Given point and line

We need the image of the point $P(3,5)$ in the line $x-y+1=0.$

Formula for reflection of a point across a line

For a line $ax+by+c=0,$ the reflection of point $(x_1,y_1)$ is $\left(x_1-\frac{2a(ax_1+by_1+c)}{a^2+b^2},\; y_1-\frac{2b(ax_1+by_1+c)}{a^2+b^2}\right).$

Here, $a=1,\quad b=-1,\quad c=1,\quad (x_1,y_1)=(3,5).$

Compute the required quantity

$ax_1+by_1+c = 1\cdot 3 + (-1)\cdot 5 + 1 = 3-5+1=-1.$

Also, $a^2+b^2 = 1^2+(-1)^2=2.$

Find the reflected point

$x' = 3 - \frac{2\cdot 1\cdot (-1)}{2} = 3+1=4,$ $y' = 5 - \frac{2\cdot (-1)\cdot (-1)}{2} = 5-1=4.$

So the image of $(3,5)$ is $P'(4,4).$

Check which option contains the point $(4,4)$

• A: $(x-4)^2+(y-4)^2=8$ Substituting $(4,4)$: $(4-4)^2+(4-4)^2=0\neq 8$ Not correct.

• B: $(x-4)^2+(y+2)^2=16$ Substituting $(4,4)$: $0^2+6^2=36\neq 16$ Not correct.

• C: $(x-2)^2+(y-2)^2=12$ Substituting $(4,4)$: $2^2+2^2=8\neq 12$ Not correct.

• D: $(x-2)^2+(y-4)^2=4$ Substituting $(4,4)$: $2^2+0^2=4$ Correct.

Conclusion

The reflected point lies on option D.