Prove that the area of the parallelogram formed by the lines `xcosalpha+ysinalpha=p ,xcosalpha+ys inalpha=q ,xcosbeta+ysinbeta=ra n dx cosbeta+ysinbeta=si s+-(p-q)(r-s)cos e c(alpha-beta)dot`

Options

A. `|((p-q)(r-s))/(cos(alpha-beta))|`
B. `|((p-q)(r-s))/(sin(alpha+beta))|`
C. `|((p-q)(r-s))/(sin(alpha-beta))|`
D. `|((p-q)(r-s))/(cos(alpha+beta))|`

Answer

C

Explanation

The equations of the sides of the parallelogram are :
`xcos alpha+ysin alpha-p=0`
`xcos alpha+ysin alpha-q=0`
`xcos beta + ysin beta -r =0`
and, `xcos beta+ysin beta-s=0`
`therefore ` Area of the parallelogram `=|(({-p)-(-q)}{(-r)-(-s)})/(|{:(cos alpha,sin alpha),(cos beta, sin beta):}|)|`
`implies` Area of the parallelogram `=|((q-p)(s-r))/((cos alpha sin beta-sin alpha cos beta))|`
`implies ` Area of parallelogram `=|((p-q)(r-s))/(sin (alpha-beta))|`