The image of the point (3,5)

Question:

The image of the point $(3,5)$ in the line $x-y+1=0$, lies on :

  1. $(x-2)^{2}+(y-2)^{2}=12$

  2. $(x-4)^{2}+(y+2)^{2}=16$

  3. $(x-4)^{2}+(y-4)^{2}=8$

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


Correct Option: , 4

Solution:

$\frac{x-3}{1}=\frac{y-5}{-1}=-2\left(\frac{3-5+1}{1+1}\right)$

$\mathrm{So}, x=4, y=4$

Hence, $(x-2)^{2}+(y-4)^{2}=4$

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