Two vertices of a triangle are (5, -1) and (- | vedclass

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(C) Let the vertices of the triangle be $A(h, k)$,$B(5, -1)$,and $C(-2, 3)$. The origin $O(0, 0)$ is the orthocentre. Since $AO \perp BC$,the slope of

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$(-4, -7)$

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(C) Let the vertices of the triangle be $A(h, k)$,$B(5, -1)$,and $C(-2, 3)$. The origin $O(0, 0)$ is the orthocentre. Since $AO \perp BC$,the slope of $AO 	imes$ slope of $BC = -1$. Slope of $BC = \frac{3 - (-1)}{-2 - 5} = \frac{4}{-7}$. Slope of $AO = \frac{k - 0}{h - 0} = \frac{k}{h}$. Thus,$\frac{k}{h} 	imes \left(-\frac{4}{7}ight) = -1$ $\Rightarrow \frac{k}{h} = \frac{7}{4}$ $\Rightarrow 7h - 4k = 0$ (Eq. $1$). Since $BO \perp AC$,the slope of $BO 	imes$ slope of $AC = -1$. Slope of $BO = \frac{-1 - 0}{5 - 0} = -\frac{1}{5}$. Slope of $AC = \frac{k - 3}{h - (-2)} = \frac{k - 3}{h + 2}$. Thus,$\left(-\frac{1}{5}ight) 	imes \left(\frac{k - 3}{h + 2}ight) = -1$ $\Rightarrow k - 3 = 5(h + 2)$ $\Rightarrow 5h - k + 13 = 0$ (Eq. $2$). From Eq. $1$,$k = \frac{7h}{4}$. Substituting this into Eq. $2$: $5h - \frac{7h}{4} + 13 = 0$ $\Rightarrow \frac{20h - 7h}{4} = -13$ $\Rightarrow 13h = -52$ $\Rightarrow h = -4$. Then $k = \frac{7(-4)}{4} = -7$. Therefore,the third vertex is $(-4, -7)$.

Two vertices of a triangle are $(5, -1)$ and $(-2, 3)$. If the origin $(0, 0)$ is the orthocentre of this triangle,then the coordinates of the third vertex of that triangle are

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