JEE MAIN - Mathematics (2023 - 1st February Morning Shift - No. 12)
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 :
$$x^2-20x+99=0$$
$$x^2-22x+120=0$$
$$x^2-19x+90=0$$
$$x^2-18x+80=0$$
Explanation
_1st_February_Morning_Shift_en_12_1.png)
$$ \mathrm{m}=-\frac{1}{2} $$
Here $\mathrm{m_BH} \times \mathrm{m_AC}=-1$
$$ \begin{array}{ll} & \left(\frac{\beta-3}{\alpha-2}\right)\left(\frac{1}{-2}\right)=-1 \\\\ & \beta-3=2 \alpha-4 \\\\ & \beta=2 \alpha-1 \\\\ & \mathrm{~m}_{\mathrm{AH}} \times \mathrm{m}_{\mathrm{BC}}=-1 \\\\ \Rightarrow & \left(\frac{\beta-2}{\alpha-1}\right)(-2)=-1 \\\\ \Rightarrow & 2 \beta-4=\alpha-1 \\\\ \Rightarrow & 2(2 \alpha-1)=\alpha+3 \\\\ \Rightarrow & 3 \alpha=5 \\\\ & \alpha=\frac{5}{3}, \beta=\frac{7}{3} \Rightarrow \mathrm{H}\left(\frac{5}{3}, \frac{7}{3}\right) \\\\ & \alpha+4 \beta=\frac{5}{3}+\frac{28}{3}=\frac{33}{3}=11 \\\\ & \beta+4 \alpha=\frac{7}{3}+\frac{20}{3}=\frac{27}{3}=9 \\\\ & \mathrm{x}^2-20 \mathrm{x}+99=0 \end{array} $$
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