generally i know that to find symmetric equation of function along line $y=x$,we should exchange $x$ and $y$ and solve back,but what about $y=-x$?should i repeat again the same procedure,but instead of $x$,should i take $-x$?let us consider following problem consider equation of circle
$(x+3)^2+(y-3)^2=9$
find it's symmetric circle equation along line $y=-x$
so should i put $-x$ instead of $x$?thanks in advance
On
Let $A \subset \mathbb{R}^2$ be a set of points given by equation $f(x,y) = 0$, then for any invertible linear transformation $T : \mathbb{R}^2 \to \mathbb{R}^2$ set $T(A)$ is given by $(f \circ T^{-1})(x,y) = 0$.
This is because $A = \{(x,y) \mid f(x,y) = 0\}$, and \begin{align} T(A) &= \Big\{(x',y')\ \Big|\ (x',y') = T(x,y) \land f(x,y) = 0\Big\} \\ &= \Big\{(x',y')\ \Big|\ T^{-1}(x',y') = (x,y) \land f(x,y) = 0\Big\} \\ &= \Big\{(x',y')\ \Big|\ f(T^{-1}(x',y')) = 0\Big\} \\ &= \Big\{(x',y')\ \Big|\ (f \circ T^{-1})(x',y') = 0\Big\}. \end{align}
For example, if $T = \left[\begin{array}{rr}0&-1\\-1&0\end{array}\right]$, that is the symmetry along $y = -x$, then it happens that $T^{-1} = T$ (this is true for all symmetries) and appropriate equation would look like $$f(-y,-x) = 0.$$
In your case $f(x,y) = (x+3)^2 + (y-3)^2 -9$, so the symmetrical figure could be described by $$(-y+3)^2+(-x-3)^2 = 9.$$
I hope this helps ;-)
When we reflect a point $(x_0,y_0)$ across the line $y=-x,$ our new $x$-coordinate will be $-y_0$ and our new $y$- coordinate will be $-x_0.$ Consequently, our general transformation in this case is to make $x\mapsto-y,$ $y\mapsto-x$.
What about the more general case of reflecting about a line $\ell$ through the origin, though? Well, first, see where the point $(1,0)$ is reflected to--say $(x_1,y_1)$--and where the point $(0,1)$ is reflected to--say $(x_2,y_2).$ Then in general, a point $(x,y)$ will be reflected about $\ell$ to $$(x_1x+x_2y,y_1x+y_2y).$$ How can we see this, though? It comes down to the fact that a reflection about a line through the origin is a linear transformation, and looking at it in terms of matrices shows us that $$\left[\begin{array}{c}x\\ y\end{array}\right]\mapsto \left[\begin{array}{cc}x_1 & x_2\\ y_1 & y_2\end{array}\right] \left[\begin{array}{c}x\\ y\end{array}\right].$$ That may be beyond what you'll encounter anytime soon. Consider it a sneak preview.