If the orthocentre of the triangle, whose vertices are (1,2),(2,3) and (3,1) is (,), then …

Question

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:

Accepted Answer

$$
\mathrm{m}=-\frac{1}{2}
$$

Here $\mathrm{m_BH} \times \mathrm{m_AC}=-1$

$$
\[\begin{array}{ll} \left(\frac{\beta-3}{\alpha-2}\right)\left(\frac{1}{-2}\right)=-1 \  \ \beta-3=2 \alpha-4 \  \ \beta=2 \alpha-1 \  \ \mathrm{~m}_{\mathrm{AH}} \times \mathrm{m}_{\mathrm{BC}}=-1 \  \ \Rightarrow  & \left(\frac{\beta-2}{\alpha-1}\right)(-2)=-1 \  \ \Rightarrow  & 2 \beta-4=\alpha-1 \  \ \Rightarrow  & 2(2 \alpha-1)=\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 \  \ \mathrm{x}^2-20 \mathrm{x}+99=0 \end{array}$$
\]

Solutions · 1