JEE MATHEMATICS MCQ: STRAIGHT LINES

1 Easy

1 of 1

Application +4 | -1

The centroid of the triangle formed by the pair of straight lines

12{x^2} - 20xy + 7{y^2} = 0

and the line

2x - 3y + 4 = 0

View answer and explanation

Correct option: C

\left( {\frac{8}{3},\frac{8}{3}} \right)

Option C (

\left( {\frac{8}{3},\frac{8}{3}} \right)

) is correct because it is the centroid calculated from the vertices of the triangle formed by the intersection of the given lines.

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