Equation of line AD is
y + 2 \(=\frac{-1}{\text{Slope of line BC}} (x - 5)\)
\(\Rightarrow\) y + 2 \(=\cfrac{-1}{\frac{4 - 2}{1 + 1}}(x - 5)\)
\(\Rightarrow\) y + 2 = -2/2 (x - 5)
\(\Rightarrow\) x + y = 5 - 2
\(\Rightarrow\) x + y = 3 ....(1)
Equation of line CF is
y - 4 \(=\frac{-1}{\text{Slope of line AB}} (x - 1)\)
\(\Rightarrow\) y - 4 \(=\cfrac{-1}{\frac{2 + 2}{1 + 1}}(x - 1)\)
\(\Rightarrow\) y - 4 = 6/4 (x - 1)
\(\Rightarrow\) 2y - 8 = 3x - 3
\(\Rightarrow\) -3x + 2y = 8 - 3
\(\Rightarrow\) -3x + 2y = 5 ......(2)
Equation (1) × 3 + Equation (2), we get
5y = 14
\(\Rightarrow\) y = 14/5
put y = 14/5 in equation (1), we get
x = 3 - 14/5 = 1/5.
Hence, orthocenter of the triangle is (1/5, 14/5).