Question

If the equation of one side of a rectangle is $3\cos \theta +4\sin \theta =\frac{7}{r}$ and its two vertices are $A(\sqrt{2},\frac{\pi }{4})$, and $B(\sqrt{50},{\tan }^{-1}\frac{1}{7})$, then the equation of the diagonal of the rectangle which passes through the vertex $B$, is

• A. $4\cos \theta -3\sin \theta =\frac{1}{r}$
• B. $4\cos \theta -3\sin \theta =\frac{\sqrt{2}}{r}$
• C. $\sin \theta =\frac{1}{r}$
• D. $\cos \theta =\frac{1}{r}$

Answer 1

The correct option is C $\sin \theta =\frac{1}{r}$

$3\cos \theta +4\sin \theta =\frac{7}{r}$
or, $3x+4y=7$
Slope of this line is $\frac{-3}{4}$
Given points are $A(\sqrt{2},\frac{\pi }{4}),B(\sqrt{50},{\tan }^{-1}\frac{1}{7})$
Cartesian form of $A(\sqrt{2},\frac{\pi }{4})$ is $x=1,y=1;i.e.A(1,1)$
Cartesian form of $B(\sqrt{50},{\tan }^{-1}\frac{1}{7})$ is $x=7,y=1;i.e.B(7,1)$
Slope of the line joining $A$ and $B$ is $0$.
So, $AB$ is the diagonal of the rectangle, not the side of rectangle.
Equation of $AB$ is
$y-1=0\Rightarrow y=1$
$r\sin \theta =1$

$\Rightarrow \sin \theta =\frac{1}{r}$

Hence, option C is the correct answer.