JEE MainMathematicsEasyStraight Lines

The area of the parallelogram contained by the lines 4y - 3x +
1 = 0, 4y - 3x - 1 = 0, 3y - 4x + 1 = 0 and 3y - 4x + 2 = 0 is
$\frac{p}{q}$ , where p and q are prime, then the value of
$\left| p^{2}\ –\ q^{2} \right|$ is.

Show answer and solution

Area

= \frac{2}{7} = \frac{p}{q}

\therefore

p + q = 9

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

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