Straight Line Question 230
Question: Let $ A( \alpha ,\frac{1}{\alpha } ),B( \alpha ,\frac{1}{\beta } ),C( \gamma ,\frac{1}{\gamma } ) $ be the vertices of a $ \Delta ABC, $ where $ \alpha ,\beta $ are the roots of the equation $ x^{2}-6p_1x+2=0,\beta ,\gamma $ $ x^{2}-6p_1x+2=0,\beta ,\gamma $ are the roots of the equation $ x^{2}-6p_2x+3=0 $ and $ \gamma ,\alpha $ are the roots of the equation $ x^{2}-6p_3x+6=0,p_1,p_2,p_3 $ being positive. Then, the coordinates of the centroid of $ \Delta ABC $ is
Options:
A) $ ( 1,\frac{11}{18} ) $
B) $ ( 0,\frac{11}{8} ) $
C) $ ( 2,\frac{11}{18} ) $
D) None of these
Correct Answer: C $ \therefore \alpha +\beta =6p_1,\alpha \beta =2 $? (i)
$ \beta ,\gamma $ are the roots of the equation $ x^{2}-6p_2x+3=0. $ $ \therefore \beta +\gamma =6p_2,\beta \gamma =3 $ ?. (ii)
$ \gamma ,\alpha $ are the roots of the equation $ x^{2}-6p_3x+6=0. $ $ \therefore \gamma +\alpha =6p_3,\gamma \alpha =6 $ ?. (iii)
From Eqs. (i), (ii) and (iii), we get $ \Rightarrow \alpha \beta \gamma =6 $ $ [\therefore \alpha ,\beta ,\gamma >0] $
Now, $ \alpha \beta =2 $ and $ \alpha \beta \gamma =6 $ $ \Rightarrow \gamma =3 $ $ \beta \gamma =3 $ and $ \alpha \beta \gamma =6 $ $ \alpha =3 $ $ \alpha =6\alpha \beta \gamma =6 $ $ \Rightarrow \beta =1 $ $ \therefore \alpha +\beta =6p_1\Rightarrow 3=6p_1 $ $ \Rightarrow p_1=\frac{1}{2} $ $ \beta +\gamma =6p_2\Rightarrow 4=6p_2 $ $ \Rightarrow p_2=\frac{2}{3} $
and $ \gamma +\alpha =6p_3\Rightarrow 5=6p_3 $ $ \Rightarrow p_3=\frac{5}{6} $
The coordinates of the centroid of triangle are
$ ( \frac{\alpha +\beta +\gamma }{3},\frac{1}{3}( \frac{1}{\alpha }+\frac{1}{\beta }+\frac{1}{\gamma } ) ) $ or
$ ( \frac{6}{3},\frac{1}{3}( \frac{1}{2}+1+\frac{1}{3} ) )or( 2,\frac{11}{18} ) $Show Answer
Answer:
Solution:
BETA
