Question

two points in the xy plane have cartesian coordinates (2.00, #4.00) m and (#3.00, 3.00) m. determine (a) the distance between these points and (b) their polar coordinates

Answer 1

two points (2, 4) and (3, 3) are given in xy plane.
so, the distance between them is given by
$s=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$

so, distance = √{(2 - 3)² + (4 - 3)²} = √(1² + 1²) = √2

Let Cartesian coordinate is written as (x ,y)
distance of point from origin is r and $\theta$ is the angle made by line joining point (x,y) and (0,0)
r = √(x² + y²) and $\theta$ = tan^-1(y/x)

polar coordinate of (2,4)
r = √(2² + 4²) = 2√5
$\theta$ = tan^-1(4/2) = tan^-1(2)

hence, polar coordinate is {2√5, tan^-1(2)}


similarly, polar coordinate of (3,3)
r = √(3² + 3²) = 3√2
$\theta$ = tan^-1(3/3) = 45° or π/4
so, (3√2, π/4) is polar coordinate.

Answer 2

Given:

Point ( 2, 4 )

Point ( 3, 3 )

xy plane.

To find:

• The distance between these points.
• Polar coordinates

Solution:

By formula,

Distance = √ ( x1 + x2 )^2 + ( y1 + y2 )^2

Here,

x1 = 2

x2 = 3

y1 =4

y2 = 3

Substituting the values,

√ { ( 2 - 3 )^2 + ( 4 - 3 )^2 }

√ ( 1^2 + 1^2 )

√2

Hence, Distance = √2

To find the distance of point from origin,

Distance = √ ( x^2 + y^2 )

tan^-1 ( y / x )

Polar coordinate = ( Distance , tan^-1 ( y / x ) )

To find the polar coordinate of ( 2, 4 )

Polar coordinate = √ ( 2^2 + 4^2 )

2√5

tan^-1 ( 4 / 2 )

tan^-1 ( 2 )

Hence, polar coordinate of ( 2, 4 ) is ( 2√5, tan^-1 ( 2 ) )

To find the polar coordinate of ( 3, 3 )

Polar coordinate = √ ( 3^2 + 3^2 )

3√2

tan^-1 ( 3 / 3 )

π/4

Hence, the polar coordinate of  ( 3, 3 ) is (3√2, π/4)