Answer:
The distance between two points (x₁, y₁) and (x₂, y₂) is given by:
[tex]d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}[/tex]
Now we know that point D, which we can write as (x, y), is at a distance of 8 units from the origin.
Where the origin is written as (0, 0)
We also know that point D is 6 units away along the y-axis.
Then point D could be:
(x, 6)
or
(x, -6)
Now, let's find the x-value for each case, we need to solve:
[tex]8 = \sqrt{(x - 0)^2 + (\pm6 - 0)^2}[/tex]
notice that because we have an even power, we will get the same value of x, regardless of which y value we choose.
[tex]8 = \sqrt{x^2 + 36} \\\\8^2 = x^2 + 36\\64 - 36 = x^2\\28 = x^2\\\pm\sqrt{28} = x\\\pm 5.29 = x[/tex]
So we have two possible values of x.
x = 5.29
and
x = -5.29
Then the points that are at a distance of 8 units from the origin, and that are 6 units away along the y-axis are:
(5.29, 6)
(5.29, -6)
(-5.29, 6)
(-5.29, -6)
