Given the value of M (2m + 3, m), n (3m-1,1), the inclination angle of the straight line Mn is a right angle acute angle obtuse angle

Given the value of M (2m + 3, m), n (3m-1,1), the inclination angle of the straight line Mn is a right angle acute angle obtuse angle

Right angle:
2m+3=3m-1 m=4
Acute angle:
tana=(1-m)/(3m-1-2m-3)
=(1-m)/(m-4)
=-(m-4)-3]/(m-4)
=-1-3/(m-4)>0
The solution is 0

Given m (M + 3, m) n (m-2,1), when what is the value, the inclination angle of the straight line Mn is an acute angle?

Let the slope be K. since it is an acute angle, we can set the slope
k=tanθ=（m-1）/(m+3-m+2)=(m-1)/5
I we know that the tangent of the acute angle is > 0, so we only need m > 1. However, m cannot be infinite. If the infinity is close to 90 degrees, it is not an acute angle. The range of M is (1, + ∞)

Given that the line L passes through the point P (- 1,2) and intersects the line segment with a (- 2, - 3), B (3,0) as the end point, the value range of the slope of the line L is obtained

The slope of the straight line AP is k = - 3 − 2 − 2 + 1 = 5, and the slope of the straight line BP is k = 0 − 23 + 1 = - 12. Let L and line AB intersect at point m, and M moves from a to B, and the slope becomes larger and larger. At a certain point, am will be parallel to the Y axis, and there is no slope. That is, K ≥ 5. After this point, the slope increases from - ∞ to - 12 of the straight line BP

Let the straight line L pass through the point a (0,3) B (m, 1) and its inclination angle be α. Then, the slope k of the straight line l can be obtained

①
If M = 0, then α = 90 ° and the slope of line L does not exist
If M is not equal to 0, then the slope k = (1-3) / (M-0) = - 2 / m
②
If 30 ° 1 / 3
m^2