Question

Two points in the Cartesian plane are A(2.00 m, −4.00 m) and B(−3.00 m, 3.00 m). Find the distance between them and their polar coordinates.

Answer 1

The distance between the two points is $d=8.6m$

The polar coordinate of A is $\left(4.47,296.57\right)$

The polar coordinate of B is $\left(4.24,135\right)$

Explanation:

The two points are $A(2,-4)$ and $B(-3,3)$

The distance between two points is given by,

$d=\sqrt{(2+3)^{2}+(-4-3)^{2}}\\d=\sqrt{(5)^{2}+(-7)^{2}}\\d=\sqrt{25+49}\\d=8.6$

Thus, the distance between the two points is $d=8.6m$

The polar coordinates of A can be written as $(Distance, tan^{-1} \frac{y}{x} )$

Distance = $\sqrt{x^{2} +y^{2} }$

Substituting $A(2,-4)$, we get,

Distance = $\sqrt{2^{2} +(-4)^{2} }=\sqrt{4+16 }=4.47$

$tan^{-1} \frac{y}{x} =tan^{-1} \frac{-4}{2}=-63.43$

To make the angle positive, let us add 360,

$\theta=360-63.43=296.57$

The polar coordinate of A is $\left(4.47,296.57\right)$

Similarly, The polar coordinate of B can be written as $(Distance, tan^{-1} \frac{y}{x} )$

Distance = $\sqrt{x^{2} +y^{2} }$

Substituting $B(-3,3)$, we get,

Distance = $\sqrt{(3)^{2}+(3)^{2}}=4.24$

$tan^{-1} \frac{y}{x} =tan^{-1} \frac{3}{-3}=-45$

To make the angle positive, let us add 360,

$\theta=180-45=135^{\circ}$

The polar coordinate of B is $\left(4.24,135\right)$