Question

Find the area of the parallelogram whose sides are given by \[x\cos \alpha +y\sin \alpha =p\] , \[x\cos \alpha +y\sin \alpha =q\] , \[x\cos \beta +y\sin \beta =r\] and \[x\cos \beta +y\sin \beta =s\].
(a) \[\pm \left( p-q \right)\left( r-s \right)\operatorname{cosec}\left( \alpha -\beta \right)\]
(b) \[\left( p+q \right)\left( r-s \right)\operatorname{cosec}\left( \alpha +\beta \right)\]
(c) \[\left( p+q \right)\left( r+s \right)\operatorname{cosec}\left( \alpha -\beta \right)\]
(d) None of these

Answer 1

Hint: The only formula we are going to use is the area of a parallelogram formed by 4 straight lines: a1x+b1y+c1=0,
a1x+b1y+d1=0, a2x+b2y+c2=0 and a2x+b2y+d2=0.
Area of the parallelogram = \[\dfrac{|{{c}_{1}}-{{d}_{1}}||{{c}_{2}}-{{d}_{2}}|}{\left| \det \left( \begin{matrix}
   {{a}_{1}} & {{b}_{1}} \\
   {{a}_{2}} & {{b}_{2}} \\
\end{matrix} \right) \right|}\]

Complete step-by-step solution -
Let us begin our solution with these things in mind.
Now the given lines are \[x\cos \alpha +y\sin \alpha =p\] , \[x\cos \alpha +y\sin \alpha =q\] , \[x\cos \beta +y\sin \beta =r\] and \[x\cos \beta +y\sin \beta =s\].
Let’s bring all the terms to one side.
\[x\cos \alpha +y\sin \alpha -p=0\] , \[x\cos \alpha +y\sin \alpha -q=0\] , \[x\cos \beta +y\sin \beta -r=0\] and
\[x\cos \beta +y\sin \beta -s=0\]
Comparing the given lines with the lines a1x+b1y+c1=0, a1x+b1y+d1=0, a2x+b2y+c2=0 and a2x+b2y+d2=0, we get,
a1 = \[\cos \alpha \] , b1 = \[sin\alpha \] , a2 = \[cos\beta \] , b2 = \[sin\beta \] , c1 = -p, d1 = -q, c2 = -r and d2 = -s
Now substituting all these terms in the formula we saw earlier we get,
Area = \[\dfrac{|-p-(-q)||-r-(-s)|}{\left| \det \left( \begin{matrix}
   \cos \alpha & \sin \alpha \\
   \cos \beta & \sin \beta \\
\end{matrix} \right) \right|}\]
Area = \[\dfrac{|-(p-q)||-(r-s)|}{\left| \cos \alpha \sin \beta -\sin \alpha \cos \beta \right|}\]
Modulus eats up the negative sign. Also \[\cos \alpha \sin \beta -\sin \alpha \cos \beta \] = \[\sin (\beta -\alpha )\]
So finally we have,
Area = \[\dfrac{(p-q)(r-s)}{\left| \sin (\beta -\alpha ) \right|}\]
Now let us write \[\sin (\beta -\alpha )=-\sin (\alpha -\beta )\] as sin of a negative angle is actually the negative sine of that angle taken as positive.
Area = \[\dfrac{(p-q)(r-s)}{\left| -\sin (\alpha -\beta ) \right|}\]
Modulus eats up the negative sign again. We also know that \[\dfrac{1}{\sin (\alpha -\beta )}=\operatorname{cosec}(\alpha -\beta )\] .So finally we have to get,
Area = \[(p-q)(r-s)\operatorname{cosec}(\alpha -\beta )\]
Thus the option(a) is satisfied.

Note: You might get really confused with the answer. How is it option(a)?
Have a look here.
But mathematically \[\pm \left( p-q \right)\left( r-s \right)\operatorname{cosec}\left( \alpha -\beta \right)\] actually means
+ \[\left( p-q \right)\left( r-s \right)\operatorname{cosec}\left( \alpha -\beta \right)\] or - \[\left( p-q \right)\left( r-s \right)\operatorname{cosec}\left( \alpha -\beta \right)\] .But we know that the area is always a positive quantity. So we can mark option(a) without any confusion as it has a “or” in between and not “and”.
“Or” means any one of the given options. So you can go with option(a).