JEE MATHEMATICS MCQ: STRAIGHT LINES

1 Easy

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Analyzing +4 | -1

If the two lines

x + (a - 1)y = 1\,\mathrm{and}{\kern 1pt} \,2x + {a^2}y = 1,(a \in R - \{ 0,1\} )

are perpendicular, then the distance of their point of intersection from the origin is

View answer and explanation

Correct option: D

\sqrt {\frac{2}{5}}

Option D (

\sqrt{\frac{2}{5}}

) is the correct answer because after finding the value of 'a' using the perpendicularity condition, we solve for the intersection point and calculate its distance from the origin.

More Questions from STRAIGHT LINES

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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