JEE MainMathematicsModerateStraight Lines

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

Show answer and solution

4^2 + 3^2 = 5^2, 5^2 + 12^2 = 13^2, 7^2 + 24^2 = 25^2
Hence, the distance of (0, 0) from the three sides are

\frac{10}{5},\frac{26}{13},\frac{50}{25}

i.e. each 2

Source: Grass-G / M-11 Mathematics Paper-2 1Nov.docx

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