- Home
- Standard 11
- Mathematics
Area of the parallelogram formed by the lines ${a_1}x + {b_1}y + {c_1} = 0$,${a_1}x + {b_1}y + {d_1} = 0$and ${a_2}x + {b_2}y + {c_2} = 0$, ${a_2}x + {b_2}y + {d_2} = 0$is
$\frac{{({d_1} - {c_1})({d_2} - {c_2})}}{{{{[(a_1^2 + b_1^2)(a_2^2 + b_2^2)]}^{1/2}}}}$
$\frac{{({d_1} - {c_1})({d_2} - {c_2})}}{{{a_1}{a_2} - {b_1}{b_2}}}$
$\frac{{({d_1} + {c_1})({d_2} + {c_2})}}{{{a_1}{a_2} + {b_1}{b_2}}}$
$\frac{{({d_1} - {c_1})({d_2} - {c_2})}}{{{a_1}{b_2} - {a_2}{b_1}}}$
Solution
(d) Area $ = \frac{{{p_1}{p_2}}}{{\sin \theta }} = {p_1}{p_2}{\rm{cosec}}\theta $ ,
where ${p_1} = \frac{{{d_1} – {c_1}}}{{\sqrt {(a_1^2 + b_1^2)} }},{p_2} = \frac{{{d_2} – {c_2}}}{{\sqrt {(a_2^2 + b_2^2)} }}$
Also $\tan \theta = \frac{{{a_1}{b_2} – {a_2}{b_1}}}{{{a_1}{a_2} + {b_1}{b_2}}}$. Since ${\rm{cose}}{{\rm{c}}^2}\theta = 1 + {\cot ^2}\theta $
or ${\rm{cose}}{{\rm{c}}^2}\theta = \frac{{{{({a_1}{a_2} + {b_1}{b_2})}^2} + {{({a_1}{b_2} – {a_2}{b_1})}^2}}}{{{{({a_1}{b_2} – {a_2}{b_1})}^2}}}$
$ = \frac{{(a_1^2 + b_1^2)(a_2^2 + b_2^2)}}{{{{({a_1}{b_2} – {a_2}{b_1})}^2}}}$
Putting for ${p_1},{p_2}$ and ${\rm{cosec}}\theta $, we get
Area =$\frac{{({d_1} – {c_1})({d_2} – {c_2})}}{{{a_1}{b_2} – {a_2}{b_1}}}$.
