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

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

|4m-9+1|/√(4²+(-3)²)=4
|4m-8|=20
|m-2|=5
m-2=-5,5
m-2=-3,m=7
2x+y

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 plane region represented by the inequality 2x + y < 3, then M = 0___ ．

∵ the distance from point m (m, 3) to the line 4x-3y + 1 = 0 is 4, ∵ d = | 4m-3 × 3 + 1 | 42 + (- 3) 2 = 4, the solution is m = 7 or M = - 3. When m = 7, 2 × 7 + 3 ＜ 3 does not hold; when m = - 3, 2 × (- 3) + 3 ＜ 3 holds

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 plane region represented by the inequality 2x + Y > 3, then M=

The slope of M = 7 straight line is four-thirds, and the slope of passing point perpendicular to the straight line 4x-3y + 1 = 0 is negative three-quarters. Correspondingly, two possibilities of M can be found, but the plane region represented by 2x + Y > 3 should be satisfied, so only 7