Answer:

The point (5, 2 + √6) and (5, 2 - √6) are equidistant from the point (-1, 2) and (3, 0)

Step-by-step explanation:

We want to find the value of k such that the point (5, k) is equidistant from the points (-1, 2) and (3, 0).

We can use the distance formula for this purpose:

The distance between two points in a Cartesian plane is given by the Pythagorean theorem, which states that the distance between two points (x1, y1) and (x2, y2) is the square root of ((x2 - x1)^2 + (y2 - y1)^2).

So, we have to find k such that:

sqrt((5 - (-1))^2 + (k - 2)^2) = sqrt((5 - 3)^2 + (k - 0)^2)

Squaring both sides and simplifying we can have:

(k-2)^2 = 2^2 + 2^2

k^2 - 4k + 4 = 8

k^2 - 4k - 4 = 0

Now we can use quadratic formula to find k,

k = (4 ± √(4^2 - 41(-4)))/2*1

k = (4 ± √24)/2

k = 2± √6

So we have two solutions for k: k = 2 + √6 , k = 2 - √6

Therefore, the point (5, 2 + √6) and (5, 2 - √6) are equidistant from the point (-1, 2) and (3, 0)