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रेखाओं ${a_1}x + {b_1}y + {c_1} = 0$,${a_1}x + {b_1}y + {d_1} = 0$ व ${a_2}x + {b_2}y + {c_2} = 0$, ${a_2}x + {b_2}y + {d_2} = 0$ से निर्मित समान्तर चतुभुज का क्षेत्रफल होगा
$\frac{{({d_1} - {c_1})({d_2} - {c_2})}}{{{{[(a_1^2 + b_1^2)(a_2^2 + b_2^2)]}^{1/2}}}}$
$\frac{{({d_1} - {c_1})({d_2} - {c_2})}}{{{a_1}{a_2} - {b_1}{b_2}}}$
$\frac{{({d_1} + {c_1})({d_2} + {c_2})}}{{{a_1}{a_2} + {b_1}{b_2}}}$
$\frac{{({d_1} - {c_1})({d_2} - {c_2})}}{{{a_1}{b_2} - {a_2}{b_1}}}$
Solution
(d)क्षेत्रफल $ = \frac{{{p_1}{p_2}}}{{\sin \theta }} = {p_1}{p_2}{\rm{cosec}}\theta $ ,
जहाँ ${p_1} = \frac{{{d_1} – {c_1}}}{{\sqrt {(a_1^2 + b_1^2)} }},{p_2} = \frac{{{d_2} – {c_2}}}{{\sqrt {(a_2^2 + b_2^2)} }}$
साथ ही, $\tan \theta = \frac{{{a_1}{b_2} – {a_2}{b_1}}}{{{a_1}{a_2} + {b_1}{b_2}}}$
${\rm{cose}}{{\rm{c}}^2}\theta = 1 + {\cot ^2}\theta $
$\therefore $ ${\rm{cose}}{{\rm{c}}^2}\theta = \frac{{{{({a_1}{a_2} + {b_1}{b_2})}^2} + {{({a_1}{b_2} – {a_2}{b_1})}^2}}}{{{{({a_1}{b_2} – {a_2}{b_1})}^2}}}$
$ = \frac{{(a_1^2 + b_1^2)(a_2^2 + b_2^2)}}{{{{({a_1}{b_2} – {a_2}{b_1})}^2}}}$
${p_1},{p_2}$ व ${\rm{cosec}}\theta $ का मान रखने पर,
अभीष्ट क्षेत्रफल =$\frac{{({d_1} – {c_1})({d_2} – {c_2})}}{{{a_1}{b_2} – {a_2}{b_1}}}$.
