This calculator finds all the main triangle parameters, such as area, medians, altitudes, centroid and incenter. The calculator shows a formula and an explanation for each parameter of a triangle.
solution
The circumcenter (center of circumscribed circle) of the triangle $ ABC $ is point:
$$ \left(2,~4\right) $$explanation
explanation
The circumcenter of a triangle is given by:
$$ C = \left( ~ \frac{\begin{vmatrix} A_x^2 + A_y^2 & A_y & 1 \\ B_x^2 + B_y^2 & B_y & 1 \\ C_x^2 + C_y^2 & C_y & 1 \\ \end{vmatrix}} {2 \cdot \begin{vmatrix} A_x & B_y & 1 \\ B_x & B_y & 1 \\ C_x & C_y & 1 \\ \end{vmatrix}}~,~ \frac{\begin{vmatrix} A_x & A_x^2 + A_y^2 & 1 \\ B_x & B_x^2+B_y^2 & 1 \\ C_x & C_x^2+C_y^2 & 1 \\ \end{vmatrix}} {2 \cdot \begin{vmatrix} A_x & B_y & 1 \\ A_x & B_y & 1 \\ C_x & C_y & 1 \\ \end{vmatrix}}~ \right) $$where $ A_x $ and $ A_y $ are $ x $ and $ y $ coordinates of the point $ A $ , $ B_x $ and $ B_y $ are $ x $ and $ y $ coordinates of the point $ B $ etc..
In this example we have : $ A_x = 1 $ , $ A_y = 5 $ , $ B_x = 3 $ , $ B_y = 3 $ , $ C_x = 3 $ and $ C_y = 5 $ .
When we insert these values into the formula, we obtain the given result.
The area of a triangle whose vertices are $A(x_A, y_A), B(x_B, y_B)$ and $C(x_C, y_C)$ is given by :
$$ {\color{blue}{ K = \frac12|x_A(y_B - y_C) + x_B(y_C - y_A) + x_C(y_A-y_B)| }} $$Example:
Find the area of the triangle whose vertices are $A(2, 4), B(3, -1)$ and $C(-3, 3)$.
Solution:
In this example we have: $ x_A = 2,~~ y_A = 4,~~ x_B = 3,~~ y_B = -1, x_C = -3,~~ y_C = 3$. So we have:
The centroid of a triangle whose vertices are $A(x_A, y_A), B(x_B, y_B)$ and $C(x_C, y_C)$ is given by :
$$ {\color{blue}{ (x,y) = \left(\frac{x_A + x_B + x_C}{3}, \frac{y_A + y_B + y_C}{3}\right) }} $$Example:
Find the centroid of the triangle whose vertices are $A(2, 4), B(3, -1)$ and $C(-3, 3)$.
Solution:
Using the same $x_A, y_A, x_B, y_B, x_C, y_C$, as in previous example we have:
$$ \begin{aligned} (x,y) & = \left(\frac{x_A + x_B + x_C}{3}, \frac{y_A + y_B + y_C}{3}\right) \\ (x,y) & = \left(\frac{2 + 3 - 3}{3}, \frac{4 - 1 + 3}{3}\right) \\ (x,y) & = \left(\frac{2}{3}, 2\right) \\ \end{aligned} $$Please tell me how can I make this better.