Tag: reflection w.r.t a line

Questions Related to reflection w.r.t a line

If $\displaystyle \left ( -2, 6 \right )$ is the image of the point $\displaystyle \left ( 4,2 \right )$ with respect to the line $\displaystyle L=0$, then $\displaystyle L=$

maths position and movement reflection w.r.t a line transformation transformation and symmetry in geometrical shapes

  1. $\displaystyle 6x-4y-7=0$

  2. $\displaystyle 2x-3y-5=0$

  3. $\displaystyle 3x-2y+5=0$

  4. $\displaystyle 3x-2y+10=0$

Show answer
Correct Option: C
Explanation:

Slope of line joining the image $Q(-2,6)$ and the point $P(4,2)$ is $\displaystyle -\frac{2}{3}$

So, the slope of mirror $L$ is $\displaystyle \frac{3}{2}$

Mid-point of $PQ$ is $(1,4)$

Since, the image and point are equidistant from mirror. So, this point $(1,4)$ lies on the mirror.

So, the equation of mirror is
$y-4=\displaystyle \frac{3}{2} (x-1)$

$\Rightarrow 3x-2y+5=0$

The equation of image of pair of lines $y=|x-1|$ with respect to y-axis is 

maths position and movement reflection w.r.t a line transformation transformation and symmetry in geometrical shapes

  1. ${x^2} - {y^2} - 2x + 1 = 0$

  2. ${x^2} - {y^2} - 4x + 4 = 0$

  3. $4{x^2} - 4x - {y^2} + 1 = 0$

  4. ${x^2} - {y^2} + 2x + 1 = 0$

Show answer
Correct Option: D
Explanation:

We have $y=|x-1|$


$\Rightarrow y^2=(x-1)^2$

Change $x$ by $-x$, then the required image is 

$y^2=(-x-1)^2$

$\Rightarrow y^2=x^2+2x+1$

$\Rightarrow x^2-y^2+2x+1=0$