JEE MATHEMATICS MCQ: STRAIGHT LINES

1 Easy

1 of 1

Analyzing +4 | -1

The lines

x+y=|a|

and

a x-y=1

intersect each other in the first quadrant. Then the set of all possible values of a is the interval:

View answer and explanation

Correct option: A

[1, \infty)

Option A (

[1, \infty)

) is correct because the intersection point lies in the first quadrant only if

a \ge 1

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\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

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