If a variable line passes through the point of intersection of the lines x+2y-1=0 and 2x-y-1=0 and meets the coordinate axes in A and B, then what is the locus of the mid-point of AB?

Explanation

First find the intersection of x+2y=1 and 2x-y=1. Multiplying the second by 2: 4x-2y=2. Adding them: 5x=3 \implies x=\frac{3}{5}. Then y = 2x-1 = \frac{1}{5}. The intersection point is (\frac{3}{5}, \frac{1}{5}). Let the mid-point of AB be (h,k). Since A is on the x-axis and B is on the y-axis, A is (2h, 0) and B is (0, 2k). The equation of AB using intercepts is \frac{x}{2h} + \frac{y}{2k} = 1. Since the line passes through (\frac{3}{5}, \frac{1}{5}), we substitute: \frac{3/5}{2h} + \frac{1/5}{2k} = 1 \implies \frac{3}{10h} + \frac{1}{10k} = 1. Multiplying by 10hk, we get 3k + h = 10hk. Therefore, the locus is x+3y=10xy.

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