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(D) Let the vertices be $A(1,2), B(2,3)$ and $C(3,1)$.Let the orthocentre be $H(\alpha, \beta)$.The slope of $AC$ is $m_{AC} = \frac{1-2}{3-1} = -\fra

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$x^2-20x+99=0$

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(D) Let the vertices be $A(1,2), B(2,3)$ and $C(3,1)$.Let the orthocentre be $H(\alpha, \beta)$.The slope of $AC$ is $m_{AC} = \frac{1-2}{3-1} = -\frac{1}{2}$.Since $BH \perp AC$,the slope of $BH$ is $m_{BH} = 2$. Thus,$\frac{\beta-3}{\alpha-2} = 2$ $\Rightarrow \beta-3 = 2\alpha-4$ $\Rightarrow \beta = 2\alpha-1$.The slope of $AB$ is $m_{AB} = \frac{3-2}{2-1} = 1$.Since $CH \perp AB$,the slope of $CH$ is $m_{CH} = -1$. Thus,$\frac{\beta-1}{\alpha-3} = -1$ $\Rightarrow \beta-1 = -\alpha+3$ $\Rightarrow \beta = -\alpha+4$.Equating the two expressions for $\beta$: $2\alpha-1 = -\alpha+4$ $\Rightarrow 3\alpha = 5$ $\Rightarrow \alpha = \frac{5}{3}$.Then $\beta = 2(\frac{5}{3})-1 = \frac{7}{3}$.The roots of the quadratic equation are $p = \alpha+4\beta = \frac{5}{3} + \frac{28}{3} = 11$ and $q = 4\alpha+\beta = \frac{20}{3} + \frac{7}{3} = 9$.The quadratic equation is $(x-p)(x-q) = 0$ $\Rightarrow (x-11)(x-9) = 0$ $\Rightarrow x^2-20x+99 = 0$.

If the orthocentre of the triangle,whose vertices are $(1,2), (2,3)$ and $(3,1)$ is $(\alpha, \beta)$,then the quadratic equation whose roots are $\alpha+4\beta$ and $4\alpha+\beta$ is:

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