What is the distance between (4 ,( 7 pi)/6 )(4,7π6) and (-1 ,( 3pi )/2 )(1,3π2)?

1 Answer
Mar 23, 2018

The distance between the two points is sqrt(3)3 units

Explanation:

To find the distance between these two points, first convert them into regular coordinates. Now, if (r,x)(r,x) are the coordinates in polar form, then the coordinates in regular form are (rcosx,rsinx)(rcosx,rsinx).
Take the first point (4,(7pi)/6)(4,7π6).
This becomes (4cos((7pi)/6),4sin((7pi)/6))(4cos(7π6),4sin(7π6))

=(-2sqrt(3),-2)(23,2)

The second point is (-1,(3pi)/2)(1,3π2)
This becomes (-1cos((3pi)/2),-1sin((3pi)/2))(1cos(3π2),1sin(3π2))

=(0,1)(0,1)

So now the two points are (-2sqrt(3),-2)(23,2) and (0,1)(0,1). Now we can use the distance formula
d=sqrt((-2sqrt(3)-0)^2 - (-2-1)^2)d=(230)2(21)2

=sqrt(12-9)129

=sqrt(3)3