JEE MATHEMATICS MCQ: STRAIGHT LINES

1 Medium

1 of 1

Application +4 | -1

P\left(\frac{1+t}{\sqrt{2}}, \frac{2+t}{\sqrt{2}}\right)

is any point on a line, then the range of values of

t

for which the point

P

lies between the parallel lines

x+2 y=1

and

2 x+4 y=15

, is

View answer and explanation

Correct option: A

-\frac{4 \sqrt{2}}{3} < t < \frac{5 \sqrt{2}}{6}

Option A (

-\frac{{4 \sqrt{{2}}}}{{3}} < t < \frac{{5 \sqrt{{2}}}}{{6}}

) is correct because it defines the range of values for

t

where the point P lies between the two given parallel lines.

More Questions from STRAIGHT LINES

Practice All Chapter MCQs →

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