Math distance problem solve

Find k such that the distance from (2,5) to (4,k) is 17.

distances on the x-y plane are always the sqrt of the
sum of the sq of the x and y differences.

distance = sqrt((4 - 2)^2 + (k - 5)^2)
we want this equal to 17
so solve for k

sqrt(2^2 + k^2 - 10k + 25) = 17
square both sides
4 + k^2 - 10k + 25 = 17^2 = 289
k^2 - 10k - 260 = 0
k = (10 +- sqrt(100 + 4(260)))/2
= (10 +- sqrt(1140))/2
= (10 +- 33.764)/2
= 21.882
or
= -11.882

So you have a triangle with hypotenuse 17, horizontal dimension of 4,
and a vertical dimension of 16.882 which is the distance away from 5 either
upward or downward.

The distance is sqrt( 2^2 + (k-5)^2 ) As it is 17 .. then it is the equation...

sqrt( 4 + k^2 -10k + 25) = 17 ... now take power 2 in both side....

it is 4 + k^2 - 10k + 25 = 289 ==> k^2 -10k - 260 = 0... it is a quadratic equation...

the roots are k = 5 + sqrt(285) or k = 5 - sqrt(285) ... the values of k. OK!

√(5-2)^2 + (k-4)^2 = 17

25 + 4 + k^2 + 16 = 17

45 + k^2 = 17

45 - 17 = -k^2

28 = -k^2

√28

Therefore = -5.292