Hello, magentarita!
The orthocenter is the intersection of the altitudes.
Did you make a sketch ??
Find the orthocenter of a triangle with vertices:
G(2,5), H(6,5), J(4,1) Code:

(2,5)  (6,5)
A o * * * * * o B
*  : *
*  : *
* : *
    + *  : *      
 * :*
 o C
 (4,1)

Recall that an altitude is perpendicular to the opposite side.
Since $\displaystyle AB$ is horizontal,
. . we see the altitude from $\displaystyle C$ is the vertical line $\displaystyle x = 4$
The slope of $\displaystyle AC$ is: .$\displaystyle m_1 \:=\:\frac{\text{}15}{4(\text{}2)} \:=\:1$
Hence, the altitude from $\displaystyle B$ has slope: .$\displaystyle m_2 \:=\:1$
The equation of altitude from $\displaystyle B(6,5)$ with slope $\displaystyle 1$
. . $\displaystyle y  5 \:=\:1(x6) \quad\Rightarrow\quad y \:=\:x1$
The intersection of this altitude with the vertical line $\displaystyle x = 4$
. . is: .$\displaystyle \begin{Bmatrix}y \:=\:x1 \\ x \:=\:4\end{Bmatrix} \quad\Rightarrow\quad \begin{Bmatrix}x \:=\:4\\y\:=\:3 \end{Bmatrix} $
Therefore, the orthocenter is $\displaystyle (4,3)$