Geometry Question (distance formula)?

Question

A=(3k,k-1) and B= (1,4) and AB = squareroot(34) and k>0.

find k

Answer 1

AB = √[ (x2 - x1)² + (y2 - y1)² ]
==> plug in values
√34 = √[ (1 - 3k)² + (4 - (k-5))² ]
==> simplify inside parentheses
√34 = √[ (1 - 6k + 9k²) + (5 - k)² ]
==> simplify further
√34 = √[ (1 - 6k + 9k²) + (25 - 10k + k²) ]
==> combine like terms
√34 = √[ 10k² - 16k + 26 ]
==> square both sides to cancel root
34 = 10k² - 16k + 26
==> set equal to zero
10k² - 16k - 8 = 0
==> divide everything by 2
5k² - 8k - 4 = 0
==> factor by grouping
(5k² - 10k) + (2k - 4)
==> pull out GCF's
5k(k - 2) + 2(k - 2)
==> regroup
(5k + 2)(k - 2) = 0
==> two answers
k = -5/2, k = 2
==> take only positive root
k = 2 ... ANSWER

Answer 2

A=(3k,k-1) and B= (1,4) and AB = sqrt.(34), with k > 0

AB = d = sqrt.(34) ==> d^2 = 34
And,
d^2 = 34 = (3k-1)^2 + (k-1-4)^2
d^2 = 34 = 9k^2 - 6k + 1 + k^2 - 10k + 25
So,
34 = 10k^2 -16k + 26
10k^2 - 16k - 8 = 0
<==>
5k^2 - 8k - 4 = 0, with k > 0 (a condition)

So, k = [8 + sqrt.(144)]/10 = [8 + 12]/10 = 2

So k = 2

Answer 3

Given: A(3k, k - 1), B(1, 4)

Length Of AB --> sqrt[(3k - 1)^2 + (k - 1 - 4)^2] = sqrt(34)
9k^2 - 6k+ 1 + k^2 - 10k + 25 = 34
10k^2 - 16k - 8 = 0
5k^2 - 8k - 4 = 0
(k - 2)(5k + 2) = 0
k = 2 or k = -2/5

Since k > 0, k = -2/5 rejected.
Therefore, k = 2.

Answer 4

AB^2 = (3k-1)^2 +(k-5)^2 =34
10k^2-16k-8=0 solve the 2nd degree equation and take the positive solution

Answer 5

k = 2

A = (6, 1)

34 = 5*5+3*3