The number of points on the line $3x + 4y = 5$, which are at a distance of ${\rm{se}}{{\rm{c}}^2}\;{\rm{\theta }} + 2{\rm{cosse}}{{\rm{c}}^2}{\rm{\;\theta }},{\rm{\;\theta }} \in R$, from the point (1, 3) is

JEE MATHEMATICS MCQ: STRAIGHT LINES

1 Medium

1 of 1

Analyzing +4 | -1

The number of points on the line

3x + 4y = 5

, which are at a distance of

\mathrm{se}{\mathrm{c}^2}\;\mathrm{\theta } + 2\mathrm{cosse}{\mathrm{c}^2}\mathrm{\;\theta },\mathrm{\;\theta } \in R

, from the point (1, 3) is

View answer and explanation

Correct option: B

Option B (2) is correct because there are two points on the line at the specified distance from the given point.

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