JEE MATHEMATICS MCQ: STRAIGHT LINES

1 Easy

1 of 1

Application +4 | -1

The area of the parallelogram formed by the lines

2 x-3 y+a=0, \quad 3 x-2 y-a=0,2 x-3 y+3 a=0

and

3 x-2 y-2 a=0

in square units, is

View answer and explanation

Correct option: B

\frac{2 a^2}{5}

Option B (

\frac{2 a^2}{5}

) is correct because it is the area of the parallelogram calculated using the formula involving the constants and determinant of coefficients.

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