The image of the point (3,8) in the line $x + 3y = 7,$ is

JEE MATHEMATICS MCQ: STRAIGHT LINES

1 Easy

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Application +4 | -1

The image of the point (3,8) in the line

x + 3y = 7,

View answer and explanation

Correct option: C

\left( { - 1, - 4} \right)

The correct answer is C ((-1, -4)) because it is the reflection of the point (3,8) across the given line, found using perpendicularity and midpoint conditions.

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If the three lines

x-3 y=p, a x+2 y=q

and

a x+y=r

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\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\}

is
The distance between the parallel lines

y = 2\;x + 4

and

6x = 3y + 5

is The distance between the pair of lines given by

{x^2} + {y^2} + 2xy - 8ax - 8ay - 9{a^2} = 0

is Find the image of the point

(3,8)

with respect of the line

x+3 y=7

assuming the line to be plane mirror: The centroid of the triangle whose three sides are given by the combined equation

\left( {{x^2} + 7xy + 2{y^2}} \right)\left( {y - 1} \right) = 0