Given that the image of a function y = KX + B passes through points m (- 1,1) and (0,2), let the image intersect with the x-axis at point a and with the y-axis at point B: Q: on the x-axis Given that the image of a function y = KX + B passes through point m (- 1,1) and point (0,2), let the image intersect with X axis at point a and Y axis at point B: Q: is there a point P on X axis, so that the triangle ZBP is an isosceles triangle? If it exists, the coordinates of all the points P that meet the conditions can be obtained. If not, please give reasons

Given that the image of a function y = KX + B passes through points m (- 1,1) and (0,2), let the image intersect with the x-axis at point a and with the y-axis at point B: Q: on the x-axis Given that the image of a function y = KX + B passes through point m (- 1,1) and point (0,2), let the image intersect with X axis at point a and Y axis at point B: Q: is there a point P on X axis, so that the triangle ZBP is an isosceles triangle? If it exists, the coordinates of all the points P that meet the conditions can be obtained. If not, please give reasons

I think your triangle ZBP should be ABP
Existence point p;
First of all, according to the oblique section of the first-order function y = KX + B, through M (- 1,1) and (0,2), the point is substituted to obtain the expression of the first-order function: y = x + 2; point a is (- 2,0), point B is (0,2)
There are three cases: 1. The isosceles triangle ABP satisfies AB = AP, and the point P is on the left side of point A. at this time, the distance between point P and the origin is op = 2 + AP = 2 + AB, because AB is obtained by Pythagorean theorem: ab = 2 √ 2, substituting to get OP = 2 + 2 √ 2, because point P is on the negative half axis, so point P is (- 2-2 √ 2,0)
2. The isosceles triangle ABP satisfies AB = BP, and at this time P and a are symmetric about y axis. So p is (2,0)
3. Isosceles triangle ABP, satisfy AP = BP, then P point is at the origin, namely (0,0)