Respuesta :
Answer:
(5,-5) can be represented in polar form by
[tex](2\sqrt{5},315^\circ)[/tex] and [tex](-2\sqrt{5},135^\circ)[/tex]
Step-by-step explanation:
polar coordinates use a distance and an angle
it would be like (x,y) but x is distance from origin to point and y is the angle measured counterclockwise from the positive x-axis.
for (5,-5)
first find the distance to that point using distance formula
distance from (0,0) to (5,-5) is
[tex]D=\sqrt{(0-5)^2+(0-(-5))^2}[/tex]
[tex]D=\sqrt{25+25}[/tex]
[tex]D=5\sqrt{2}[/tex]
so our point has to be in the form [tex](x,y)[/tex] where [tex]\mid x\mid=5\sqrt{2}[/tex]
now finding the degree
using inverse tangent
[tex]tan^{-1}(\frac{-5}{5})=-45^\circ[/tex]
if we look on the graph, it is also 360-45=315 degrees from positive x axis
so one polar coordiante is [tex](2\sqrt{5},315^\circ)[/tex]
the other one is in the oposite side
we add or subtract 180 degrees and make the sign of x negative to go in the oposite direction
subtraction 180 to get 135 degrees
so the other point is [tex](-2\sqrt{5},135^\circ)[/tex]
(5,-5) can be represented in polar form by
[tex](2\sqrt{5},315^\circ)[/tex] and [tex](-2\sqrt{5},135^\circ)[/tex]
Answer:
Step-by-step explanation:
Alright, lets get started.
The given co-ordinates are (5,-5)
This is the form of (x,y).
We need to find this co-ordinate into polar form means in (r,Θ) form.
The formula for r is :
[tex]r=\sqrt{x^2+y^2}[/tex]
[tex]r=\sqrt{5^2+(-5)^{2} }[/tex]
[tex]r=\sqrt{25+25}[/tex]
[tex]r=\sqrt{50}[/tex]
[tex]r=5\sqrt{2}[/tex]
Now the Θ part
The formula for Θ is : Θ = [tex]tan^{-1}(\frac{y}{x} )[/tex]
Θ=[tex]tan^{-1}(\frac{-5}{5})[/tex]
Θ=[tex]tan^{-1}(-1)[/tex]
There are two angles within the intervals where tan is -1. That angles are 135 and 315.
So two polar co-ordinates are : [tex](5\sqrt{2}, 135)[/tex] and [tex](5\sqrt{2},315)[/tex] : Answer
Hope it will help :)
