Equation of line BD is
\(y + 1 = \cfrac{-1}{\frac{5 + 1}{2 + 2}}(x - 6)\)
(∵ slope of line BD \(=\frac{-1}{\text{slope of line AC}}\))
y + 1 = -4/6 (x - 6)
y + 1 = -2/3 (x - 6)
\(\Rightarrow\) 3y + 3 = -2x + 12
\(\Rightarrow\) 2x + 3y = 9 .....(1)
Now, equation of line AF is
\(y + 1 = \cfrac{-1}{\frac{5 + 1}{2 - 6}}(x + 2)\) (slope of line AF \(=\frac{-1}{\text{slope of line BC}}\))
\(\Rightarrow\) y + 1 = 4/6 (x + 2)
\(\Rightarrow\) 3y + 3 = 2x + 4
\(\Rightarrow\) -2x + 3y = 1 ....(2)
Adding equation (1) and (2), we get
6y = 10
\(\Rightarrow\) y = 10/6 = 5/3
By putting y = 5/3 in equation (1), we get
2x = 9 - 5 = 4
\(\Rightarrow\) x = 4/2 = 2
Hence, orthocenter of the triangle is (2, 5/3).