If the point P (- 1-2a, 2a-4) is symmetric about the origin in the first quadrant, then the integer solution of a has () A. 1 B. 2 C. 3 d. 4

If the point P (- 1-2a, 2a-4) is symmetric about the origin in the first quadrant, then the integer solution of a has () A. 1 B. 2 C. 3 d. 4


∵ the point P (- 1-2a, 2a-4) is symmetrical about the origin. In the first quadrant, a ∵ 1 − 2A ∵ 0, a ∵ 2A − 4 ∵ 0, a ∵ 12, a ∵ 2, a = 1 or 0



If the point P (- 1-2a, 2a-4) is symmetric about the origin in the first quadrant, then the integer solution of a has ()
A. 1 B. 2 C. 3 d. 4


∵ the point P (- 1-2a, 2a-4) is symmetrical about the origin. In the first quadrant, a ∵ 1 − 2A ∵ 0, a ∵ 2A − 4 ∵ 0, a ∵ 12, a ∵ 2, a = 1 or 0



If the point P (1-2a, A-1) symmetric to the origin is the point of the first quadrant, then the value range of a is ()
A. a>12B. a<12C. 12<a<1D. 12≤a≤1


∵ the point P (1-2a, A-1) is symmetrical about the origin, which is the point of the first quadrant, ∵ P is in the third quadrant, ∵ 1 − 2A < 0A − 1 < 0, the solution is: 12 < a < 1, so choose: C



Given that the symmetric point of point m (a, A-2) about the origin is in the third quadrant, then the value range of a is


The symmetric point of point m (a, A-2) about the origin is in the third quadrant
So point m must be in the first quadrant
That is: a > 0, A-2 > 0
The solution is: a > 2