Respuesta :
The distance between the two points is [tex]d=8.6m[/tex]
The polar coordinate of A is [tex]\left(4.47,296.57\right)[/tex]
The polar coordinate of B is [tex]\left(4.24,135\right)[/tex]
Explanation:
The two points are [tex]A(2,-4)[/tex] and [tex]B(-3,3)[/tex]
The distance between two points is given by,
[tex]d=\sqrt{(2+3)^{2}+(-4-3)^{2}}\\d=\sqrt{(5)^{2}+(-7)^{2}}\\d=\sqrt{25+49}\\d=8.6[/tex]
Thus, the distance between the two points is [tex]d=8.6m[/tex]
The polar coordinates of A can be written as [tex](Distance, tan^{-1} \frac{y}{x} )[/tex]
Distance = [tex]\sqrt{x^{2} +y^{2} }[/tex]
Substituting [tex]A(2,-4)[/tex], we get,
Distance = [tex]\sqrt{2^{2} +(-4)^{2} }=\sqrt{4+16 }=4.47[/tex]
[tex]tan^{-1} \frac{y}{x} =tan^{-1} \frac{-4}{2}=-63.43[/tex]
To make the angle positive, let us add 360,
[tex]\theta=360-63.43=296.57[/tex]
The polar coordinate of A is [tex]\left(4.47,296.57\right)[/tex]
Similarly, The polar coordinate of B can be written as [tex](Distance, tan^{-1} \frac{y}{x} )[/tex]
Distance = [tex]\sqrt{x^{2} +y^{2} }[/tex]
Substituting [tex]B(-3,3)[/tex], we get,
Distance = [tex]\sqrt{(3)^{2}+(3)^{2}}=4.24[/tex]
[tex]tan^{-1} \frac{y}{x} =tan^{-1} \frac{3}{-3}=-45[/tex]
To make the angle positive, let us add 360,
[tex]\theta=180-45=135^{\circ}[/tex]
The polar coordinate of B is [tex]\left(4.24,135\right)[/tex]
