Question

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

Answer 1

The formula of a distance between two points:

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

We have:

$(2,\ k)\to x_1=2,\ y_1=k\\(0,\ -6)\to x_2=0,\ y_2=-6\\d=5-\sqrt5$

Substitute:

$\sqrt{(0-2)^2+(-6-k)^2}=5-\sqrt5\\\\\sqrt{(-2)^2+[-(6+k)]^2}=5-\sqrt5\\\\\sqrt{4+(6+k)^2}=5-\sqrt5\ \ \ \ \ |^2\\\\4+(6+k)^2=(5-\sqrt5)^2\ \ \ \ |use\ (a-b)^2=a^2-2ab+b^2$

$4+(6+k)^2=5^2-2\cdot5\cdot\sqrt5+(\sqrt5)^2\\\\4+(6+k)^2=25-10\sqrt5+5\ \ \ \ |-4\\\\(6+k)^2=26-10\sqrt5\to6+k=\pm\sqrt{26-10\sqrt5}\ \ \ \ |-6\\\\\boxed{k=-6-\sqrt{26-10\sqrt5}\ \vee\ k=-6+\sqrt{26-10\sqrt5}}$

If $d=\sqrt5$, then:

$\sqrt{4+(6+k)^2}=\sqrt5\ \ \ \ |^2\\\\(\sqrt{4+(6+k)^2})^2=(\sqrt5)^2\\\\4+(6+k)^2=5\ \ \ \ |-4\\\\(6+k)^2=1\to 6+k=\pm\sqrt1\\\\6+k=\pm1\ \ \ \ |-6\\\\\boxed{k=-7\ \vee\ k=-5}$