JEE MATHEMATICS MCQ: STRAIGHT LINES

1 Easy

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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

, is

View answer and explanation

Correct option: C

\left|\frac{(p-q)(r-s)}{\sin (\alpha-\beta)}\right|

Option C is correct. The area of the parallelogram formed by the given lines is found using the formula involving the determinant of the coefficients and the difference of the constants, resulting in

\left|\frac{{(p-q)(r-s)}}{{\sin (\alpha -\beta )}}\right|

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