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在解平面解析几何题时,常常会遇到过两曲线交点求一新曲线方程的问题,使用曲线系方程解这类问题是一种比较好的方法,此方法具有思路清晰、运算简捷等优点。下面用几个例子说明以上观点。例1.求过两直线x-2y+3=0和x+2y-9=0的交点和原点的直线方程。解:过交点的直线系为 x-2y+3+λ(x+2y-9)=0。∴ (1+λ)x+(2λ-2)y+3-9λ=0。∵直线过原点(0,0),故得3-9λ=0,∴λ=1/3。∴直线方程为(1+1/3)x+(2·1/3-2)y+3-9·1/3=0, ∴ x-y=0为所求。
When solving geometric problems in the solution plane, we often encounter the problem of finding a new curve equation at the intersection of two curves. It is a better method to solve such problems by using the curve system equation. This method has the advantages of clear thinking and simple operation. . Here are a few examples to illustrate the above points. Example 1. Find the linear equation of the intersection and origin of two straight lines x-2y+3=0 and x+2y-9=0. Solution: The straight line passing through the intersection is x-2y+3+λ(x+2y-9)=0. ∴ (1+λ)x+(2λ-2)y+3-9λ=0. The straight line passes through the origin (0,0), so 3-9λ=0 and ∴λ=1/3. ∴ The linear equation is (1+1/3)x+(2·1/3-2)y+3-9·1/3=0, and ∴ x-y=0 is required.