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A point equidistant from the lines 4x + 3y + 10 = 0, 5x – 12y + 26 = 0 and 7x + 24y – 50 = 0 is

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A point equidistant from the lines 4x + 3y + 10 = 0, 5x – 12y + 26 = 0 and 7x + 24y – 50 = 0 is

(A) (1, –1)

(B) (1, 1) 

(C) (0, 0) 

(D) (0, 1)

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Answer is (c)

Clearly distance of each of three lines from (0, 0) is 2 units

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