The length of perpendicular from the point $\left( {a\cos {\rm{\alpha }},\;a\sin \alpha } \right)$ upon the straight line $y = x\tan \alpha + c,c > 0$ is

$c\;{\rm{si}}{{\rm{n}}^2}{\rm{\alpha }}$

$c\;{\rm{se}}{{\rm{c}}^2}{\rm{\alpha }}$

JEE MATHEMATICS MCQ: STRAIGHT LINES

1 Easy

1 of 1

Application +4 | -1

The length of perpendicular from the point

\left( {a\cos \mathrm{\alpha },\;a\sin \alpha } \right)

upon the straight line

y = x\tan \alpha + c,c > 0

View answer and explanation

Correct option: C

c\cos \mathrm{\alpha }

Option C (

c\cos \mathrm{\alpha }

) is correct because it is the perpendicular distance calculated from the point to the line using the perpendicular distance formula.

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