# Locus of the Orthocenter of the Traingle

Coordinates of $\Delta ABC$ are $A(3,4)$, $B(5 \cos\theta, 5 \sin\theta)$ and $C(5 \sin\theta,-5 \cos\theta)$. Find the locus of its orthocenter.

My idea: It is clear that $(0,0)$ is equidistant from the three coordinates. So $S(0,0)$ is the circumcenter of $\Delta ABC$. Also the centroid of $\Delta ABC$ is

$$G\left( \frac{5 \cos\theta+5\sin\theta+3}{3},\frac{5\sin\theta-5\cos\theta+4}{3} \right)$$

Now if the orthocenter is $H(h,k)$, we have $G$ divides $H$ and $S$ in the ratio 2:1. So $G$ is given by G\left(\frac{h}{3},\frac{k}{3}\right) \implies \\ \begin{align} h &= 5 \cos\theta+5\sin\theta+3 \\ k &= 5 \sin\theta-5\cos\theta+4 \end{align}

Eliminating $\theta$ we get the locus as:

$$(x-3)^2+(y-4)^2=50$$

Let me know whether my approach is logical .. if there is an alternate way I will be happy to know.

Yes, its completely logical. Alternative approach would be to find the point of intersection of two perpendiculars drawn from the vertices onto opposite sides, by finding solution of two linear equations but it'll be long of course.

Modeling all with linear algebra. The lines through the sides are: \begin{align} g_a(t) &= u_B + (u_C - u_B) t \\ g_b(t) &= u_C + (u_A - u_C) t \\ g_c(t) &= u_A + (u_B - u_A) t \end{align} One then goes for the normals, looking for solutions to

\begin{align} n_a \cdot (u_C-u_B) = 0 \\ n_b \cdot (u_A-u_C) = 0 \\ n_c \cdot (u_B-u_A) = 0 \end{align}

In two dimension given a vector $v = (x, y)$, the vector $w = (y, -x)$ is orthogonal, as $v \cdot w = xy - yx = 0$. The normal to $v$ is then $n = w / |w|$, where $|v| = \sqrt{x^2 + y^2}$ is the norm of the vector $v$. With this recipe you can find the tree normals to the given difference vectors.

The normal lines through the triangle corners are then: \begin{align} h_a(t) &= u_A + n_a t \\ h_b(t) &= u_B + n_b t \\ h_c(t) &= u_C + n_c t \end{align}

Then one would solve for an intersection of those normal lines, considering two lines should be enough: $$h_b(t_b) = h_c(t_c) \iff \\ u_B + n_b t_b = u_C + n_c t_c \iff \\ n_b t_b - n_c t_c = u_C - u_B \iff \\ L x = y \\ x = L^{-1} y$$ with the matrix $L = (n_b,-n_c)$, solution vector $t = (t_b, t_c)$ and inhomogenity $y = u_C - u_B$.

For $2\times 2$-matrices the inverse can be calculated with this: $$L = \left( \begin{matrix} \alpha & \beta \\ \gamma & \delta \end{matrix} \right) \Rightarrow L^{-1} = \frac{1}{\det L} \left( \begin{matrix} \delta & -\beta \\ -\gamma & \alpha \end{matrix} \right)$$ where $\det L = \alpha\delta - \beta\gamma$ is the determinant, which needs to be not zero.

Finally you would use one of the found solutions $(t_b, t_c)$ and put it into the normal line to get the coordinates of the orthocenter, for example

$$h_b(t_b) = u_B + n_b t_b$$