The orthocentre of the triangle with vertices | vedclass

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(C) Let $A(-2, -6)$,$B(-2, 4)$,and $C(1, 3)$ be the vertices of the triangle.Slope of $BC = \frac{3 - 4}{1 - (-2)} = \frac{-1}{3}$.The slope of the al

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$(1, 3)$

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(C) Let $A(-2, -6)$,$B(-2, 4)$,and $C(1, 3)$ be the vertices of the triangle.Slope of $BC = \frac{3 - 4}{1 - (-2)} = \frac{-1}{3}$.The slope of the altitude through $A$ is the negative reciprocal of the slope of $BC$,which is $3$.The equation of the altitude through $A(-2, -6)$ is $y - (-6) = 3(x - (-2))$,which simplifies to $y + 6 = 3x + 6$,or $y = 3x$ $(i)$.Slope of $AC = \frac{3 - (-6)}{1 - (-2)} = \frac{9}{3} = 3$.The slope of the altitude through $B$ is the negative reciprocal of the slope of $AC$,which is $-\frac{1}{3}$.The equation of the altitude through $B(-2, 4)$ is $y - 4 = -\frac{1}{3}(x - (-2))$,which simplifies to $3y - 12 = -x - 2$,or $x + 3y = 10$ $(ii)$.Substituting $y = 3x$ from $(i)$ into $(ii)$: $x + 3(3x) = 10 \implies 10x = 10 \implies x = 1$.Then $y = 3(1) = 3$.Thus,the orthocentre is $(1, 3)$.

The orthocentre of the triangle with vertices $(-2, -6)$,$(-2, 4)$ and $(1, 3)$ is

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