The equation for the straight line x 2y 10 symmetry about the straight line x 1 is

Solution: Your answer is wrong. Solution 1: The point (x,y) is any point on the symmetry equation, and it is easy to know that the symmetry point of the line x=1 is (2-x,y), and the straight line equation x+2y+8=0 is also brought into the line x-2y=10

Solution 2: From the straight line to the symmetry of the straight line, there is such a conclusion: the product of the slope of the two straight lines k1k2=-1 can be set to the straight line equation as: x+

2y+b=0, and we know that the two straight lines are symmetrical with respect to x=1 and both pass the point (1,-9 2), and bring in b=8, that is, the linear equation is x+2y+8=0

Let the equation for the straight line be y=kx+b

This line is symmetrical with the line x-2y=10 with respect to the line x=1 and satisfies the following conditions: 1) passes through the intersection of the line x-2y=10 and x=1.

2) The angle between the x-axis and the straight line x-2y=10 is complementary to the angle of the x-axis.

The intersection of the straight line x-2y=10 and x=1 is (1,-9 2)The tangent of the angle between the line x-2y=10 and the x-axis tan =1 2, then the tangent of the angle between the line y=kx+b and the x-axis tan =tan(180 - =-tan =-1 2

Slope k = -1 2

Therefore, the straight line is y+9 2=-1 2(x-1), i.e. x+2y+8=0

x-2y=10

x=1 gives x=1 y=-9 2

So the intersection is (1,-9 2).

With respect to x=1 symmetry x=1 and the y-axis is parallel.

So the slope of the symmetrical line = -1 2

So y=-k 2+b

Substitute (1,-9 2).

9/2=-1/2+b b=-4

The symmetrical line is y=-x 2-4

2y+x+8=0

Answer That doesn't seem right.

The straight line x-2y+1=0 intersects x+y+1=0 at (-1,0), then the symmetrical straight pure line of x-2y+1=0 also passes through this. The slope of x-2y+1=0 is, the slope of x+y+1=0 is -1, and the slope of the symmetrical straight line is k.

Let any point p(x,y) on the line y 2x 1 be symmetrical with respect to the point (2, 1) and the symmetrical point coordinates are p0(x0, y0), then x x0 2 2 4, and y y0 1 2 2 is x 4 x0, y 2 y0 and p(x, y) on the line y 2x 1, so 2 y0 2 (4 wang nai x0) 1 is sleepy spring y0 2x0 11, that is, the straight line cha orange.

The intersection of x+y-1=0 and x=2 is (2,-1).

He is also in a symmetrical straight line.

Take one point on x+y-1=0.

For example, (0,1).

His symmetry point about x=2 is (a,b).

Then (a+0) 2=2

b = 1 so the disadvantage a = 4

So straight through (2,-1), (4,1).

Rented by two points.

y+1) Xunpei (1+1) = (x-2) (4-2) so it is x-y-3 = 0

Find the intersection first: get (-10 3, -8 3).

1-1 2 1-1 2*1 = k+1 Take the cherry 1-k and get the nucleus k=2 or k=1 2 (rounded).

Therefore, the equation for finding the straight line is: y=2(x+10 3)-8 3, i.e., 6x-3y+12=0

The angle formula k-k1 1+k*k1 = k2-k 1+k*k2

Analysis: Find two special points on the line x-2y+1=0, determine the coordinates of the symmetry points of the two points about the line x=1, and then determine the linear equation of the symmetry of the line x-2y+1=0 about the line x=1.

Method 1: The line x-2y+1=0 intersects the x-axis at the point (-1,0) and the line x=1 at the point (1,1).

The point of symmetry of the point (-1,0) with respect to the line x=1 is the point (3,0) and the point of symmetry with respect to the line x=1 is the point (1,1).

The straight line x-2y+1=0 is symmetrical about the straight line x=1 passing point (3,0) and the point (1,1).

Substituting (3,0) and (1,1) into the undetermined analytic equation yields:

The linear equation is: x+2y-3=0.

Method 2: Set the point (x1,y1) on a straight line, find it about the symmetry point (2-x1,y1) and write it as (x2,y2).

x1-2y1+1=0

y2=y1, x2=2-x1, so x1=2-x2

So there is (2-x2)-2y2+1=0

This gives -x2-2y2+3=0

x+2y-3=0

Find two points on the line (these two points are easy to find with respect to the symmetry point of the line x=1), and then find the symmetry points of these two points with respect to the line x=1, and the line from these two points is the straight line.

For example; The line x-2y+1=0 intersects the x-axis at the point (-1,0) and the line x=1 at the point (1,1).

The point of symmetry of the point (-1,0) with respect to the line x=1 is the point (3,0) and the point of symmetry with respect to the line x=1 is the point (1,1).

The straight line x-2y+1=0 is symmetrical about the straight line x=1 passing point (3,0) and the point (1,1).

Substituting (3,0) and (1,1) into the undetermined analytic equation yields:

The linear equation is: x+2y-3=0.

Regular method (more difficult to understand)] Let a(x,y) be the point on the straight line sought, then the symmetry point of point A with respect to the line x=1 must be on the line x-2y+1=0, and because the symmetry point of point A with respect to x=1 is a'(2-x,y), so a'On the straight line x-2y+1=0, bring a' into (2-x)-2(y)+1=0 and simplify x+2y-3=0

Set (x. ,y。) is a point on the original straight line, and its symmetry point with respect to x=1 is (x,y) then there is, x. +

x=1×2y。=y

x。=2-x

Substitute x. -2y。+1=0

That is, the straight line is obtained: x+2y-3=0

y=1 2x symmetrical straight line with respect to straight line x=1.

y=1 2x passes through the (0,0) and (2,1) points of Cayen, and its pairs of conglomerates with respect to x=1 lines are called (2,0) and infiltration (0,1).

Substituting (2,0) and (0,1) into y=kx+b gives y=-1 2x+1