The vertices of a triangle are at -hat{i}+3 h | vedclass

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(B) Let the vertices be $A=(-1, 3)$,$B=(2, 5)$,and $C=(a, b)$. The orthocenter is $H=(1, 2)$.Since $AH \perp BC$,the slope of $AH$ $(m_{AH})$ multipli

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$\left(\frac{5}{7}, \frac{17}{7}\right)$

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(B) Let the vertices be $A=(-1, 3)$,$B=(2, 5)$,and $C=(a, b)$. The orthocenter is $H=(1, 2)$.Since $AH \perp BC$,the slope of $AH$ $(m_{AH})$ multiplied by the slope of $BC$ $(m_{BC})$ is $-1$.$m_{AH} = \frac{2-3}{1-(-1)} = \frac{-1}{2}$.$m_{BC} = \frac{b-5}{a-2}$.Since $m_{AH} 	imes m_{BC} = -1$,we have $\left(\frac{-1}{2}ight) 	imes \left(\frac{b-5}{a-2}ight) = -1 \Rightarrow b-5 = 2(a-2) \Rightarrow b-5 = 2a-4 \Rightarrow 2a-b = -1$ ... $(i)$Similarly,since $BH \perp AC$,the slope of $BH$ $(m_{BH})$ multiplied by the slope of $AC$ $(m_{AC})$ is $-1$.$m_{BH} = \frac{2-5}{1-2} = \frac{-3}{-1} = 3$.$m_{AC} = \frac{b-3}{a-(-1)} = \frac{b-3}{a+1}$.Since $m_{BH} 	imes m_{AC} = -1$,we have $3 	imes \left(\frac{b-3}{a+1}ight) = -1 \Rightarrow 3b-9 = -a-1 \Rightarrow a+3b = 8$ ... (ii)Solving equations $(i)$ and (ii):From $(i)$,$b = 2a+1$. Substituting into (ii): $a + 3(2a+1) = 8 \Rightarrow a + 6a + 3 = 8 \Rightarrow 7a = 5 \Rightarrow a = \frac{5}{7}$.Then $b = 2(\frac{5}{7}) + 1 = \frac{10}{7} + \frac{7}{7} = \frac{17}{7}$.Thus,$(a, b) = \left(\frac{5}{7}, \frac{17}{7}ight)$.

The vertices of a triangle are at $-\hat{i}+3 \hat{j}$ and $2 \hat{i}+5 \hat{j}$ and its orthocenter is at $\hat{i}+2 \hat{j}$. If the position vector of the third vertex is $a \hat{i}+b \hat{j}$,then $(a, b)=$

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