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

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(D) Let the vertices be $A(5, -1)$,$B(-2, 3)$,and $C(h, k)$. The orthocentre $H$ is $(0, 0)$.Since $CH \perp AB$,the slope of $AB$ is $m_{AB} = \frac{

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$(-4, -7)$

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(D) Let the vertices be $A(5, -1)$,$B(-2, 3)$,and $C(h, k)$. The orthocentre $H$ is $(0, 0)$.Since $CH \perp AB$,the slope of $AB$ is $m_{AB} = \frac{3 - (-1)}{-2 - 5} = \frac{4}{-7} = -\frac{4}{7}$.The slope of the altitude $CH$ is $m_{CH} = -\frac{1}{m_{AB}} = \frac{7}{4}$.Since $CH$ passes through $(0, 0)$ and $(h, k)$,its slope is $\frac{k}{h} = \frac{7}{4}$,which implies $7h - 4k = 0$  ---$(1)$.Similarly,since $AH \perp BC$,the slope of $BC$ is $m_{BC} = \frac{k - 3}{h - (-2)} = \frac{k - 3}{h + 2}$.The slope of the altitude $AH$ is $m_{AH} = \frac{0 - (-1)}{0 - 5} = -\frac{1}{5}$.Since $AH \perp BC$,$m_{BC} 	imes m_{AH} = -1$,so $\left(\frac{k - 3}{h + 2}ight) 	imes \left(-\frac{1}{5}ight) = -1$.$\frac{k - 3}{h + 2} = 5$ $\Rightarrow k - 3 = 5h + 10$ $\Rightarrow 5h - k + 13 = 0$ ---$(2)$.Solving equations $(1)$ and $(2)$: From $(1)$,$k = \frac{7h}{4}$. Substituting into $(2)$: $5h - \frac{7h}{4} + 13 = 0$.$\frac{20h - 7h}{4} + 13 = 0$ $\Rightarrow 13h = -52$ $\Rightarrow h = -4$.Then $k = \frac{7(-4)}{4} = -7$.Thus,the third vertex is $(-4, -7)$.

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

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