Respuesta :

So, the distance between two points (x1,y1) and (x2,y2) is:

d = sqrt( (x2-x1)^2 + (y2-y)^2 )

Some times is better the square of the distance:

d^2 = (x2-x1)^2 + (y2-y)^2

Now, in your case:

5^2 = (0-2)^2 + ( -6-k)^2 = (-2)^2 + (6+k)^2 = 4 + (6+k)^2

Notice the little thing about (-6-k)^2 = (6+k)^2 (less '-') Finally:

25 = 4 + (6+k)^2 ===> (6+k)^2 = 21,

6+k  = sqrt(21)

Notice we did not expand (6+k)^2! No need this time. But now, the tricky part!

6+k=+sqrt(21) and 6+k=-sqrt(21), because both work (never forget that sqrt come with +/- when looking for solutions!)

k = -6 + sqrt(21) and k=-6-sqrt(21).

Both are valid answers. Some times one is not good because you need the solution be positive (for instance), but here, both are good.


facundo3141592

Given the points (2,k) and (0,−6) for which values of k would the distance between the points be  √5 ?

We will find that there are two possible values of k;

k = -5 and k = -7.

We know that 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]

Here we have the points (2,k) and (0,−6), then the distance will be given by:

[tex]d = \sqrt{(2 - 0)^2 + (k - (-6))^2} = \sqrt{4 + (k + 6)^2}[/tex]

Now we want to find the value of k such that the distance is equal to √5, then we need to solve:

[tex]\sqrt{5} = \sqrt{4 + (k + 6)^2}[/tex]

removing the square root in both sides we get:

[tex]5 = 4 + (k + 6)^2\\\\5 - 4 = (k + 6)^2\\\\1 = (k + 6)^2[/tex]

Then we can see that (k + 6) = ± 1, solving these two equations (one for each sign) we get:

k + 6 = 1

k = 1 - 6 = -5

and

k + 6 = -1

k = -1 - 6 = -7

Then we found two possible values of k:

k = -5

k = -7

Below you can see the graph of the points A = (2, -5), B = (0, -6) and C = (2, -7)

If you want to learn more about distances, you can read:

https://brainly.com/question/5057237

Ver imagen facundo3141592

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