Two vertices of a triangle are (5, -1) and (- | vedclass

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(A) Let the third vertex be $A(\alpha, \beta)$. Let $B = (5, -1)$ and $C = (-2, 3)$. The orthocenter $O$ is $(0, 0)$.The slope of $BC = \frac{3 - (-1)

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$(-4, -7)$

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(A) Let the third vertex be $A(\alpha, \beta)$. Let $B = (5, -1)$ and $C = (-2, 3)$. The orthocenter $O$ is $(0, 0)$.The slope of $BC = \frac{3 - (-1)}{-2 - 5} = \frac{4}{-7} = -\frac{4}{7}$.Since $AD \perp BC$,the slope of $AD$ is the negative reciprocal of the slope of $BC$,which is $\frac{7}{4}$.Since $AD$ passes through $A(\alpha, \beta)$ and $O(0, 0)$,the slope of $AO$ is $\frac{\beta - 0}{\alpha - 0} = \frac{\beta}{\alpha}$.Thus,$\frac{\beta}{\alpha} = \frac{7}{4} \Rightarrow 7\alpha = 4\beta$  --- $(1)$.The slope of $AC = \frac{3 - \beta}{-2 - \alpha} = \frac{\beta - 3}{\alpha + 2}$.Since $BE \perp AC$,the slope of $BE$ is the negative reciprocal of the slope of $AC$.The slope of $BO = \frac{0 - (-1)}{0 - 5} = \frac{1}{-5} = -\frac{1}{5}$.So,$\frac{\beta - 3}{\alpha + 2} 	imes (-5) = -1$ $\Rightarrow \frac{\beta - 3}{\alpha + 2} = \frac{1}{5}$ $\Rightarrow 5\beta - 15 = \alpha + 2$ $\Rightarrow \alpha - 5\beta = -17$  --- $(2)$.From $(1)$,$\beta = \frac{7\alpha}{4}$. Substituting into $(2)$:$\alpha - 5(\frac{7\alpha}{4}) = -17$ $\Rightarrow 4\alpha - 35\alpha = -68$ $\Rightarrow -31\alpha = -68$ $\Rightarrow \alpha = \frac{68}{31}$.Wait,let us re-evaluate the slopes.Slope of $AO = \frac{\beta}{\alpha}$. Slope of $BC = -\frac{4}{7}$. $AO \perp BC$ $\Rightarrow \frac{\beta}{\alpha} 	imes (-\frac{4}{7}) = -1$ $\Rightarrow 4\beta = 7\alpha$.Slope of $BO = \frac{0 - (-1)}{0 - 5} = -\frac{1}{5}$. Slope of $AC = \frac{3 - \beta}{-2 - \alpha}$. $BO \perp AC$ $\Rightarrow (-\frac{1}{5}) 	imes \frac{3 - \beta}{-2 - \alpha} = -1$ $\Rightarrow \frac{3 - \beta}{2 + \alpha} = 5$ $\Rightarrow 3 - \beta = 10 + 5\alpha$ $\Rightarrow 5\alpha + \beta = -7$.Substitute $\beta = \frac{7\alpha}{4}$ into $5\alpha + \beta = -7$:$5\alpha + \frac{7\alpha}{4} = -7$ $\Rightarrow 20\alpha + 7\alpha = -28$ $\Rightarrow 27\alpha = -28$ $\Rightarrow \alpha = -\frac{28}{27}$.Re-checking the calculation: $B(5, -1), C(-2, 3), O(0, 0)$.Slope $BC = \frac{3 - (-1)}{-2 - 5} = \frac{4}{-7}$. Altitude from $A$ to $BC$ has slope $\frac{7}{4}$. Line $AO$ passes through $(0,0)$ and $A(\alpha, \beta)$,so $\frac{\beta}{\alpha} = \frac{7}{4} \Rightarrow 4\beta = 7\alpha$.Slope $BO = \frac{0 - (-1)}{0 - 5} = -\frac{1}{5}$. Altitude from $B$ to $AC$ has slope $5$. Line $AC$ passes through $C(-2, 3)$ and $A(\alpha, \beta)$,so $\frac{\beta - 3}{\alpha - (-2)} = 5$ $\Rightarrow \beta - 3 = 5\alpha + 10$ $\Rightarrow \beta - 5\alpha = 13$.Substitute $\beta = \frac{7\alpha}{4}$ into $\beta - 5\alpha = 13$:$\frac{7\alpha}{4} - 5\alpha = 13$ $\Rightarrow 7\alpha - 20\alpha = 52$ $\Rightarrow -13\alpha = 52$ $\Rightarrow \alpha = -4$.Then $\beta = \frac{7(-4)}{4} = -7$.The third vertex is $(-4, -7)$.

Two vertices of a triangle are $(5, -1)$ and $(-2, 3)$. If the orthocenter of the triangle is the origin $(0, 0)$,find the coordinates of the third vertex.

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