Expert: Sherman D. - 7/24/2004

Question
Hi, thank you for your reply, I understood the problem. However I was struck up with these 2 problems.

1) Find all possible values of k so that (3,5),(5,-1) and (k,8) are the verticies of a right triangle.


2) Find all points with -x+2y=0 so that (x,y) is 5 units from (-1,4)

For the first one I got 15.2 and 0.8 which were wrong and the second one to be sqrt40/5=y and sqrt80/5=x which was also wrong.

waiting for your early reply,
Avinash

Answer
AB = (3,5) and (5,-1)
BC = (5,-1) and (k,8)
AC = (3,5) and (k,8)

AB = (3,5) and (5,-1)
AB = sqrt((-1 - 5)^2 + (5 - 3)^2)
AB = sqrt((-1 + (-5))^2 + (2)^2)
AB = sqrt((-6)^2 + 4)
AB = sqrt(36 + 4)
AB = sqrt(40)

BC = (5,-1) and (k,8)
BC = sqrt((8 - (-1))^2 + (k - 5)^2)
BC = sqrt((8 + 1)^2 + (k - 5)^2)
BC = sqrt((9)^2 + (k - 5)^2)
BC = sqrt(81 + (k - 5)^2)
BC = sqrt(81 + ((k - 5)(k - 5)))
BC = sqrt(81 + (k^2 - 5k - 5k + 25))
BC = sqrt(81 + (k^2 - 10k + 25))
BC = sqrt(k^2 - 10k + 25 + 81)
BC = sqrt(k^2 - 10k + 106)

AC = (3,5) and (k,8)
AC = sqrt((8 - 5)^2 + (k - 3)^2)
AC = sqrt((3)^2 + (k - 3)^2)
AC = sqrt(9 + (k - 3)^2)
AC = sqrt(9 + ((k - 3)(k - 3)))
AC = sqrt(9 + (k^2 - 3k - 3k + 9))
AC = sqrt(9 + (k^2 - 6k + 9))
AC = sqrt(9 + k^2 - 6k + 9)
AC = sqrt(k^2 - 6k + 18)

now just use the pythagorean thereom

(AB)^2 + (BC)^2 = (AC)^2
(AB)^2 + (AC)^2 = (BC)^2
(BC)^2 + (AC)^2 = (AB)^2

and see which works out

(sqrt(40))^2 + (sqrt(k^2-10k+106))^2 = (sqrt(k^2-6k+18)

In short and here on out, we can just say that the sqrts cancel out because of the exponent 2, so you can say

40 + k^2 - 10k + 106 = k^2 - 6k + 18
40 + k^2 - 6k + 18 = k^2 - 10k + 106
k^2 - 10k + 106 + k^2 - 6k + 18 = 40

lets start with the first one first

40 + k^2 - 10k + 106 = k^2 - 6k + 18
k^2 - 10k + 146 = k^2 - 6k + 18
-10k + 146 = -6k + 18
-4k = -128k
k = 32

40 + k^2 - 6k + 18 = k^2 - 10k + 106
k^2 - 6k + 58 = k^2 - 10k + 106
-6k + 58 = -10k + 106
4k = 48
k = 12

k^2 - 6k + 18 + k^2 - 10k + 106 = 40
2k^2 - 16k + 124 = 40
k^2 - 8k + 62 = 20
k^2 - 8k + 42 = 0

Using the quadratic formula

x = (-b ± sqrt(b^2 - 4ac))/2a
x = (-(-8) ± sqrt((-8)^2 - 4(42)(1)))/2(1)
x = (8 ± sqrt(64 - 168))/2
x = (8 ± sqrt(-104))/2
x = (8 ± sqrt(-4 * 26))/2
x = (8 ± 2isqrt(26))/2
x = 4 ± isqrt(26)

so i get the "k" to be 12, 32, and 4 ± isqrt(26). If 4 ± isqrt(26) isn't exceptable, then i would go with 12 and 32. Also to let you know, ± means - and + so they are 2 different solutions.

Sorry i don't think i can help you with the second problem.