JEE MATHEMATICS MCQ: STRAIGHT LINES

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Application +4 | -1

\tan \alpha, \tan \beta, \tan \lambda

are the roots of the equation

t^3-12 t^2+15 t-1=0

; then the centroid of triangle having vertices

(\tan \alpha, \cot \alpha) ;(\tan \beta, \operatorname{cotb} \beta) ;(\tan \lambda, \cot \lambda)

is given by

G(h, k) ;

then evaluate

(h+k) /(k-h)

View answer and explanation

Correct option: A

Option A (9) is correct because the centroid's coordinates are (4, 5), and (h+k)/(k-h) = (4+5)/(5-4) = 9.

More Questions from STRAIGHT LINES

Practice All Chapter MCQs →

x-3 y=p, a x+2 y=q

a x+y=r

\left\{ {a{t_1}{t_2},\;a\left( {{t_1} + {t_2})\} ,\{ a{t_2}{t_3},a({t_2} + {t_3}} \right)} \right\},\;\left\{ {a{t_3}{t_1},\;a\left( {{t_3} + {t_1}} \right)} \right\}

y = 2\;x + 4

6x = 3y + 5

{x^2} + {y^2} + 2xy - 8ax - 8ay - 9{a^2} = 0

(3,8)

x+3 y=7

\left( {{x^2} + 7xy + 2{y^2}} \right)\left( {y - 1} \right) = 0