If the co-ordinates of orthocenter of the triangle formed by the straight linesL1:y=m1x+2m1 L2:y=m2x+2m2L3:y=m3x+2m3 As sides be (α,β) where m1,m2,m3 are the roots of the cubic equation x3−9x2+26x−24=0, then 4β+α=______

Orthocentre is (−2,2(1m1+1m2+1m3+1m1m2m3)) =(−2,2(2624+124))=(−294) So 4β+α=9−2=7

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If the co-ordinates of orthocenter of the triangle formed by the straight lines
$L_{1} : y = m_{1} x + \frac{2}{m_{1}}$
$L_{2} : y = m_{2} x + \frac{2}{m_{2}}$
$L_{3} : y = m_{3} x + \frac{2}{m_{3}}$
As sides be $(α, β)$ where $m_{1}, m_{2}, m_{3}$ are the roots of the cubic equation $x^{3} - 9 x^{2} + 26 x - 24 = 0,$ then $4 β + α =______$

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answer is 7.

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Detailed Solution

$Orthocentre is (- 2, 2 (\frac{1}{m_{1}} + \frac{1}{m_{2}} + \frac{1}{m_{3}} + \frac{1}{m_{1} m_{2} m_{3}}))$
$= (- 2, 2 (\frac{26}{24} + \frac{1}{24})) = (- 2 \frac{9}{4})$
So $4 β + α = 9 - 2 = 7$

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