Let the orthocenter be H(h,k).
Slope of `BC =(2-1)/(5-(-2))=(1)/(7)`
Slope of `AC =(2-0)/(5-1)=(1)/(2)`
`AHbotBC`
`therefore` Slope of `AH=-7`
or `=(k-0)/(h-1)=-7`
`therefore -7h+7=k` or `7h+k=7`
Also, `BHbotAC `
`therefore` Slope of `BH=-2`
or `=(k-0)/(h+2)=-2`
`therefore k-1 =-2h-4` or ` 2h+k=-3`
Solving (1) and (2), we get `h=2` and `k=-7`. Hence, the orthocenter is `(2,-7)`