JEE MainMathematicsToughStraight Lines

Locus of the points which are at equal distance from 3x + 4y -
11 = 0 and 12x + 5y + 2 = 0 and which is nearer to origin is
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2 = 0 ls cjkcj nwjh ij

Show answer and solution

\left( \frac{3x + 4y - 11}{5} \right) = - \left( \frac{12x + 5y + 2}{13} \right)

is required bisector as the angle contains origin

\left( \frac{3x + 4y - 11}{5} \right) = - \left( \frac{12x + 5y + 2}{13} \right)

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Source: Grass-G / M-25 Mathematics Paper 7 Jan.docx

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