Let us take an example.

Find the orthocenter of the triangle having coordinates are: A (3, 1)  B(2, 2) C (3, 5)

Step 1:  Find equations of the line segments AB and BC.

To find any line segment, you will need to find the slope of the line and then the corresponding y-intercept.

So, Slope of AB = (1-2)/(3-2) = -1/1 = -1

y = mx + b (substitute m = -1, x = 3, y = 1)

1 = -1*3 + b

=> 1 = -3 + b

=> b = 4

Now, Equation of AB:  y = -1*x + 4

=> y = -x + 4

Slope of BC = (2-5)/(2-3) = -3/-1 = 3

y = mx + b (substitute m = 3, x = 2, y = 2)

2 = 3*2 + b

=> 2 = 6 + b

=> b = 2 - 6

=> b = -4

Equation of BC:  y = 3x - 4 

Step 2:  Find the slope of the corresponding perpendicular lines

Slope of AB = -1

Slope of perpendicular line to AB:  -1*m = -1

=> m = 1

Slope of BC = 3

Slope of perpendicular line to BC:  3*m = -1

=> m = -1/3

Step 3:  Find the equation of the perpendicular lines

Slope of perpendicular line to AB:  m = 1

We will use the coordinate of the opposite vertex (point C) to find the equation of the line.

y = mx + b (substitute m = 1, x = 3, y = 5)

5 = 1*3 + b

=> 5 = 3 + b

=> b = 5 - 3

=> b = 2

Equation of perpendicular line to AB:  y = 1*x + 2

=> y = x + 2

Slope of perpendicular line to BC:  m = -1/3

We will use the coordinate of the opposite vertex (point A) to find the equation of the line.

y = mx + b (substitute m = -1/3, x = 3, y = 1)

1 = -1/3*(3) + b

=> 1 = -1 + b

=> b = 2

Equation of perpendicular line to AB:  y = -x/3 + 2

Step 4:  solve 2 perpendicular lines

equation 1:  y = x + 2

equation 2:  y = -x/3 + 2

Solving for x and y:

x + 2 = -x/3 + 2

=> x + x/3 = 2 - 2

=> 4x/3 = 0

=> x = 0

Now, y = 1*0 + 2

=> y = 2

Now, the coordinates are (0, 2)

This is the orthocenter of the triangle.