Given the triangle ABC, a [4,0], the absolute value BC = 2, BC moves on the y-axis, then the trajectory equation of the outer center P of △ ABC is

Given the triangle ABC, a [4,0], the absolute value BC = 2, BC moves on the y-axis, then the trajectory equation of the outer center P of △ ABC is

Let P (x, y) B (0, y + 1)
|BP|=|AP|
x^+1=(4-x)^+y^
y^-8x+15=0

Take any three points in a circle and find out the probability that the triangle formed by three points is an acute triangle
Take any three points in a circle and find out the probability that the triangle formed by the three points is an acute triangle

Take any point a in the circle, and let a be the diameter n of the diameter m and the vertical m (i.e. the circle is divided into four equal parts). Suppose it is divided into four parts: 1, 2, 3 and 4. The other two points a and B can make the triangle ABC an acute triangle or a right triangle only in the opposite part (i.e. 1, 3 or 2, 4). It is not difficult to calculate its probability as 0.25