IF A(2,3),B(7,3) AND (C(5,9) ARE THE VERTICES OF TRIANGLE ABC, THEN THE Y-COORDINATE OF THE ORTHOCENTER OF TRIANGLE ABC IS
+128460
The "orthocenter" of a triangle is defined as the point where the three altitudes intersect.....BTW....in some triangles, this point can actually occur outside the triangle.
This isn't that difficult - just a little lengthy.
We need to find some slopes between these points and then write the equations for the lines representing the altitudes. Where these lines meet is the orthocenter of the triangle.
The slope of the base represented by the side with endpoints (2,3) and (7,3) is easy.... it's 0. So the equation of the altitude from the apex, (5,9), to this base is just x = 5
And the slope of the side with endpoints (5,9) and (2,3) = 2. And a perpendicular line to this side through point (7,3) will have the equation y - 3 = -(1/2)(x - 7) → y = (-1/2)x + 7/2 + 3 → y = (-1/2)x + 13/2
And when x = 5, y = (-1/2)(5)+ 13/2 = 4 and this is the y coordinate of the orthocenter.
Let's prove that the other altitude also intersects these two lines at y =4
The slope on the side with endpoints (5,9) and (7,3) = -3. And a perpendicular line to this side through (2,3) will have the equation y - 3 = (1/3)(x - 2) → y = (1/3)x - 2/3 + 3 →y= (1/3)x + 7/3
And, when x = 5, we have y = (1/3)(5) + 7/3 = 4....just as we expected......!!!
Then, the y coordinate of the orthocenter is 4 and the orthocenter point is (5,4)
BTW....I have a feeling that at least a couple of people on this site may know a shortcut "formula" for this.....Sorry, I have to do it by "brute force"........
+128460
The "orthocenter" of a triangle is defined as the point where the three altitudes intersect.....BTW....in some triangles, this point can actually occur outside the triangle.
This isn't that difficult - just a little lengthy.
We need to find some slopes between these points and then write the equations for the lines representing the altitudes. Where these lines meet is the orthocenter of the triangle.
The slope of the base represented by the side with endpoints (2,3) and (7,3) is easy.... it's 0. So the equation of the altitude from the apex, (5,9), to this base is just x = 5
And the slope of the side with endpoints (5,9) and (2,3) = 2. And a perpendicular line to this side through point (7,3) will have the equation y - 3 = -(1/2)(x - 7) → y = (-1/2)x + 7/2 + 3 → y = (-1/2)x + 13/2
And when x = 5, y = (-1/2)(5)+ 13/2 = 4 and this is the y coordinate of the orthocenter.
Let's prove that the other altitude also intersects these two lines at y =4
The slope on the side with endpoints (5,9) and (7,3) = -3. And a perpendicular line to this side through (2,3) will have the equation y - 3 = (1/3)(x - 2) → y = (1/3)x - 2/3 + 3 →y= (1/3)x + 7/3
And, when x = 5, we have y = (1/3)(5) + 7/3 = 4....just as we expected......!!!
Then, the y coordinate of the orthocenter is 4 and the orthocenter point is (5,4)
BTW....I have a feeling that at least a couple of people on this site may know a shortcut "formula" for this.....Sorry, I have to do it by "brute force"........
+118608 Thanks Chris, I just wanted to give a visual.
I have not marked right angles as that took too much space but the intenal intervals are all altitudes (have right angles)
I have not included all the working, just the main steps. 
1) Equation of EC is x=5
2) Gradient of AC = 2
3) Gradient of DB = -1/2
4) Equation of DB m=-1/2 B(7,3)
y=(-x+13)/2
5) On linterval DB When x=5 y=4
6) Intersection of DB and EC is (5,4)
Orthocentre is (5,4)
You could check this by finding the equation of AF and checking that (5,4) satisfies this equation.
