JEE MATHEMATICS MCQ: STRAIGHT LINES

1 Easy

1 of 1

Analyzing +4 | -1

{t_1} + {t_2} + {t_3} = - {t_1}{t_2}{t_3}

, then orthocentre of the triangle formed by the points

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

and

\left[ {a{t_3}{t_1},a\left( {{t_3} + {t_1}} \right)} \right]

, lies on

View answer and explanation

Correct option: B

\left( { - a,\;0} \right)

Option B ((-a, 0)) is correct because the orthocenter of the triangle formed by the given points lies on the line x = -a, which means the orthocenter is (-a, 0).

More Questions from STRAIGHT LINES

Practice All Chapter MCQs →

If the three lines

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

and

a x+y=r

form a right-angled triangle then :
The orthocentre of the triangle whose vertices are

\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\}

is
The distance between the parallel lines

y = 2\;x + 4

and

6x = 3y + 5

is The distance between the pair of lines given by

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

is Find the image of the point

(3,8)

with respect of the line

x+3 y=7

assuming the line to be plane mirror: The centroid of the triangle whose three sides are given by the combined equation

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