The orthocentre of the triangle formed by the | vedclass

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(D) Let the vertices of the triangle be $A(1,3), B(-3,5)$,and $C(5,-1)$.Slope of $AC = \frac{-1-3}{5-1} = \frac{-4}{4} = -1$.The altitude from $B$ to

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$(19,27)$

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(D) Let the vertices of the triangle be $A(1,3), B(-3,5)$,and $C(5,-1)$.Slope of $AC = \frac{-1-3}{5-1} = \frac{-4}{4} = -1$.The altitude from $B$ to $AC$ is perpendicular to $AC$. Let its slope be $m_1$. Since $m_1 	imes (-1) = -1$,$m_1 = 1$.The equation of the altitude from $B(-3,5)$ is $(y-5) = 1(x+3) \Rightarrow x-y = -8$ (Equation $i$).Slope of $BC = \frac{-1-5}{5-(-3)} = \frac{-6}{8} = -\frac{3}{4}$.The altitude from $A$ to $BC$ is perpendicular to $BC$. Let its slope be $m_2$. Since $m_2 	imes (-\frac{3}{4}) = -1$,$m_2 = \frac{4}{3}$.The equation of the altitude from $A(1,3)$ is $(y-3) = \frac{4}{3}(x-1)$ $\Rightarrow 3y-9 = 4x-4$ $\Rightarrow 4x-3y = -5$ (Equation $ii$).To find the orthocentre,solve the system of equations:$x-y = -8 \Rightarrow 3x-3y = -24$ (Equation $iii$).Subtracting (ii) from (iii): $(3x-3y) - (4x-3y) = -24 - (-5)$ $\Rightarrow -x = -19$ $\Rightarrow x = 19$.Substituting $x=19$ into $(i)$: $19-y = -8 \Rightarrow y = 27$.Thus,the orthocentre is $(19,27)$.

The orthocentre of the triangle formed by the points $(1,3), (-3,5)$ and $(5,-1)$ is

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