Question

The equation of two sides of a rectangle are $3\cos \theta +4\sin \theta +\frac{5}{r}=0,$ $4\cos \theta -3\sin \theta +\frac{15}{r}=0,$ and its one vertex is pole, then the area of the rectangle (in sq.units) is

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is C $3$

Given sides of rectangle
$3\cos \theta +4\sin \theta +\frac{5}{r}=0$
$3r\cos \theta +4r\sin \theta +5=0$ .....(i)
$4\cos \theta -3\sin \theta +\frac{15}{r}=0$
$4r\cos \theta -3r\sin \theta +15=0$ .....(ii)
Length of perpendicular from $(0,0)$ to (i)
$p_1=\frac{|5|}{5}=1$
Length of perpendicular from $(0,0)$ to (ii)
$p_1=\frac{|15|}{5}=3$
So, Area $=3$ sq.units