If a triangle ABC has vertices A(1, -6), B(2, | vedclass

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(A) Step $1$: Find the slope of side $BC$. $m_{BC} = \frac{-2 - (-3)}{3 - 2} = \frac{1}{1} = 1$.Step $2$: Find the equation of the altitude from $A$ t

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$(-6, 1)$

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(A) Step $1$: Find the slope of side $BC$. $m_{BC} = \frac{-2 - (-3)}{3 - 2} = \frac{1}{1} = 1$.Step $2$: Find the equation of the altitude from $A$ to $BC$. The slope of the altitude is $m_1 = -\frac{1}{m_{BC}} = -1$. The equation is $y - (-6) = -1(x - 1) \implies y + 6 = -x + 1 \implies x + y = -5$.Step $3$: Find the slope of side $AC$. $m_{AC} = \frac{-2 - (-6)}{3 - 1} = \frac{4}{2} = 2$.Step $4$: Find the equation of the altitude from $B$ to $AC$. The slope of the altitude is $m_2 = -\frac{1}{m_{AC}} = -\frac{1}{2}$. The equation is $y - (-3) = -\frac{1}{2}(x - 2) \implies 2y + 6 = -x + 2 \implies x + 2y = -4$.Step $5$: Solve the system of equations $x + y = -5$ and $x + 2y = -4$. Subtracting the first from the second: $(x + 2y) - (x + y) = -4 - (-5) \implies y = 1$. Substituting $y = 1$ into $x + y = -5$, we get $x + 1 = -5 \implies x = -6$. The orthocenter is $(-6, 1)$.

If a triangle $ABC$ has vertices $A(1, -6)$, $B(2, -3)$ and $C(3, -2)$, then the coordinates of its orthocenter are....

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