Vector of P magnitude: O(0,0), A(-1,3), B(-5,4), C(7,P)

R

richiesmasher

Junior Member
Joined
Dec 15, 2017
Messages
111
Hello, I have four points O (0,0) A(-1,3) B(-5,4) and C(7,P)

I have found vector AC to be (8, (p-3))

Now the question is stating what are the values of P for which '' l Vector AC l '' (those two vertical lines are included)=10

Im guessing this is th magnitude so I sub it into the formula which would be (sqrt) (82+(p-3)2)= 10

But from here I keep failing and I am not sure what to do, because I keep ending up with (sqrt) (64 + (p-3)2) =10
And if I try to solve (p-3)2 = 0 to get values for P I would end up with P=3 and then my answer for the equation would just not give me the correct answer.

How do i solve this?
 
Last edited:
lev888 said:
Could you clarify where this came from?
Click to expand...
It didnt come from anywhere, I just figured to solve the overall equation I could solve that equation separately first.

SO I know (p-3)2 is a perfect square

So: (p-3) (p-3) =0

Each value of p would be then 3

So I used that in the equation i guess?
 
richiesmasher said:
SO I know (p-3)2 is a perfect square

So: (p-3) (p-3) =0
Click to expand...

Yes, I was asking how you figured out that the next step would be that equation. I don't quite understand the reasoning. First, I don't see any requirements that (p-3)2 should be a perfect square. And even if it is, why 0?
 
lev888 said:
Yes, I was asking how you figured out that the next step would be that equation. I don't quite understand the reasoning. First, I don't see any requirements that (p-3)2 should be a perfect square. And even if it is, why 0?
Click to expand...

Idk because it's a quadratic? :(
 
You must log in or register to reply here.
Top