The distance of point $(-1,1)$ from the line $12 x-5 y+9=0$ is

JEE MATHEMATICS MCQ: STRAIGHT LINES

1 Easy

1 of 1

Application +4 | -1

The distance of point

(-1,1)

from the line

12 x-5 y+9=0

View answer and explanation

Correct option: B

\frac{8}{13}

Option B (

\frac{{8}}{{13}}

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

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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