Question

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

• A. $(x-2{)}^2+(y-4{)}^2=4$
• B. $(x-4{)}^2+(y+2{)}^2=16$
• C. $(x-4{)}^2+(y-4{)}^2=8$
• D. $(x-2{)}^2+(y-2{)}^2=12$

Answer 1

The correct option is A $(x-2{)}^2+(y-4{)}^2=4$

Image of $P(3,5)$ on the line $x–y+1=0$ is
$\frac{x-3}{1}=\frac{y-5}{-1}=\frac{-2(3-5+1)}{2}=1$
$x=4,y=4$
$\therefore$ Image is $(4,4)$
Which lies on
$(x-2{)}^2+(y-4{)}^2=4$