The image of the point $(-1,3,4)$ in the plane $x-2 y=0$ is

$\left(-\frac{17}{3},-\frac{19}{3}, 4\right)$

$\left(-\frac{17}{3},-\frac{19}{3}, 1\right)$

JEE MATHEMATICS MCQ: STRAIGHT LINES

1 Hard

1 of 1

Application +4 | -1

The image of the point

(-1,3,4)

in the plane

x-2 y=0

View answer and explanation

Correct option: D

None of these

Option D (None of these) is correct because the calculated image point does not match any of the provided options.

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