Class did not learn! Especially the map, the most headache! There are some junior high school knowledge is also more dizzy! Four. Point about line, line about point, point about point, line about line

Class did not learn! Especially the map, the most headache! There are some junior high school knowledge is also more dizzy! Four. Point about line, line about point, point about point, line about line

I've just learned it. As far as I understand it, there are two kinds of symmetry problems,
The first is the problem of finding the symmetric point when the known point is symmetric with respect to the straight line
The second is the problem of finding a symmetrical line when a line is symmetrical about a line
For the first, the solution is simple
As long as the equations are listed: 1. The product of the slope of the straight line between the known point and the symmetrical point and the slope of the symmetrical axis is - 1 (special consideration if there is no slope)
2. The midpoint is on the axis of symmetry
The problem can be solved by establishing equations
For the second method, let P (x, y) be the point on the line, and its symmetric point about the axis of symmetry be q (x ', y')
The equations are listed as follows: 1. The product of the slope of the straight line between the known point and the symmetrical point and the slope of the symmetrical axis is - 1 (special consideration if there is no slope)
2. The midpoint is on the axis of symmetry
Solve X '=
y'=
Because Q (x ', y') is on a known straight line, the problem can be solved by substituting the solution value into the known straight line equation
This is my most intuitive understanding of symmetry. I wonder if it can help you
(sorry, this is what we have learned now, because we have learned the third and the fourth in junior high school.)
There are two other cases
The third: point about point symmetry, the problem of finding symmetric point
The coordinates of the point can be set
According to the midpoint of the line between the point and the symmetrical point, which is the symmetrical center, we can get the result
The specific methods are as follows:
The symmetry of point a (a, b) with respect to point O (m, n) is a '(2m-A, 2n-b)
The fourth: the symmetry of a line with respect to a point
The method of special points can be adopted
Set the coordinates of a point on the line
The method in the third method can be used to find the symmetric point a about the known symmetric center
And because point a is on a known line, it is substituted into the equation of a known line,
Then we can find the equation of straight line
How about this?

Straight line and equation of straight line in Senior Two
1. Find the inclination angle range of xcosa + y + B = 0 (a, B ∈ R)
2. Let u = R be the complete set, a be the solution set of inequality │ X-1 │ + A-1 > 0 (a ∈ R) about X, set B = {x │ sin (π X - π / 3) + √ 3cos (π X - π / 3) = 0}, if (CUA) ∩ B has exactly three elements, the range of a is obtained
Please give me the process,
It's a root sign-_-

I don't understand the meaning of √ in B = {x │ sin (π X - π / 3) + √ 3cos (π X - π / 3) = 0}

It is known that the equation of the line where the high line on the AB side of △ ABC is 2x-3y + 1 = 0, and the equation of the line where the high line on the AC side is x + y = 0, and the vertex a (1,2). The equation of the line where the BC side is located is obtained

Let B (- m, m), C (12 (3n-1), n) get KAC = n − 212 (3N − 1) − 1 = − 1 − 1 = 1, the solution get n = - 1, the solution get C (- 2, - 1) KAB = m − 2 − m − 1 = − 123 = - 32, the sum of solutions M = - 7, the solution get B (7, - 7). Therefore, the equation of straight line BC is y + 1 − 7 + 1 = x + 27 + 2, the simplification get 2x + 3Y + 7 = 0

It is known that the line L: 5ax-5y-a + 3 = 0. (1) it is proved that no matter what the value of a is, the line l always passes through the first quadrant; (2) if the line L does not pass through the second quadrant, the range of a is obtained

(1) It is proved that: ∵ the line L is 5ax-5y-a + 3 = 0, that is, a (5x-1) + (- 5Y + 3) = 0; ∵ 5x − 1 = 0 − 5Y + 3 = 0, the solution is x = 15y = 35; ∵ no matter what the value of a is, the line l always passes the point (15,35) in the first quadrant, that is, the line L passes the first quadrant; (2) according to the meaning of the title, draw a figure, as shown in the figure

On 1-x = MX + 1 (m ∈ R) under the root of the equation of X
1. When there is a real root, find the value range of M
2. When there are two real roots, find the value range of M
Sorry, it's 1-x2 under the root

Make function f (x) = √ (1-x ^ 2)
And G (x) = MX + 1
Note that G (x) passes through point (0,1)
F (x) = √ (1-x ^ 2) shape is actually the upper half of the circle [x ^ 2 + y ^ 2 = 1]. It passes through points (0,1), and (1,0), (- 1,0)
Point (1,0) is the highest point of F (x)
Both f (x) and G (x) pass through point (0,1)
It can be seen that,
1. If f (x) and G (x) have an intersection, then
1)
F (x) and G (x) are tangent to point (0,1)
Since point (1,0) is the highest point of F (x), the,
Then M = 0;
2)
The point of G (1) is below f (1) = 0
That is: G (1) = m + 1 < 0
→m＜-1.
3)
The point of G (1) is below f (- 1) = 0
That is: G (1) = - M + 1 < 0
→m＞1.
When there is a real root, the value range of M
M = 0 or m ＜ - 1 or m ＞ 1
two
From the conclusion of 1, we can get: 1. The complement of a real number is the value range of m when there are two real roots
Namely:
-1 ≤ m < 0 or 0 < m ≤ 1

The corresponding rule f is known: P (m, n) → P ′ (m, n) (M > 0, n > 0). The existing a (9,3) → a ′, B (3,9) → B ′. M is a moving point on the line AB, m → m ′. When m moves from a to B on the line AB, the point m ′ moves from a ′ to B ′, then the length of the route m ′ passes is___ ．

The equation of AB is: ab: x + y = 12, 3 ≤ x ≤ 9, let the coordinates of M be (x0, Y0), because m is on AB, we can get x0 + Y0 = 12, 3 ≤ x ≤ 9. From the meaning of the question, if the coordinates of m 'are (x, y), then the equation of x = x0, y = Y0, the trajectory of M' satisfies is x2 + y2 = 12, in which - 3 ≤ x ≤ 3, because x > 0, Y > 0 is required, so the two ends of M 'trajectory are a (3, 3) and a (3, 3) B (3,3) ≠ AOX = 30 ° and ∠ box = 60 °, that is to say, the track of M 'is an arc with a center angle of 30 ° and the length of the route that M' passes through is π 6 × 12 = 3 π 3, so the answer is: 3 π 3

The center of hyperbola C is at the origin, the focus is on the y-axis, and the angles of the moving chord PQ parallel to the imaginary axis are complementary
(1) Proof: hyperbola C is equiaxed hyperbola
(2) The connecting line Mn of two intersections m, n between hyperbola C and circle D: (x-4) ^ 2 + (y-6) ^ 2 = 13 is exactly the diameter of circle D. try to find the equation of hyperbola C

(1) The hyperbolic equation can be: y ^ 2 / A ^ 2-x ^ 2 / b ^ 2 = 1 (a > 0, b > 0), let P (x0, Y0), then q (- x0, Y0),
∵∠PAQ+∠PBQ=180°,∴∠PAE+∠PBE=90°
∴tan∠PAE•tan∠PBE=|x0/(y0-a)|•|x0/(y0+a)|=|x0^2/(y0^2-a^2)|
By substituting the hyperbolic equation into the above equation, we can get Tan ∠ PAE &; Tan ∠ PBE = B ^ 2 / A ^ 2 = 1
Ψ a = B (a = - B is rounded off),
The hyperbola C is an equiaxed hyperbola
(2) Let m (x1, Y1), n (X2, Y2), then Y1 ^ 2 - X1 ^ 2 = a ^ 2, Y2 ^ 2 - x2 ^ 2 = a ^ 2,
∴(y1+y2)•(y1-y2)-(x1+x2)(x1-x2)=0
The midpoint of Mn is d (4,6),
∴12(y1-y2)-8(x1-x2)=0,(y1-y2)/(x1-x2)=8/12=2/3
That is Kmn = 2 / 3
∴MN:y-6= 2(x-4)/3
Substituting into the equation of circle, we get: (x-4) ^ 2 + (x-4) ^ 2 &; 4 / 9 = 13,
Ψ x = 7 or 1,
The coordinates of point m are (7,8) or (1,4)
The hyperbolic equation gives a ^ 2 = 8 ^ 2-7 ^ 2 (or 4 ^ 2-1 ^ 2) = 15,
The hyperbolic equation is y ^ 2-x ^ 2 = 15

As shown in the figure, we know that a point P, PA ⊥ plane ABCD, e and F are the midpoint of AB and PC respectively. (1) prove: EF ⊥ plane pad; (2) prove: EF ⊥ CD

It is proved that: (1) if we take the midpoint Q of PD and connect AQ and QF, then aefq is a parallelogram ∩ EF ∩ AQ and ∩ AQ are in the planar pad, EF is not in the planar pad, EF is in the planar pad, and ∩ ad = APA in the planar pad, ad is in the planar pad, CD ⊥ PA is in the planar pad, and ∩ AQ is in the planar pad

What is the reduction number from 2011 to 2012

|-2012 | is this an absolute value?
2011-2012= -1

Simplified calculation: (1 / 1 + √ 2 + 1 / √ 2 + √ 3 + 1 / √ 3 + √ 4 +. + 1 / √ 2011 + √ 2012)

1/（1+√2)+1/(√2+√3)+1/(√3+√4)+.+1/(√2011+√2012)
=(√2-1)/(2-1)+(√2-√3)/(2-3)+(√3-√4)/(3-4)+.+(√2011-√2012)/(2011-2012)
=-1+√2-√2+√3-√3+√4-√4+.+√2011-√2011+√2012
=√2012-1