Two vertices of a triangle are (4, -3) and (- | vedclass

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(B) Let the vertices of the triangle be $A(h, k)$,$B(4, -3)$,and $C(-2, 5)$. Let the orthocentre be $O(1, 2)$.$1$. The slope of $BC$ is $m_{BC} = \fra

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$(33, 26)$

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(B) Let the vertices of the triangle be $A(h, k)$,$B(4, -3)$,and $C(-2, 5)$. Let the orthocentre be $O(1, 2)$.$1$. The slope of $BC$ is $m_{BC} = \frac{5 - (-3)}{-2 - 4} = \frac{8}{-6} = -\frac{4}{3}$.Since the altitude $AO$ is perpendicular to $BC$,the slope of $AO$ is $m_{AO} = -\frac{1}{m_{BC}} = \frac{3}{4}$.The equation of line $AO$ passing through $A(h, k)$ and $O(1, 2)$ is $\frac{k - 2}{h - 1} = \frac{3}{4}$,which simplifies to $4(k - 2) = 3(h - 1) \Rightarrow 3h - 4k + 5 = 0$ ... $(i)$.$2$. The slope of $AC$ is $m_{AC} = \frac{k - 5}{h - (-2)} = \frac{k - 5}{h + 2}$.Since the altitude $BO$ is perpendicular to $AC$,the slope of $BO$ is $m_{BO} = \frac{2 - (-3)}{1 - 4} = \frac{5}{-3} = -\frac{5}{3}$.Therefore,$m_{AC} = -\frac{1}{m_{BO}} = \frac{3}{5}$.Equating the slopes: $\frac{k - 5}{h + 2} = \frac{3}{5}$ $\Rightarrow 5(k - 5) = 3(h + 2)$ $\Rightarrow 3h - 5k + 31 = 0$ ... $(ii)$.$3$. Solving equations $(i)$ and $(ii)$:Subtracting $(ii)$ from $(i)$: $(3h - 4k + 5) - (3h - 5k + 31) = 0$ $\Rightarrow k - 26 = 0$ $\Rightarrow k = 26$.Substituting $k = 26$ into $(i)$: $3h - 4(26) + 5 = 0$ $\Rightarrow 3h - 104 + 5 = 0$ $\Rightarrow 3h = 99$ $\Rightarrow h = 33$.Thus,the third vertex is $(33, 26)$.

Two vertices of a triangle are $(4, -3)$ and $(-2, 5)$. If the orthocentre of the triangle is at $(1, 2)$,then the third vertex is

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