If (alpha, beta) is the orthocentre of the tr | vedclass

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(A) Let the vertices be $A(2,2)$,$B(5,1)$,and $C(4,4)$.To find the orthocentre,we find the equations of two altitudes.Slope of $BC = \frac{4-1}{4-5} =

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$6$

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(A) Let the vertices be $A(2,2)$,$B(5,1)$,and $C(4,4)$.To find the orthocentre,we find the equations of two altitudes.Slope of $BC = \frac{4-1}{4-5} = \frac{3}{-1} = -3$.The altitude from $A(2,2)$ to $BC$ has slope $m_1 = -\frac{1}{-3} = \frac{1}{3}$.Equation of altitude from $A$: $y-2 = \frac{1}{3}(x-2)$ $\Rightarrow 3y-6 = x-2$ $\Rightarrow x-3y = -4 \quad \dots(i)$.Slope of $AC = \frac{4-2}{4-2} = \frac{2}{2} = 1$.The altitude from $B(5,1)$ to $AC$ has slope $m_2 = -\frac{1}{1} = -1$.Equation of altitude from $B$: $y-1 = -1(x-5)$ $\Rightarrow y-1 = -x+5$ $\Rightarrow x+y = 6 \quad \dots(ii)$.Adding $(i)$ and $(ii)$ multiplied by $3$: $(x-3y) + 3(x+y) = -4 + 18$ $\Rightarrow 4x = 14$ $\Rightarrow x = \frac{7}{2}$.Substituting $x = \frac{7}{2}$ in $(ii)$: $\frac{7}{2} + y = 6 \Rightarrow y = 6 - \frac{7}{2} = \frac{5}{2}$.Thus,the orthocentre $(\alpha, \beta) = (\frac{7}{2}, \frac{5}{2})$.Therefore,$\alpha+\beta = \frac{7}{2} + \frac{5}{2} = \frac{12}{2} = 6$.

If $(\alpha, \beta)$ is the orthocentre of the triangle with the vertices $(2,2), (5,1), (4,4)$,then $\alpha+\beta=$

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