Find the coordinates of the orthocenter for a triangle with vertices (1,3) , (1,12) , (8,10)

For convenience, I'll name the three points: A(1, 3), B(1,12), C(8,10)

The orthocenter is where the three altitudes of the triangle meet. That is, where the perpendicular lines from each vertex to the opposite side meet.

the goal in this case is to find out where the perpendicular line from C to AB, the perpendicular line from B to AC, and the perpendicular line form A to BC meet.
It is enough to find two perpendicular lines.

step one: find the equation of two perpendicular lines

find the slope of AB: m=(12-3)/(1-1), notice the denominator is 0, so the slope doesn't exist, that means AB is a vertical line parallel to the y axis, so the perpendicular line has a slope of o, a horizontal line. this horizontal line goes through C(8,10), so the equation is y=10
use the same method, slope of AC: (10-3)/(8-1)=1, so the perpendicular line from B to AC has a slope of -1. y=-x+b
Now use the point B(1,12) to find b: 12=-1+b =>b=13

so the equation for this line is y=-x+13

Step two: solve the system of equations: y=10
                                                                  y=-x+13
we get x=3, y=10, which are the coordinates of the orthocenter.

I've also found the equation of the third line, which is y=(7/2)x-1/2, the point (3,10) is also on this line. So it works!

None liner bro because