Expert: Ahmed Salami - 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
Hi avinash,
Sorry about the delay.
1)Considering the problem, the slopes of the lines joining each of the other lines to (k,8)should be perpendicular. Hence the product of the slopes should be -1.
The slopes are;
5-8/3-k = -3/3-k
and
-1-8/5-k = -9/5-k
Therefore,
(-3/3-k)(-9/5-k)= -1
27/(15 - 8k + k^2)= -1
15 - 8k + k^2 = -27
k^2 - 8k + 42 = 0
The solution of this should be your answer. But don't forget, the method of solution is paramount.

2)The distance of (x,y)from (-1,4)
= sqrt((x+1)^2 + (y-4)^2)= 5
Therefore,
(x+1)^2 + (y-4)^2 = 25
But -x+2y=0
2y = x
So
(2y+1)^2 + (y-4)^2 = 25
4y^2 + 4y + 1 + y^2 - 8y + 16 = 25
5y^2 - 4y + 17 = 25
5y^2 - 4y - 8 = 0
The solution of this should be your answer. Again don't forget, the method of solution is paramount.When you have found y, you find x by 2y = x. There would be two values for each of  y and x.
Hoping you will find all these helpful.
Regards.