As shown in the figure, in the plane rectangular coordinate system, the endpoint coordinates of line AB are a (- 2,4), B (4,2), and the line y = k X-2 has an intersection with line AB, then the value range of K is

As shown in the figure, in the plane rectangular coordinate system, the endpoint coordinates of line AB are a (- 2,4), B (4,2), and the line y = k X-2 has an intersection with line AB, then the value range of K is

Substituting a (- 2,4) into y = kx-2, we get 4 = - 2k-2 and K = - 3,
When the line y = kx-2 intersects the line AB and passes the second and fourth quadrants, the condition K satisfies is k ≤ - 3;
By substituting B (4,2) into y = kx-2, we get 4k-2 = 2, and the solution is k = 1,
When the line y = kx-2 intersects the line AB and passes the first and third quadrants, K satisfies the condition that K ≥ 1
That is k ≤ - 3 or K ≥ 1
So if the line y = kx-2 intersects the line AB, then the value of K cannot be - 2
So choose B

As shown in the figure, in the plane rectangular coordinate system, the endpoint coordinates of line AB are a (- 2,4), B (4,2), and the line y = kx-2 has an intersection with line ab. please write a possible value of K______ ．

∵ the line y = kx-2 intersects with line AB, and the coordinates of point B satisfy y = kx-2, ∵ 4k-2 = 2, ∵ k = 1

In the plane rectangular coordinate system, a (- 2,2), B (3,4), C (0, - 1) are known. If the line y = KX + B passes through point C and has an intersection with line AB, then the value range of K is

The straight line y = KX + B passes through point C, that is, B = - 1,
Let f (x) = kx-1, and have the intersection point with the line ab
(f (- 2) - 2) * (f (3) - 4) ≤ 0, that is (- 2k-3) * (3k-5) ≤ 0,
The solution is k ≤ - 3 / 2, K ≥ 5 / 3
K value range (- ∞, - 3 / 2] u [5 / 3, + ∞)

In the plane rectangular coordinate system, a (1,0) B (0, - 2) rotates the line segment AB 90 degrees counterclockwise around point a to AC to find the coordinate of C

Let C (x, y) be in the fourth quadrant
The slope of AB is 2, ab ⊥ AC
So the slope of AC is - 1 / 2
Y / (x-1) = - 1 / 2 (1)
|ab|=|ac|
(x-1)²+y²=5 (2)
The simultaneous solution is x = 3, y = - 1
So C (3, - 1)