ \end{array}$$ $$B _ {x} - A _ {x}) \left(C _ {y} - A _ {y}\right) - \left(C _ {x} - A _ {x}\right) \left(B _ {y} - A _ {y}\right) = A _ {x} \left(B _ {y} - C _ {y}\right) + B _ {x} \left(C _ {y} - A _ {y}\right) + C _ {x} \left(A _ {y} - B _ {y}\right)$$ $$\begin{array}{l} \left(B _ {x} ^ {2} - A _ {x} ^ {2} + B _ {y} ^ {2} - A _ {y} ^ {2}\right) \left(C _ {y} - A _ {y}\right) - \left(C _ {x} ^ {2} - A _ {x} ^ {2} + C _ {y} ^ {2} - A _ {y} ^ {2}\right) \left(B _ {y} - A _ {y}\right) = \\ = \left(B _ {x} ^ {2} + B _ {y} ^ {2}\right) \left(C _ {y} - A _ {y}\right) - \left(A _ {x} ^ {2} + A _ {y} ^ {2}\right) \left(C _ {y} - A _ {y}\right) - \\ - \left(C _ {x} ^ {2} + C _ {y} ^ {2}\right) \left(B _ {y} - A _ {y}\right) + \left(A _ {x} ^ {2} + A _ {y} ^ {2}\right) \left(B _ {y} - A _ {y}\right) = \\ = \left(A _ {x} ^ {2} + A _ {y} ^ {2}\right) \left(B _ {y} - A _ {y} - C _ {y} + A _ {y}\right) + \left(B _ {x} ^ {2} + B _ {y} ^ {2}\right) \left(C _ {y} - A _ {y}\right) + \left(C _ {x} ^ {2} + C _ {y} ^ {2}\right) \left(A _ {y} - B _ {y}\right) = \\ = \left(A _ {x} ^ {2} + A _ {y} ^ {2}\right) \left(B _ {y} - C _ {y}\right) + \left(B _ {x} ^ {2} + B _ {y} ^ {2}\right) \left(C _ {y} - A _ {y}\right) + \left(C _ {x} ^ {2} + C _ {y} ^ {2}\right) \left(A _ {y} - B _ {y}\right) \\ \end{array}$$ $$\begin{array}{l} \boxed {\left(B _ {x} ^ {2} - A _ {x} ^ {2} + B _ {y} ^ {2} - A _ {y} ^ {2}\right) \left(C _ {y} - A _ {y}\right) - \left(C _ {x} ^ {2} - A _ {x} ^ {2} + C _ {y} ^ {2} - A _ {y} ^ {2}\right) \left(B _ {y} - A _ {y}\right) =} \\ = \left(A _ {x} ^ {2} + A _ {y} ^ {2}\right) \left(B _ {y} - C _ {y}\right) + \left(B _ {x} ^ {2} + B _ {y} ^ {2}\right) \left(C _ {y} - A _ {y}\right) + \left(C _ {x} ^ {2} + C _ {y} ^ {2}\right) \left(A _ {y} - B _ {y}\right) \\ \end{array}$$

Then,

$$U _ {x} = \frac {\left(A _ {x} ^ {2} + A _ {y} ^ {2}\right) \left(B _ {y} - C _ {y}\right) + \left(B _ {x} ^ {2} + B _ {y} ^ {2}\right) \left(C _ {y} - A _ {y}\right) + \left(C _ {x} ^ {2} + C _ {y} ^ {2}\right) \left(A _ {y} - B _ {y}\right)}{2 \left(A _ {x} \left(B _ {y} - C _ {y}\right) + B _ {x} \left(C _ {y} - A _ {y}\right) + C _ {x} \left(A _ {y} - B _ {y}\right)\right)}$$

2 step: From this system we find $U_y$

$$\left\{ \begin{array}{l} 2 U _ {x} (B _ {x} - A _ {x}) + 2 U _ {y} (B _ {y} - A _ {y}) = B _ {x} ^ {2} - A _ {x} ^ {2} + B _ {y} ^ {2} - A _ {y} ^ {2} \big | \times (C _ {x} - A _ {x}) \\ 2 U _ {x} (C _ {x} - A _ {x}) + 2 U _ {y} (C _ {y} - A _ {y}) = C _ {x} ^ {2} - A _ {x} ^ {2} + C _ {y} ^ {2} - A _ {y} ^ {2} \big | \times (B _ {x} - A _ {x}) \end{array} \right.$$ $$\left\{ \begin{array}{l} 2 U _ {x} (B _ {x} - A _ {x}) (C _ {x} - A _ {x}) + 2 U _ {y} (B _ {y} - A _ {y}) (C _ {x} - A _ {x}) = (B _ {x} ^ {2} - A _ {x} ^ {2} + B _ {y} ^ {2} - A _ {y} ^ {2}) (C _ {x} - A _ {x}) \\ 2 U _ {x} (C _ {x} - A _ {x}) (B _ {x} - A _ {x}) + 2 U _ {y} (C _ {y} - A _ {y}) (B _ {x} - A _ {x}) = (C _ {x} ^ {2} - A _ {x} ^ {2} + C _ {y} ^ {2} - A _ {y} ^ {2}) (B _ {x} - A _ {x}) \end{array} \right.$$ $$\begin{array}{l} 2 U _ {y} \left(B _ {y} - A _ {y}\right) \left(C _ {x} - A _ {x}\right) - 2 U _ {x} \left(C _ {y} - A _ {y}\right) \left(B _ {x} - A _ {x}\right) = \\ = \left(B _ {x} ^ {2} - A _ {x} ^ {2} + B _ {y} ^ {2} - A _ {y} ^ {2}\right) \left(C _ {x} - A _ {x}\right) - \left(C _ {x} ^ {2} - A _ {x} ^ {2} + C _ {y} ^ {2} - A _ {y} ^ {2}\right) \left(B _ {x} - A _ {x}\right)\rightarrow \\ \end{array}$$ $$U _ {y} = \frac {\left(B _ {x} ^ {2} - A _ {x} ^ {2} + B _ {y} ^ {2} - A _ {y} ^ {2}\right) \left(C _ {x} - A _ {x}\right) - \left(C _ {x} ^ {2} - A _ {x} ^ {2} + C _ {y} ^ {2} - A _ {y} ^ {2}\right) \left(B _ {x} - A _ {x}\right)}{2 \left(\left(B _ {y} - A _ {y}\right) \left(C _ {x} - A _ {x}\right) - \left(C _ {y} - A _ {y}\right) \left(B _ {x} - A _ {x}\right)\right)}$$

Applying similar transformations, we obtain

$$U _ {y} = \frac {\left(A _ {x} ^ {2} + A _ {y} ^ {2}\right) \left(C _ {x} - B _ {x}\right) + \left(B _ {x} ^ {2} + B _ {y} ^ {2}\right) \left(A _ {x} - C _ {x}\right) + \left(C _ {x} ^ {2} + C _ {y} ^ {2}\right) \left(B _ {x} - A _ {x}\right)}{2 \left(A _ {x} \left(B _ {y} - C _ {y}\right) + B _ {x} \left(C _ {y} - A _ {y}\right) + C _ {x} \left(A _ {y} - B _ {y}\right)\right)}$$

Conclusion,

$$\left\{ \begin{array}{c} A \big (A _ {x}; A _ {y} \big) \\ B \big (B _ {x}; B _ {y} \big) \\ C \big (C _ {x}; C _ {y} \big) \\ U \big (U _ {x}; U _ {y} \big) - \text{the circumcenter of the triangle } \Delta ABC \end{array} \right. \to$$ $$\left\{ \begin{array}{l} U _ {x} = \dfrac {\left(A _ {x} ^ {2} + A _ {y} ^ {2}\right) \left(B _ {y} - C _ {y}\right) + \left(B _ {x} ^ {2} + B _ {y} ^ {2}\right) \left(C _ {y} - A _ {y}\right) + \left(C _ {x} ^ {2} + C _ {y} ^ {2}\right) \left(A _ {y} - B _ {y}\right)}{2 \left(A _ {x} \left(B _ {y} - C _ {y}\right) + B _ {x} \left(C _ {y} - A _ {y}\right) + C _ {x} \left(A _ {y} - B _ {y}\right)\right)} \\ U _ {y} = \dfrac {\left(A _ {x} ^ {2} + A _ {y} ^ {2}\right) \left(C _ {x} - B _ {x}\right) + \left(B _ {x} ^ {2} + B _ {y} ^ {2}\right) \left(A _ {x} - C _ {x}\right) + \left(C _ {x} ^ {2} + C _ {y} ^ {2}\right) \left(B _ {x} - A _ {x}\right)}{2 \left(A _ {x} \left(B _ {y} - C _ {y}\right) + B _ {x} \left(C _ {y} - A _ {y}\right) + C _ {x} \left(A _ {y} - B _ {y}\right)\right)} \end{array} \right.$$

(More information: https://en.wikipedia.org/wiki/Circumscribed_circle)

2) Find the coordinates of the orthocenter - $H\bigl (H_x;H_y\bigr)$

We can use such properties as

$$\overline{UH} = \overline{UA} + \overline{UB} + \overline{UC}, \text{ where } U - \text{the circumcenter of the triangle } \Delta ABC$$

(More information: https://en.wikipedia.org/wiki/Altitude_(triangle))

In our case,

$$\left\{ \begin{array}{l} \overline{UH} = (H_x - U_x; H_y - U_y) \\ \overline{UA} = (A_x - U_x; A_y - U_y) \\ \overline{UB} = (B_x - U_x; B_y - U_y) \\ \overline{UC} = (C_x - U_x; C_y - U_y) \end{array} \right. \Rightarrow$$ $$H_x - U_x = (A_x - U_x) + (B_x - U_x) + (B_x - U_x) \Rightarrow$$ $$\boxed{H_x = (A_x + B_x + C_x) - 2U_x}$$ $$H_y - U_y = (A_y - U_y) + (B_y - U_y) + (B_y - U_y) \Rightarrow$$ $$\boxed{H_y = (A_y + B_y + C_y) - 2U_y}$$

As we know

\left\{ \begin{array}{l} U_x = \dfrac{(A_x^2 + A_y^2)(B_y - C_y) + (B_x^2 + B_y^2)(C_y - A_y) + (C_x^2 + C_y^2)(A_y - B_y)}{2(A_x(B_y - C_y) + B_x(C_y - A_y) + C_x(A_y - B_y))} \\ U_y = \dfrac{(A_x^2 + A_y^2)(C_x - B_x) + (B_x^2 + B_y^2)(A_x - C_x) + (C_x^2 + C_y^2)(B_x - A_x)}{2(A_x(B_y - C_y) + B_x(C_y - A_y) + C_x(A_y - B_y)) \end{array} \right.

Then,

$$\begin{array}{l} H _ {x} = \left(A _ {x} + B _ {x} + C _ {x}\right) - 2 \\ \cdot \frac {\left(A _ {x} ^ {2} + A _ {y} ^ {2}\right) \left(B _ {y} - C _ {y}\right) + \left(B _ {x} ^ {2} + B _ {y} ^ {2}\right) \left(C _ {y} - A _ {y}\right) + \left(C _ {x} ^ {2} + C _ {y} ^ {2}\right) \left(A _ {y} - B _ {y}\right)}{2 \left(A _ {x} \left(B _ {y} - C _ {y}\right) + B _ {x} \left(C _ {y} - A _ {y}\right) + C _ {x} \left(A _ {y} - B _ {y}\right)\right)} \\ \end{array}$$ $$\begin{array}{l} H _ {y} = \left(A _ {y} + B _ {y} + C _ {y}\right) - 2 \\ \cdot \frac {\left(A _ {x} ^ {2} + A _ {y} ^ {2}\right) \left(C _ {x} - B _ {x}\right) + \left(B _ {x} ^ {2} + B _ {y} ^ {2}\right) \left(A _ {x} - C _ {x}\right) + \left(C _ {x} ^ {2} + C _ {y} ^ {2}\right) \left(B _ {x} - A _ {x}\right)}{2 \left(A _ {x} \left(B _ {y} - C _ {y}\right) + B _ {x} \left(C _ {y} - A _ {y}\right) + C _ {x} \left(A _ {y} - B _ {y}\right)\right)} \\ \end{array}$$

Or

$$H _ {x} = \left(A _ {x} + B _ {x} + C _ {x}\right) - \frac {\left(A _ {x} ^ {2} + A _ {y} ^ {2}\right) \left(B _ {y} - C _ {y}\right) + \left(B _ {x} ^ {2} + B _ {y} ^ {2}\right) \left(C _ {y} - A _ {y}\right) + \left(C _ {x} ^ {2} + C _ {y} ^ {2}\right) \left(A _ {y} - B _ {y}\right)}{\left(A _ {x} \left(B _ {y} - C _ {y}\right) + B _ {x} \left(C _ {y} - A _ {y}\right) + C _ {x} \left(A _ {y} - B _ {y}\right)\right)}$$ $$H _ {y} = \left(A _ {y} + B _ {y} + C _ {y}\right) - \frac {\left(A _ {x} ^ {2} + A _ {y} ^ {2}\right) \left(C _ {x} - B _ {x}\right) + \left(B _ {x} ^ {2} + B _ {y} ^ {2}\right) \left(A _ {x} - C _ {x}\right) + \left(C _ {x} ^ {2} + C _ {y} ^ {2}\right) \left(B _ {x} - A _ {x}\right)}{2 \left(A _ {x} \left(B _ {y} - C _ {y}\right) + B _ {x} \left(C _ {y} - A _ {y}\right) + C _ {x} \left(A _ {y} - B _ {y}\right)\right)}$$

Conclusion,

$$\left\{ \begin{array}{c} A \big (A _ {x}; A _ {y} \big) \\ B \big (B _ {x}; B _ {y} \big) \\ C \big (C _ {x}; C _ {y} \big) \\ H \big (H _ {x}; H _ {y} \big) - \text{the orthocenter of the triangle } \Delta ABC \end{array} \right. \rightarrow$$ $$\left\{ \begin{array}{l} H _ {x} = \left(A _ {x} + B _ {x} + C _ {x}\right) - \frac {\left(A _ {x} ^ {2} + A _ {y} ^ {2}\right) \left(B _ {y} - C _ {y}\right) + \left(B _ {x} ^ {2} + B _ {y} ^ {2}\right) \left(C _ {y} - A _ {y}\right) + \left(C _ {x} ^ {2} + C _ {y} ^ {2}\right) \left(A _ {y} - B _ {y}\right)}{\left(A _ {x} \left(B _ {y} - C _ {y}\right) + B _ {x} \left(C _ {y} - A _ {y}\right) + C _ {x} \left(A _ {y} - B _ {y}\right)\right)} \\ H _ {y} = \left(A _ {y} + B _ {y} + C _ {y}\right) - \frac {\left(A _ {x} ^ {2} + A _ {y} ^ {2}\right) \left(C _ {x} - B _ {x}\right) + \left(B _ {x} ^ {2} + B _ {y} ^ {2}\right) \left(A _ {x} - C _ {x}\right) + \left(C _ {x} ^ {2} + C _ {y} ^ {2}\right) \left(B _ {x} - A _ {x}\right)}{2 \left(A _ {x} \left(B _ {y} - C _ {y}\right) + B _ {x} \left(C _ {y} - A _ {y}\right) + C _ {x} \left(A _ {y} - B _ {y}\right)\right)} \end{array} \right.$$

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