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The altitudes of a triangle meet in a point named orthocenter.
The altitude of triangle is the line that starts from one vertex and it is perpendicular to the opposite line.
The altitudes of the triangle are: AD, BE, CF.
You need to find the equations of two altitudes, at least. Then you need to compute the solution of the system of simultaneous equations of altitudes. This solution is the orthocenter.
You may find the equation of altitude AD using the slope intercept form.
Use the fact that the altitude is perpendicular to the opposite line.
AD`_|_` BC => `m_(BC)=-1/m_(AD)`
Find the slope of the side BC: `m_(BC)=(y_C-y_B)/(x_C-x_B)`
Equation of the altitude AD: `y-y_A=m_(AD)(x-x_A)`
`y-2=x-1` => `y=x+1`
Equation of the altitude BE: `y-y_B=m_(BE)(x-x_B)`
Use the fact that the altitude BE is perpendicular to the opposite line, AC.
Find `m_(AC)=(y_C-y_A)/(x_C-x_A)=(4-2)/(7-1)=2/6=1/3` => `m_(BE)=-1/(1/3)=-3`
Equation of the altitude BE: `y-8=-3*(x-3)`
`y=-3x+9+8` => `y=-3x+17`
Solve the system of simultaneous equations AD and BE.
`x+1=-3x+17` => `x+3x=17-1` => `4x=16` =>`x=4`
`y=4+1` => `y=5`
Answer: The orthocenter H is: H(4,5).
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