Question

What are the polar coordinates of the point with $( - 3,3)$ rectangle coordinate?
(A) $(3,135)$
(B) $( - 3,135)$
(C) $(3\sqrt 2 ,45)$
(D) $(3\sqrt 2 ,135)$
(E) $( - 3\sqrt 2 ,45)$

Answer 1

Hint : A polar coordinate system is a two dimensional coordinate system in which each point on a plane is determined by a distance from the reference point and an angle from a reference direction. Polar coordinates is a pair of coordinates locating the position of a point in a plane, first being the distance from a fixed point on a line and the second is the angle made by this line with a fixed line. In Cartesian coordinates there is exactly one set of coordinates for a given point. Cartesian coordinates are written as $(x,y)$ whereas polar coordinates are written as $(r,\theta )$ .
Here $x,y$ are the respective values of abscissa and ordinate .
So ,$r = \sqrt {{x^2} + {y^2}}$ ,
$\tan \theta = \dfrac{y}{x}$
Therefore, $\theta = {\tan ^{ - 1}}\dfrac{y}{x}$

Complete step-by-step answer :
We know Cartesian coordinates are represented by $(x,y)$ where $x$ is the abscissa and $y$ is the ordinate.
Here, in this question, the given coordinates are $( - 3,3)$ .
So the abscissa ,$x = - 3$.
And the ordinate ,$y = 3$.
Now we have to convert it to polar coordinates $(r,\theta )$ .
Where $r$ is the distance from a fixed point of a line and $\theta$ is the angle made by this line with a fixed line.
We know ,$r = \sqrt {{x^2} + {y^2}}$ .
Putting the value of $x$ and $y$ in the formula of $r$ , we get,
So, $r = \sqrt {{{( - 3)}^2} + {{(3)}^2}}$ .
$\Rightarrow r = \sqrt {9 + 9}$
$\Rightarrow r = 3\sqrt 2$
We know ,$\tan \theta = \dfrac{y}{x}$
Therefore $\theta = {\tan ^{ - 1}}\dfrac{y}{x}$
Putting the value of $x$ and $y$ in the formula of $\theta$ , we get.
So, $\theta = {\tan ^{ - 1}}\dfrac{{ - 3}}{3}$
$\Rightarrow \theta = {\tan ^{ - 1}}( - 1)$
$\Rightarrow \theta = \dfrac{{3\pi }}{4} = \dfrac{{7\pi }}{4}$ .
Since, $\tan \dfrac{{3\pi }}{4} = - 1$ and $\tan \dfrac{{7\pi }}{4} = - 1$ .
As $( - 3,3)$ lies in 2nd quadrant so, $\theta = \dfrac{{3\pi }}{4}$ ,Since in 2nd quadrant $x$ is negative and $y$ is positive .
Therefore, $r = 3\sqrt 2$
$\theta = \dfrac{{3\pi }}{4}$
$\Rightarrow \theta = \dfrac{{3 \times 180}}{4}$
hence, $\theta = 135^\circ$
So ,The polar coordinates of the point with $( - 3,3)$ rectangle coordinates are $(3\sqrt 2 ,135^\circ )$ .
So, the correct answer is “Option D”.

Note : We should be careful in calculating the value of $\theta$ as it depends on the quadrant. Avoid mistakes regarding the sign of the value of abscissa and ordinate. In the first quadrant , the value of all trigonometric identities is always positive . Whereas in the second quadrant the value of sin and cosec is always positive , in the third quadrant the value of tan and cot is always positive and in the fourth quadrant the value of cos and sec is always positive.