How to use the nature of axisymmetry to find the coordinates of the symmetry point by using the coor

With regard to y-axis symmetry, the y-value does not change, and the x-value is reversed.

m=3/2 n=1/2

With regard to x-axis symmetry, the value of x does not change, and the value of y is reversed.

q(-a 2-2,a-2) x,y are both less than 0, the third quadrant.

With regard to x-axis symmetry, the value of x does not change, and the value of y is reversed.

The symmetry point is in the third quadrant and M is in the second quadrant.

1-a<0===>a>1

2a+2>0===>a>-1

Get a>1

With regard to x-axis symmetry, the value of x does not change, and the value of y is reversed.

With regard to y-axis symmetry, the y-value does not change, and the x-value is reversed.

2x+y-3=x+3,x-2y=-(y-4)x=5,y=1,a(8,3)With respect to the y-axis, the symmetry point is b(-8,3) Set a1 in any quadrant, and after four transformations, that is, a5 coincides with a1 and then recycles.

A1 is in the second quadrant and A97 coincides with A1.

A98 is in the third quadrant, A99 is in the fourth quadrant, and A100 is in the first quadrant!

1: The y-axis is symmetrical, the y-coordinate does not change, and the x-coordinate becomes the opposite number. Conversely, the x-axis is symmetrical, the x-coordinate does not change, and the y-coordinate becomes the opposite number.

2: a<0, then -a -2<0, 2-a>0 p point is in the second quadrant, and the symmetric point about the x-axis is in the third quadrant.

3: For the x-axis symmetry point in the third quadrant, explain that the m point is in the second quadrant, (x<0, y>0)4: The same method as in the first question.

5:100 4=25, so a100 is in quadrant 1.

1. When the straight line is perpendicular to the x-axis.

Symmetrical by axis. The property shows that y=b,aa' is on the line x=k, then,a+x) 2=k,x=2k-a

So it is easy to find the coordinates of a' (2k-a,b).

2. When the straight line is perpendicular to the y-axis.

From the properties of axisymmetry, it can be seen that the midpoint of x=a, bb' is on the line y=k, then, y+b) 2=k, y=2k-b

So it's easy to find the coordinates of b' (a,2k-b).

3. When the straight line is a general straight line, that is, its general form can be expressed as y=kx+b, and it is transformed into the form of a straight hidden answer line ax+by+c=0.

a, b) The coordinates of the symmetry point with respect to the line ax+by+c=0 are .

Resolve geometry from a plane.

, a straight line on a plane is defined by a Cartesian coordinate system of the plane.

is a binary linear equation in .

The graph represented.

1. The coordinates of the midpoint c of the two points a(x1,y1),b(x2,y2) are [(x1+ x2) 2,(y1+ y2) 2];

2. If two points are symmetrical with respect to a certain line, the midpoint of the two points is in this line (axis of symmetry.

, if the line y=k1x+b1 is perpendicular to the line y=k2x+b2, then k1 k2=-1.

3. Point symmetry dot drawing method for straight lines: cross the point to make the perpendicular line of the straight line.

and extended to a'so that they are at an equal distance from the straight line.

The known points a(2,-3)b(-1,2)c(-6,-5)d( ,1)e(4,0).

The symmetry points on the x-axis a(2,3)b(1,-2)c(-6, 5)d(1,2,-1)e (4,0).

The abscissa of the point of symmetry with respect to the x-axis is unchanged, and the ordinate is its opposite.

The symmetry points on the y-axis a (-2 ,-3 ) b (1 , 2) c (6 ,-5 ) d (-1 2 ,1 ) e (-4 ,0).

With respect to the point of symmetry on the y-axis, the abscissa is its opposite, and the ordinate is unchanged.

2.Write the coordinates of the following points with respect to the x- and y-symmetrical points:

Points of symmetry on the y-axis (3,-6) (7,-9) (6,1) (3,5) (0,-10).

points of symmetry on the y-axis; (3,6) (7,9) （6，-1） (3,-5) (0,10)

With respect to the point of symmetry on the x-axis: the abscissa is unchanged, and the ordinate is the opposite of the original number;

About the symmetry point of the y-axis: the ordinate is unchanged, and the abscissa is the opposite of the original.

The symmetry points on the x-axis a(2,3)b(1,-2)c(-6, 5)d(1,2,-1)e (4,0).

The symmetry points on the y-axis a (-2 ,-3 ) b (1 , 2) c (6 ,-5 ) d (-1 2 ,1 ) e (-4 ,0).

About the symmetry point of the x-axis (3,-6)(-7,-9)(6,1)(-3,5)(0,-10).

Points of symmetry on the y-axis (3,-6) (7,-9) (6,1) (3,5) (0,-10).

Pass the point D as DG CA to the point G, connect DB and DC

AD divides GAC, DE AB, DG AG DE DG

DF bisects BC perpendicularly

db dc in rt dbe and rt dcg, de dg, db dc rt dbe rt dcg(hl).

Be CG in RT DGA and RT DEA, DE DG, DA da RT DGA RT DEA (HL).

ae＝agcg-ac＝ag

be-ac＝ae

It is to pass the point d as dg ca to the point g, connect db and dc because ad bisects gac, de ab, dg ag so de dg

Because DF bisects BC perpendicularly

So db dc

Because in RT DBE and RT DCG, de dg, db DC, rt dbe rt dcg(hl).

Because be cg

So in RT DGA and RT DEA, de dg, da da, so rt dga rt dea(hl).

So ae ag

Because of CG-AC AG

So be-ac ae

This kind of problem will be much easier as long as you remember to move the line segment that is not in a straight line to a straight line. For example, in this problem, just move AC and AE to a straight line, and I believe you will do it below.

A is about a straight line, and the point of symmetry of l, such as cherry blossoms, is a'Hu Sleepy, Lian A'b crosses l to c, then ac=a'c (vertical bisector line slag to do cluster management).

Because the line segment between two points is the shortest, a'b=a'C+CB=AC+CB is the shortest.

The principle is that the straight line between two points is the shortest. Symmetrical past, connect to know the route.

Diagram Problem Analysis: Make the symmetry point C about the river bank of the point A, connect CB, intersect L at the point P, connect AP, then AP+BP is the shortest route of the day Answer: Solution:

This problem uses the property of axial symmetry, the analytical formula of the function is determined by the method of undetermined coefficients, and the property of the shortest line segment between two points is solved