The orthocentre of the triangle whose vertice | vedclass

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(B) Let the vertices be $A(0, 0)$,$B(2, -1)$,and $C(1, 3)$.$1$. Find the equation of altitude from $A$ to $BC$:Slope of $BC = \frac{3 - (-1)}{1 - 2} =

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$\left( -\frac{4}{7}, -\frac{1}{7} \right)$

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(B) Let the vertices be $A(0, 0)$,$B(2, -1)$,and $C(1, 3)$.$1$. Find the equation of altitude from $A$ to $BC$:Slope of $BC = \frac{3 - (-1)}{1 - 2} = \frac{4}{-1} = -4$.The slope of the altitude $AD$ is the negative reciprocal of the slope of $BC$,which is $\frac{1}{4}$.The equation of $AD$ passing through $A(0, 0)$ is $y - 0 = \frac{1}{4}(x - 0)$,which simplifies to $x - 4y = 0$ ..... $(i)$.$2$. Find the equation of altitude from $B$ to $AC$:Slope of $AC = \frac{3 - 0}{1 - 0} = 3$.The slope of the altitude $BE$ is the negative reciprocal of the slope of $AC$,which is $-\frac{1}{3}$.The equation of $BE$ passing through $B(2, -1)$ is $y - (-1) = -\frac{1}{3}(x - 2)$,which simplifies to $3(y + 1) = -(x - 2)$ $\Rightarrow 3y + 3 = -x + 2$ $\Rightarrow x + 3y + 1 = 0$ ..... $(ii)$.$3$. Solve the system of equations $(i)$ and $(ii)$:From $(i)$,$x = 4y$. Substitute this into $(ii)$:$4y + 3y + 1 = 0$ $\Rightarrow 7y = -1$ $\Rightarrow y = -\frac{1}{7}$.Then $x = 4(-\frac{1}{7}) = -\frac{4}{7}$.Thus,the orthocentre is $\left( -\frac{4}{7}, -\frac{1}{7} \right)$.

The orthocentre of the triangle whose vertices are $(0, 0)$,$(2, -1)$,and $(1, 3)$ is

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