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

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(C) Let the vertices of the triangle be $A(5, -1)$,$B(-2, 3)$,and $C(x, y)$. The orthocenter $H$ is $(0, 0)$.Slope of $AH = \frac{-1-0}{5-0} = -\frac{

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$(-4, -7)$

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(C) Let the vertices of the triangle be $A(5, -1)$,$B(-2, 3)$,and $C(x, y)$. The orthocenter $H$ is $(0, 0)$.Slope of $AH = \frac{-1-0}{5-0} = -\frac{1}{5}$.Since $BC \perp AH$,the slope of $BC$ is $5$.The equation of line $BC$ passing through $B(-2, 3)$ is $y - 3 = 5(x + 2) \Rightarrow y = 5x + 13$ (Equation $i$).Slope of $BH = \frac{3-0}{-2-0} = -\frac{3}{2}$.Since $AC \perp BH$,the slope of $AC$ is $\frac{2}{3}$.The equation of line $AC$ passing through $A(5, -1)$ is $y + 1 = \frac{2}{3}(x - 5)$ $\Rightarrow 3y + 3 = 2x - 10$ $\Rightarrow 3y = 2x - 13$ (Equation $ii$).Substitute $y = 5x + 13$ into Equation $ii$:$3(5x + 13) = 2x - 13$$15x + 39 = 2x - 13$$13x = -52 \Rightarrow x = -4$.Substitute $x = -4$ into Equation $i$:$y = 5(-4) + 13 = -20 + 13 = -7$.Thus,the third vertex is $(-4, -7)$.

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

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