这道题如何解---反比例函数面积问题 【4】如图，一次函数 y₁ = kx + b (k ≠ 0) 的图象与反比例函数 y₂ = m/x (m ≠ 0) 的图象相交于 A (1, 3), B (n, -1) 两点. (1) 求一次函数和反比例函数的表达式; (2) 根据图象，直接写出 y₁ > y₂ 时，x 的取值范围; (3) 过点 B 作直线 OB, 交反比例函数图象于点 C, 连接 AC, 求 △ABC 的面积. **Graph Description:** * **Type:** Coordinate plane graph showing a linear function and an inverse proportion function. * **Coordinate Axes:** X-axis and Y-axis are shown, intersecting at the origin O, labeled (0, 0). Arrows indicate the positive direction of both axes. * **Functions:** * A straight line representing the linear function y₁ = kx + b. It passes through points A and B. * A curve representing the inverse proportion function y₂ = m/x. It consists of two branches, one in the first quadrant and one in the third quadrant. The line intersects this curve at points A and B. * **Points:** * Point A is labeled in the first quadrant, located at the intersection of the line and the curve. Its coordinates are given as (1, 3). * Point B is labeled in the third quadrant, located at the intersection of the line and the curve. Its coordinates are given as (n, -1). * Point C is labeled in the first quadrant, located on the curve. It is the intersection point of the line OB (passing through O and B) and the inverse proportion function graph. * Point O is the origin (0, 0). * **Lines/Segments:** * The straight line passes through A and B. * The curve is the graph of the inverse proportion function. * A line segment connects A and C. * A line passes through O and B and intersects the curve at C. This implies O, B, and C are collinear.