Question

The position vector of p (relative to the origin) is op = (x, y). if the magnitude of op is 5 units, find the set of all possible values of (x, y), with x, y E z​

Answer 1

S = {(0, 5), (0, - 5), (3, 4), (3, -4), (-3, 4), (-3, -4), (4, 3), (4, -3), (-4, 3), (-4, -3), (5,0), (-5, 0) }

Explanation:

Remember that the distance between two points (a, b) and (c, d) is given by:

`distance = √((a - c)^2 + (b - d)^2)`

So, here we have that the distance between the point (x, y) and the origin, (0, 0) must have a magnitude of 5 units, then we want to solve:

`5 = √((x - 0)^2 + (y - 0)^2) \\5 = √(x^2 + y^2)`

If we square both sides, we get

`5^2 = (√(x^2 + y^2) )^2\\25 = x^2 + y^2`

Now we want to find all the points (x, y) that meet this condition, suc that:

x ∈ Z

y ∈ Z

So both x and y must be integers.

So here we can just play with different values of x and y.

For example, if we define:

x = 0 we get:

25 = 0^2 + y^2

25 = y^2

√25 = y

Then we can have y = 5 or y = -5

from this we got two points:

(0, 5) and (0, - 5)

if x = 1 we have:

25 = 1^2 + y^2

25 - 1 = y^2

24 = y^2

There is no integer such that its square is equal to 24, so we can stop here.

if x = 2 or - 2, we have:

25 = 2^2 + y^2

25 = 4 + y^2

25 -4 = 21 = y^2

Again, there is no integer such that its square is equal to 21, so we can stop here.

if x = +3 or -3, we have:

25 = 3^2 + y^2

25 = 9 + y^2

25 - 9 = 16 = y^2

√16 = y

then we can have y = 4 or y = -4

from this we got four points:

(3, 4)

(-3, 4)

(3, -4)

(-3, -4)

And for symmetry, if x = 4 or -4 we have the points:

(-4, 3)

(4, 3)

(-4, -3)

(4, -3)

finally, again for symmetry, if we take x = 5 or x = -5 we have the points:

(5,0)

(-5, 0)

Concluding, the set of all possible values (x, y) is:

S = {(0, 5), (0, - 5), (3, 4), (3, -4), (-3, 4), (-3, -4), (4, 3), (4, -3), (-4, 3), (-4, -3), (5,0), (-5, 0) }