If there is no solution to the linear equation of one variable | m | x-2x = m-2 about X, find the value of M

If there is no solution to the linear equation of one variable | m | x-2x = m-2 about X, find the value of M

When m > 0, mx-2x = m-2 has no solution
（m-2)x=m-2 x=1
When m = 0, x = 1
When m

Let y = m (a1x + B1) + n (a2x + B2) (where M + n = 1) be the generating functions of x-linear functions y = a1x + B1 and y = a2x + B2. (1) please write an analytic expression of the generating functions of y = x + 1 and y = 3x-1; (2) when x = C, find the function values of the generating functions of y = x + C and y = 3x-c; (3) if the intersection of the images of y = a1x + B1 and y = a2x + B2 is p (a, 5) ）When a 1B1 = a 2B2 = 1, find the value of the algebraic formula m (a 12a2 + B 12) + n (a 22a2 + B 22) + 2mA + 2Na

(1) For example, when m = 2, n = - 1. The generating function is y = 2 (x + 1) - (3x-1) = - x + 3, that is, y = - x + 3 When x = C, y = m (x + C) + n (3x-c) = 2C (M + n) (2 points) ∵ m + n = 1, ∵ y = 2C (M + n) = 2C (3 points) (3) ∵ point & nbsp; P & nbsp; (a, 5) on the image of y = a1x + B1 and y = a2x + B2, ∵ A1A + B1 = 5, A2A + B2 = 5 As & nbsp; a12a2 + B12 = (& nbsp; A1A + B1) 2-2 & nbsp; aa1b1 = 52-2 & nbsp; aa1b1, a22a2 + B22 = (A2A + B2) 2-2aa2b2 = 52-2aa2a2b2. When & nbsp; A1B1 = a2b2 = 1, m (a12a2 + B12) + n & nbsp; (A1A + B1; a; a1 + B1) 2) 2-2-2 & nbsp; aa1b1 = 52-2 & nbsp; aa1b1, aa1b1, a22a2a2a2, a22a2a2a2 + B22 + 2mA + 2Na + 2Na = M & nbsp; (52-2a & nbsp; (52-2a & nbsp; (52-2a & nbsp; (52-2a & nbsp; (52-2a & nbsp; (52-2a & nbsp; (52-2a & nbsp; (52-2a & nbsp;) + n; (a22a & nbsp;) + n (a22a) + n (52-2a + n (52-2a + n (52-2a 2Na = 25 (M + n) = 25 (6 points)

In the linear function y = - 2x + B, when x = 1, Y0. Then the value range of B is

x=1;y=-2+b＜1;
∴b＜3;
x=-1;y=2+b＞0;
∴b＞-2;
∴-2＜b＜3;
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If three straight lines ax + 2Y + 8 = 0,4x + 3Y = 0,2x-y = 10 intersect at a point, then a =?

Solution
4x+3y=0 A
2x-y=10 B
have to
3B 6x-3y=30
A+3B 10x=30
If x = 3 is substituted by 4x + 3Y = 0, then
y=-4
Substituting (3, - 4) into ax + 2Y + 8 = 0
3a-8+8=0
a=0

If three straight lines ax + 2Y + 8 = 0, 4x + 3Y = 10, 2x-y = 10 intersect at a point, then the value of a is ()
A. -2B. -1C. 0D. 1

By solving the equations 4x + 3Y = 10, 2x-y = 10, the coordinates of the intersection point are (4, - 2), and substituting ax + 2Y + 8 = 0, a = - 1 is obtained

If three straight lines ax + 2Y + 8 = 0, 4x + 3Y = 10 and 2x-y = 10 intersect at a point, find a

ax+2y+8=0,
4x+3y=10
2x-y=10,
Solving the equations, we get x = 4, y = - 2, a = - 1

If there are only two points on the circle (x-1) 2 + (y + 1) 2 = R2 and the distance from the two points to the straight line 4x + 3Y = 11 is equal to 1, then the value range of radius R is______ ．

The distance from the center of a circle to the straight line is 2, and there are only two points on the circle (x-1) 2 + (y + 1) 2 = R2. The distance from the center of the circle to the straight line 4x + 3Y = 11 is equal to 1, satisfying | R - | 4-3-11 | 42 + 32 | < 1, that is: | R-2 | < 1, the solution is 1 | r < 3. Therefore, the value range of radius R is 1 | r < 3 (drawing), so the answer is: (1, 3)

Given two points a (4, - 3), B (2, - 1) and a straight line L: 4x + 3y-2 = 0, find a point P such that | PA | = | Pb |, and the distance from point P to straight line L is equal to 2

Let P (x0, Y0), the midpoint of AB (3, - 2) ∫ PA | = | Pb |, the equation of the vertical line of AB is y = X-5 ∫ point P is on the vertical line of AB, and the distance to L is 2 ∫ Y0 = x0 − 52 = | 4x0 + 3y0 − 2 | 5 ∫ x0 = 277y0 = − 87 or x0 = 1y0 = − 4. ∫ P (277, & nbsp; − 87) or (1, & nbsp; − 4)

It is known that a (4, - 3) and B (2, - 1) are symmetric with respect to the line L. there is a point P on L, where the distance from P to the line 4x + 3y-2 = 0 is equal to 2
Find the coordinates of point P

Set point P (x, y)
∵|PA|=|PB|
∴(x-4)²+(y+3)²=(x-2)²+(y+1)²
The result is x = y + 5
Then we can get it from the problem
|4(y+5)+3y-2|/5=2
Ψ y = - 8 / 7 or y = - 4
Combining with x = y + 5, we can get
P (27 / 7, - 8 / 7) or P (1, - 4)

It is known that a (4, - 3) and B (2, - 1) are symmetric with respect to the line L. there is a point P on L, where the distance from P to the line 4x + 3y-2 = 0 is equal to 2
Finding the coordinates of point P

If I is perpendicular to line AB and passes through the midpoint, the equation of I can be easily obtained: x + Y-1 = 0
Let the above point a be (a, 1-A) and use the formula from point to straight line