Question

What is the distance between `(3 ,( 5 pi)/12 )` and `(-2 , ( 3 pi )/2 )`?

Answer 1

The distance between the two points is approximately `1.18` units.

Explanation:

You can find the distance between two points using the Pythagorean theorem `c^2 = a^2 + b^2`, where `c` is the distance between the points (this is what you're looking for), `a` is the distance between the points in the `x` direction and `b` is the distance between the points in the `y` direction.

To find the distance between the points in the `x` and `y` directions, first convert the polar co-ordinates you have here, in form `(r,\theta)`, to Cartesian co-ordinates.

The equations that transform between polar and Cartesian co-ordinates are:

`x = r cos\theta`
`y = r sin \theta`

Converting the first point
`x = 3 cos(\frac{5\pi}{12})`
`x = 0.77646`

`y = 3 sin (\frac{5\pi}{12})`
`y = 2.8978`

Cartesian co-ordinate of first point: `(0.776, 2.90)`

Converting the second point
`x = -2 cos(\frac{3\pi}{2})`
`x = 0`

`y = -2 sin (\frac{3\pi}{2})`
`y = 2`

Cartesian co-ordinate of first point: `(0, 2)`

Calculating `a`
Distance in the `x` direction is therefore `0.776-0 = 0.776`

Calculating `b`
Distance in the `y` direction is therefore `2.90-2 = 0.90`

Calculating `c`
Distance between the two points is therefore `c`, where
`c^2 = a^2 + b^2`
`c^2 = 0.776^2 + 0.9^2`
`c^2 = 1.4122`
`c = 1.1884`
`c \approx 1.18`

The distance between the two points is approximately `1.18` units.

The diagrams about halfway down this page, in the section 'Vector addition using components' might be useful in understanding the process just performed.