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

• A. $x^2-19 x+90=0$
• B. $x^2-20 x+99=0$
• C. $x^2-18 x+80=0$
• D. $x^2-22 x+120=0$

Answer 1

Answer: (B)

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

Answer 2

The correct answer is option (B) : x 2−20 x +99=0

$(\frac{β−3}{α−2})(\frac{1}{−2})=−1$
$β−3=2α−4$
$β=2α−1$
$m_{AH}\times m_{BC}=−1$
$⇒(\frac{β−2}{α−1})(−2)=−1$
$⇒2β−4=α−1$
$⇒2(2α−1)=α+3$
$⇒3α=5$
$α=\frac{5}{3}, β=\frac{7}{3} ​$
$⇒H(\frac{5}{3}, \frac{7}{3})$
$α+4β= \frac{5}{3} + \frac{28}{3} ​= \frac{33}{3} ​=11$
$β+4α= \frac{7}{3} + \frac{20}{3} = \frac{27}{3} = 9$
$x^2 −20x+99=0$