JEE Main & Advanced Mathematics Straight Line Image of a Point in Different Cases

Image of a Point in Different Cases

Category : JEE Main & Advanced

(1) The image of a point with respect to the line mirror  The image of \[A({{x}_{1}},\,{{y}_{1}})\] with respect to the line mirror \[ax+by+c=0\] be \[B\,(h,\,\,k)\] is given by,

 

 

\[\frac{h-{{x}_{1}}}{a}=\frac{k-{{y}_{1}}}{b}=\frac{-2(a{{x}_{1}}+b{{y}_{1}}+c)}{{{a}^{2}}+{{b}^{2}}}\]

 

 

 

(2) The image of a point with respect to x-axis : Let \[P(x,y)\] be any point and \[{P}'({x}',\,{y}')\] its image after reflection in the x-axis, then \[{x}'=x\]

 

 

\[{y}'=-y,\] (\[\because \] \[{O}'\] is the mid point of P and \[{P}'\])

 

 

(3) The image of a point with respect to y-axis : Let \[P(x,\,y)\] be any point and \[{P}'({x}',\,{y}')\] its image after reflection in the y-axis,  then \[{x}'=-x\]     

 

 

\[{y}'=y\],    (\[\because \] \[{O}'\] is the mid point of P and  \[{P}'\])

 

 

(4) The image of a point with respect to the origin : Let \[P(x,y)\] be any point and \[{P}'({x}',\,{y}')\] be its image after reflection through the origin, then \[{x}'=-x\]                           

 

\[{y}'=-y\],(\[\because \] \[O\] is the mid point of P,\[{P}'\]).

 

 

(5) The image of a point with respect to the line \[y=x\] : Let \[P(x,\,y)\] be any point and \[{P}'({x}',\,{y}')\] be its image after reflection in the line \[y=x\], then \[{x}'=y\]

 

 

\[{y}'=x\], (\[\because \]\[{O}'\]is the mid point of P and \[{P}'\]).

 

 

(6) The image of a point with respect to the line   \[y=x\tan \theta \] : Let \[P(x,\,y)\] be any point and \[{P}'({x}',\,{y}')\] be its image after reflection in the line \[y=x\tan \theta \],  then

 

\[{x}'=x\cos 2\theta +y\sin 2\theta \] 

 

\[{y}'=x\sin 2\theta -y\cos 2\theta \], (\[\because \] \[{O}'\] is the mid point of P and \[{P}'\])

 

 

 


You need to login to perform this action.
You will be redirected in 3 sec spinner