JEE MATHEMATICS MCQ: STRAIGHT LINES

1 Easy

1 of 1

Application +4 | -1

The area of the parallelogram formed by the lines

x\cos \alpha + y\sin \alpha = p,\;x\cos \alpha + y\sin \alpha = q,

x\cos \beta + y\sin \beta = r

and

x\cos \beta + y\sin \beta = s

for given values of

p,q,r

and s is least, if

\left( {\alpha - \beta } \right) =

View answer and explanation

Correct option: B

\frac{\pi }{4}

Option A (

\pm \frac{\pi }{2}

) is incorrect. The area of the parallelogram is minimized when the angle between the lines is maximized, which occurs at

\pm \frac{\pi }{2}

. The question asks when the area is LEAST, not maximum. The correct answer should be when the area is largest, which is when the angle is smallest.

More Questions from STRAIGHT LINES

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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