If there is only one point between the origin and point (1,1) in the plane region represented by inequality 2x-y + a > 0, then the value range of a is larger

From the meaning of the title
（0＋a)(2－1＋a)

If point B (1, - 1) is not in the inequality MX + Y-2

Analysis: point B is not in the plane region represented by the inequality, which indicates that the coordinates of point B do not satisfy the inequality,
That is to say, if the left side of the inequality is greater than or equal to 0, the coordinates of point B will hold,
So m-1-2 ≥ 0, m ≥ 3

Given that at least one point P (1,2) and its symmetric point about the origin is in the region represented by the inequality Y > KX + 2, then the value of K is
The value range is

For example, P (1,2) is in the region represented by the inequality Y > KX + 2,
Then: 2 > K + 2
k

If the distance between the point P (m, 3) and the line 4x-3y + 1 = 0 is 4, and the point P is in the inequality 2x + y

Answer: the distance from point P (m, 3) to line 4x-3y + 1 = 0 is 4. According to the distance formula from point to line: D = | 4m-3 × 3 + 1 | / √ (4 & # 178; + 3 & # 178;) = | 4m-8 | / 5 = 4, so: | 4m-8 | = 4 × 54 | m-2 | = 20 | m-2 | = 5, so: m-2 = 5 or m-2 = - 5, the solution is: M = 7 or M = - 3, because point P is 2x +

If there is only one point between the origin and point (1,1) in the plane region represented by inequality 2x-y + a > 0, then the value range of a is larger