To find the orthocenter of a triangle, we first need to find the slopes of the lines that connect the vertices of the triangle.
Let's take the vertices A(-3,-7), B(5,1) and C(6,-4).
We first find the slope of line AB, which is (1 + 7)/(5 + 3) = 1. Then, we calculate the y-intercept of line AB, which is -7 - (-3 * 1) = -4.
Next, we'll find the slope of line BC, which is (-4 - 1)/(6 - 5) = -5. The y-intercept of line BC will be 1 - (5*-5) = 26.
Now, we're going to find the equations for the lines that are perpendicular to AB and BC and pass through points B and C respectively. The slope of a line perpendicular to another is the negative reciprocal of that line's slope. Thus, the slope of the line perpendicular to AB is -1/1 = -1 and the slope of the line perpendicular to BC is -1/(-5) = 1/5.
The y-intercept of the line perpendicular to AB through B is 1 - 5*-1 = 6.
For the perpendicular to BC through C, the y-intercept is -4 - 6*(1/5) = -5.2.
Finally, to find the orthocenter, we solve for the intersection of these two lines. Upon solving, we find that the x-coordinate for the orthocenter is solution of equation x = (6 + 5.2) / (-1 - (-1/5)) = 9.33 and the y-coordinate is -1* 9.33 + 6 = -3.33. Thus, the orthocenter of the triangle formed by these vertices is located at (9.33, -3.33).