Mathematics - Point Question with Solution | TestHub

We have,

Here $m_{BH}\times m_{AC}=-1$

$\Rightarrow \left(\frac{\beta -3}{\alpha -2}\right)\left(\frac{1}{-2}\right)=-1$

$\Rightarrow \beta -3=2\alpha -4$

$\Rightarrow \beta =2\alpha -1$        ....(1)

$m_{AH}\times m_{BC}=-1$

$\Rightarrow \left(\frac{\beta -2}{\alpha -1}\right)\left(-2\right)=-1$

$\Rightarrow 2\beta -4=\alpha -1$

Using (1), we can write:

$\Rightarrow 2(2\alpha -1)-4=\alpha -1$

$\Rightarrow 2\left(2\alpha -1\right)=\alpha +3$

$\Rightarrow 3\alpha =5$

$\alpha =\frac{5}{3}, \beta =\frac{7}{3}\Rightarrow \mathrm{H}\left(\frac{5}{3},\frac{7}{3}\right)$

$\alpha +4\beta =\frac{5}{3}+\frac{28}{3}=\frac{33}{3}=11$

$\beta +4\alpha =\frac{7}{3}+\frac{20}{3}=\frac{27}{3}=9$

Required equation is

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