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# Choose the correct answer from the given four options. A point equidistant from the lines $4x + 3y + 10 = 0, 5x – 12y + 26 = 0$ and $7x+24y-50=0$ is

$\begin {array} {1 1} (A)\;(1, -1) & \quad (B)\;(1,1) \\ (C)\;(0,0) & \quad (D)\;(0,1) \end {array}$