If a triangle ABC has vertices A(-1, 7),B(-7, | vedclass

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(C) Let the orthocentre be $H(x_0, y_0)$.The slope of $BC$ is $m_{BC} = \frac{-5-1}{5-(-7)} = \frac{-6}{12} = -\frac{1}{2}$.Since $AH \perp BC$,the sl

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$(-3, 3)$

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(C) Let the orthocentre be $H(x_0, y_0)$.The slope of $BC$ is $m_{BC} = \frac{-5-1}{5-(-7)} = \frac{-6}{12} = -\frac{1}{2}$.Since $AH \perp BC$,the slope of $AH$ is $m_{AH} = -\frac{1}{m_{BC}} = 2$.The equation of altitude $AH$ passing through $A(-1, 7)$ is $y - 7 = 2(x + 1)$,which simplifies to $2x - y + 9 = 0$ ... $(1)$.The slope of $AC$ is $m_{AC} = \frac{-5-7}{5-(-1)} = \frac{-12}{6} = -2$.Since $BH \perp AC$,the slope of $BH$ is $m_{BH} = -\frac{1}{m_{AC}} = \frac{1}{2}$.The equation of altitude $BH$ passing through $B(-7, 1)$ is $y - 1 = \frac{1}{2}(x + 7)$,which simplifies to $x - 2y + 9 = 0$ ... $(2)$.Solving equations $(1)$ and $(2)$:From $(1)$,$y = 2x + 9$. Substituting into $(2)$:$x - 2(2x + 9) + 9 = 0$$x - 4x - 18 + 9 = 0$$-3x - 9 = 0 \Rightarrow x = -3$.Then $y = 2(-3) + 9 = 3$.Thus,the orthocentre is $(-3, 3)$.

If a $\triangle ABC$ has vertices $A(-1, 7)$,$B(-7, 1)$ and $C(5, -5)$,then its orthocentre has coordinates

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