The image of $P(a, b)$ on $y=-x$ is $Q$ and the image of $Q$ on the line $y=x$ is $R$. Then the mid point of $R$ is

$\left(\frac{a+b}{2}, \frac{b+a}{2}\right)$

JEE MATHEMATICS MCQ: STRAIGHT LINES

1 Easy

1 of 1

Application +4 | -1

The image of

P(a, b)

y=-x

Q

and the image of

Q

on the line

y=x

R

. Then the mid point of

R

View answer and explanation

Correct option: D

(0,0)

Option D ((0,0)) is correct because the image of P(a, b) on y = -x is Q(-b, -a), and the image of Q on y = x is R(-a, -b). The midpoint of PR is then ((a - a)/2, (b - b)/2) = (0, 0).

More Questions from STRAIGHT LINES

Practice All Chapter MCQs →

If the three lines

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

and

a x+y=r

form a right-angled triangle then :
The orthocentre of the triangle whose vertices are

\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