在学习直线知识的过程中,经常涉及到直线与线段相交的问题.例1已知两点A(-2,2),B(2,-2)与直线l:y=kx+4,若l与线段AB有交点,求k的取值范围.解法1线段AB的方程为y-2-2-2=x+22+2,即y=-x(-2≤x≤2),由y=kx+4,y=-x解得x=-4k+1,令-2≤-4k+1≤2,解得k≤-3或k≥1.图1解法2图解法2如图1 In the process of learning linear knowledge, it often involves the intersection of a straight line and a line segment. Example 1 Known two points A(-2,2), B(2,-2) and a straight line l:y=kx+4, if l There is an intersection with the line segment AB and the range of the value of k is found. The equation of the line AB of the solution 1 line is y-2-2-2=x+22+2, that is, y=-x (-2≤x≤2). y=kx+4,y=-x solution x=-4k+1, let -2≤-4k+1≤2, solve k≤-3 or k≥1. Figure 1 Solution 2 Graphic Method 2 1