JEE MATHEMATICS MCQ: STRAIGHT LINES

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Application +4 | -1

The point

\left(a^2, a+1\right)

lies in the angle between the lines

3 x-y+1=0

and

x+2 y-5=0

containing the origin, if

View answer and explanation

Correct option: A

a \in(-3,0) \cup\left(\frac{1}{3}, 1\right)

Option A (

a \in(-3,0) \cup(\frac{1}{3}, 1)

) is correct because the point and the origin must lie on the same side of both lines, leading to two inequalities that define the range of 'a'.

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