1, let point P(x, y) be p ′ (x ′, y ′) relative to point (a, b),

x′= 2a-x

According to the midpoint coordinate formula, y' = 2b-y.

2. the symmetry point of point P(x, y) about the straight line l: ax+by+c = o is

x′= x-(Ax+By+C)

Then P'(x', y')

y′= y-(AX+BY+C)

In fact, the midpoint of ∵PP'⊥L and PP' is on the straight line L, and we can get: Ax'+By'=-Ax-By-2C.

We can draw a conclusion by solving this system of equations.

(-)=- 1(B≠0)

In particular, the point P(x, y) is approximately

1, the symmetry points of the x axis and the y axis are (x, -y) and (-x, y) respectively.

2. The punctuation marks of straight lines x=a and y=a are (2a-x, y) and (x, 2a-y) respectively.

3. The symmetry points of straight lines y=x and y=-x are (y, x) and (-y, -x) respectively.

Example: 1 the light is emitted from a (3,4), reflected by the straight line x-2y=0, and then reflected by the y axis. The reflected light passes through point B (1, 5), and the linear equation after the reflected light enters the Y axis is found.

Solution: As shown in the figure, the symmetrical point of A about the straight line x-2y=0 can be obtained from the formula.

A ′ (5,0), the axis symmetry point of B about Y is (-1 5), and the equation of straight line A ′ b ′ is 5x+6y-25=0.

` C(0，)

The equation of BC line is 5x-6y+25=0.

Secondly, the problem of curve symmetry of a known point or a known straight line.

To find the symmetric curve equation of a known curve F(x, y)=0 about a known point or a known straight line, we only need to replace the coordinates of any point (x, y) on the symmetric point F(x, y)=0 of a known point or a known straight line with the corresponding symmetry in equation F(x, y)= 0, and we can draw the following conclusions.

1, curve F(x, y)=0 The equation of the symmetric curve about point (a, b) is F(2a-x, 2b-y)=0.

2. Curve F(x, y)=0 The curve equation that is symmetrical about the straight line Ax+By+C=0 is F(x-(Ax+By+C), y-(Ax+By+C))=0.

In particular, the curve F(x, y)=0 is about

The curve equations of (1)X axis and y axis are F(x, -y) and F(-x, y)=0, respectively.

(2) The symmetric curve equations about straight lines x=a and y=a are F(2a-x, y)=0 and F(x, 2a-y)=0, respectively.

(3) The symmetric curve equations about straight lines y=x and y=-x are F(y, x)=0 and F(-y, -x)=0, respectively.

In addition, there are two conclusions as follows: for the image whose function is y=f(x), the image on the left side of the y axis is removed, and the image on the right side of the y axis is kept, and the image with y=f(|x|) is obtained; Keep the image above the X-axis, and fold the image below the X-axis to get an image with y=|f(x)|.

Example 2 (National College Entrance Examination) Let the equation of curve C be y = x3-X. Curve C 1 is obtained by moving C in parallel by t and s unit lengths along the positive direction of X axis and Y axis respectively:

1) Write the equation of curve C 1.

2) Prove that curves c and C 1 are symmetrical about point a (,).

(1) knows that the equation of C 1 is y = (x-t) 3-(x-t)+s.

(2) It is proved that any point B(a, b) on the curve C, let B 1(a 1, B 1) be the symmetric point of B about A, and let a=t-a 1, and B = s-b/kloc-.

s-b 1 =(t-a 1)3-(t-a 1)

` b 1 =(a 1-t)3-(a 1-t)+s

B 1 (A 1, b1) satisfies the equation of C 1

B 1 is on curve C 1, and it is easy to prove that the point on curve C 1 is on curve C.

Curves c and C 1 are symmetric about a.

We use the previous conclusion to prove that the symmetry point of point P(x, y) about A is P 1(t-x, s-y). In order to get the symmetric curve of C about A, we substitute its coordinates into the equation of C and get: s-y=(t-x)3-(t-x).

` y=(x-t)3-(x-t)+s

This is the equation of C 1, and the symmetric curve of `c about a is C 1.

Third, the symmetry of the curve itself

The necessary and sufficient condition for the curve F(x, y)=0 to be a (central or axisymmetric) symmetrical curve is that the equation remains unchanged after the coordinates of the symmetrical point of any point P(x, y) on the curve F(x, y)=0 are replaced by the corresponding coordinates in the curve equation.

For example, the coordinates of any point p(x, y) on the parabola y2=-8x and the symmetrical point p'(x, -y) on the X axis, that is, y=0, also satisfy the equation y2=-8x, and `Y2 =-8x is symmetrical about X. ..

Example 3 curve represented by equation xy2-x2y=2x:

A, symmetric about y axis b, symmetric about straight line x+y=0.

C, symmetry d about the origin and symmetry about the straight line x-y=0.

Solution: in the equation, x is replaced by -x, and y is replaced by-y.

(-x)(-y)2-(-x)2(-y)=-2x, that is, the equation of xy2-x2y=2x remains unchanged.

"The curve is symmetrical about the origin.

Regarding the symmetry of the function image itself about lines and points, we have the following important conclusions:

1, the straight line defined by function f(x) is r, and a is constant. If f(a+x)=f(a-x) exists for any x∈R, then the image of y=f(x) is symmetric about x = a. ..

This is because two points, a+x and a-x, are listed on the left and right sides of A respectively and are symmetrical about A, and their function values are equal, which shows that these two points are symmetrical about the straight line x=a, and a conclusion can be drawn from the arbitrariness of X.

For example, for f(x), if both t∈R have f(2+t)=f(2-t), then the image of f(x) is symmetric about x=2. What if the condition is changed to f( 1+t)=f(3-t) or f(t)=f(4-t)? In the first formula, let t= 1+m to get f (2+m) = f (2-m); In the second formula, let t=2+m and also get f(2+m)=f(2-m), then we still have the same conclusion that x=2 is symmetrical, from which we can draw the following more general conclusions:

2. The domain of function f(x) is r, and a and b are constants. If f(a+x)=f(b-x) exists for any x∈R, its image is symmetric about the straight line x=.

Let's discuss the following question again: What if the condition is changed to f(2+t)=-f(2-t)? Imagine f(t)=-f(t) if 2 is replaced by 0. This is a odd function, and the image is symmetrical about the center of (0,0). Now the translation is caused by f(2+t)=-f(2-t), so we guess that the image is centrosymmetric about m (2 2,0). As shown in the figure, take point A (2+t, f(2+t)), and its symmetrical point about m (2 2,0) is A'(2-x, -f(2+x)).

∫-f(2+x)= f (2-x)' a' is obviously (2-x, f(2-x)) on the image.

The image is centrosymmetric about m (2 2,0).

If the condition is changed to f(x)=-f(4-x), the following important conclusions can be drawn:

3. the domain of f (x) is r, and a and b are constants. If f(a+x)=-f(b-x) exists for any x∈R, its image is centrosymmetric about point m (0).

I hope it helps you.