Question

Area of parallelogram whose sides are $x\cos \alpha +y\sin \beta =p,x\cos \beta +y\sin \alpha =q,x\cos \beta +y\sin \beta =r$ and $x\cos \beta +y\sin \beta =s$ is

• A. $\pm (p-q)(r-s)co\sec (\alpha -\beta )$
• B. $(p-q)(r-s)co\sec (\alpha +\beta )$
• C. $(p-q)(r+s)co\sec (\alpha -\beta )$
• D. None of these

Answer 1

The correct option is A $\pm (p-q)(r-s)co\sec (\alpha -\beta )$

The equation of lines should be

$x\cos \alpha +y\sin \alpha =p$

$x\cos \alpha +y\sin \alpha =q$

$x\cos \beta +y\sin \beta =r$

$x\cos \beta +y\sin \beta =s$

So that pair of parallel lines are formed and parallelogram can be possible

now,

Let $\gamma$ is the required between the parallel lines

and $\tan \gamma =\frac{m_1-m_2}{1-m_1{m}_2}$

where ,

$m_1=-\frac{\cos \alpha }{\sin \alpha }$

&

$m_2=-\frac{\cos \beta }{\sin \beta }$

putting these values

$\tan \gamma =\frac{-\frac{\cos \alpha }{\sin \alpha }+\frac{\cos \beta }{\sin \beta }}{1+\frac{\cos \alpha }{\sin \alpha }\frac{\cos \beta }{\sin \beta }}$

$\tan \gamma =\frac{\frac{-\cos \alpha \sin \beta +\cos \beta \sin \alpha }{\sin \alpha \sin \beta }}{\frac{\sin \alpha \sin \beta +\cos \alpha \cos \beta }{\sin \alpha \sin \beta }}$

$\tan \gamma =\frac{\sin \alpha -\beta }{\cos \alpha -\beta }$

$\tan \gamma =\tan \alpha -\beta$

hence,

$\gamma =\alpha -\beta$

let's say $p_1$ & $p_2$ are the distance between two parallel lines

Area of parallelogram = $\frac{p_1\times p_2}{\sin \gamma }$

$p_1=\frac{|p-q|}{{\sqrt{\cos \alpha }}^2+\sin {\alpha }^2}$

i.e.,

$p_1=|p-q|$

similarly , $p_2=|r-s|$

Applying the formula of area

Area = $\frac{|p-q||r-s|}{\sin \alpha -\beta }$

$Area={+}_{-}(p-q)(r-s)co\sec \alpha -\beta$