(-2, -1) and (2, 5) are two vertices of a tri | vedclass

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(C) Let the vertices be $A(-2, -1)$,$B(2, 5)$,and $C(m, n)$. Let $H\left(2, \frac{5}{3}\right)$ be the orthocenter.Slope of $AH = \frac{\frac{5}{3} -

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$5$

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(C) Let the vertices be $A(-2, -1)$,$B(2, 5)$,and $C(m, n)$. Let $H\left(2, \frac{5}{3}\right)$ be the orthocenter.Slope of $AH = \frac{\frac{5}{3} - (-1)}{2 - (-2)} = \frac{8/3}{4} = \frac{2}{3}$.Since $AH \perp BC$,the slope of $BC = -\frac{1}{2/3} = -\frac{3}{2}$.Equation of $BC$: $y - 5 = -\frac{3}{2}(x - 2)$ $\Rightarrow 2y - 10 = -3x + 6$ $\Rightarrow 3x + 2y = 16 \quad ...(i)$Slope of $BH = \frac{\frac{5}{3} - 5}{2 - 2} = \frac{-10/3}{0}$,which is undefined. This means $BH$ is a vertical line $x = 2$.Since $BH \perp AC$,$AC$ must be a horizontal line. Since $A$ is $(-2, -1)$,the equation of $AC$ is $y = -1$.Since $C(m, n)$ lies on $AC$,$n = -1$.Substitute $n = -1$ into equation $(i)$: $3m + 2(-1) = 16$ $\Rightarrow 3m = 18$ $\Rightarrow m = 6$.Thus,the third vertex is $(6, -1)$.Therefore,$m + n = 6 + (-1) = 5$.

$(-2, -1)$ and $(2, 5)$ are two vertices of a triangle and $\left(2, \frac{5}{3}\right)$ is its orthocenter. If $(m, n)$ is the third vertex of that triangle,then $m+n=$

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