Question

The line passing through \[( - 1,\pi /2)\] and perpendicular to \[\sqrt 3 \sin \theta + 2\cos \theta = \dfrac{4}{r}\] be
A. \[2 = \sqrt 3 r\cos \theta - 2r\sin \theta \]
В. \[5 = - 2\sqrt 3 r\sin \theta + 4r\cos \theta \]
C. \[2 = \sqrt 3 r\cos \theta + 2r\cos \theta \]
D. \[5 = 2\sqrt 3 r\sin \theta + 4r\cos \theta \]

Answer 1

Hint: In the example above, we need to write the equation of the line perpendicular to the supplied line because it is in parametric form. Additionally, we have provided the point to make it simple to locate the equation's constant term.

Formula Used: To do this, we will apply the idea that by substituting \[\theta \] for
\[\left( {{{90}^o} + \theta } \right)\]
We will obtain the equation of the line in parametric form perpendicular to it.

Complete step by step solution: We have been given the equation that it is,
Passing through the point
\[( - 1,\pi /2)\]
And it is perpendicular to
\[\sqrt 3 \sin \theta + 2\cos \theta = \dfrac{4}{r}\]
Now, we can write the given equation as,
\[\sqrt 3 {\rm{r}}\sin \theta + 2{\rm{r}}\cos \theta = 4\]
As, we know that
\[r\sin \theta = y,r\cos \theta = x\]
Now, we can write the equation as,
\[ \Rightarrow \sqrt 3 {\rm{y}} + 2{\rm{x}} = 4\]
Hence, from the above equation, the slope may be
\[ = \tan \theta = \dfrac{{ - 2}}{{\sqrt 3 }}\]
As we already knew that the lines are perpendicular, we have
\[{{\rm{m}}_1}\;{{\rm{m}}_2} = - 1\]
Then, the slope of the line perpendicular to it will become,
\[ = \dfrac{{ - 1}}{{\tan \theta }} = \dfrac{{\sqrt 3 }}{2}\]
Now, we have the given point as,
\[(r,\theta ) \equiv \left( { - 1,\dfrac{\pi }{2}} \right)\]
Hence, from the above expression, we get
\[r = - 1,\theta = \dfrac{\pi }{2}\]
Now, we have the below equations
\[{\rm{x}} = {\rm{r}}\cos \theta = - 1*0 = 0\]
\[{\rm{y}} = {\rm{r}}\sin \theta = - 1*1 = - 1\]
Now, the equation of line with slope \[\dfrac{{\sqrt 3 }}{2}\] through \[(0, - 1)\] is
\[(y + 1) = \dfrac{{\sqrt 3 }}{2}(x - 0)\]
On simplifying the above equation to make it less complicated, we get
\[ \Rightarrow \sqrt 3 x - 2y = 2\]
In polar form, we obtained as
\[ \Rightarrow \sqrt 3 {\rm{r}}\cos \theta - 2{\rm{r}}\sin \theta = 2\]
Therefore, the line passing through \[( - 1,\pi /2)\] and perpendicular to \[\sqrt 3 \sin \theta + 2\cos \theta = \dfrac{4}{r}\] be \[2 = \sqrt 3 r\cos \theta - 2r\sin \theta \]

Option ‘A’ is correct

Note: By changing the equation from parametric to coordinate form, we can answer the question. The given point is in the form of \[(r,\theta )\] and can also be converted into coordinates by replacing x and y with their respective \[x \to rcos\theta \] and \[y \to rsin\theta \] values.