1. Originally Posted by LllamaGlama
Huh... i thought i was gonna find the distance from the offshre connecting to the onshore then add the distance from the refinery to the harbour?
that may of been what u just said cuz idk wat a pythagoras is (is that (x2-x2)2+ (y2-y1)2 square rooted?

lol my math typing skills are pretty bad
I'm not trying to be rude at all when I ask this question:

Do you honestly not know what Pythagoras is?

2. Originally Posted by topher0805
I'm not trying to be rude at all when I ask this question:

Do you honestly not know what Pythagoras is?
Nah i dont take that rude at all. As u can see math isnt ny strongest... could it be a2+b2=c2?

3. Originally Posted by LllamaGlama
Nah i dont take that rude at all. As u can see math isnt ny strongest... could it be a2+b2=c2?
Yes.

$\displaystyle a^2 = b^2 + c^2$

So, in the diagram that I drew, find the equation of the line connecting A and B in the form y = mx + b

Then, take the negative reciprocal of the slope of AB. (flip the fraction and multiply it by -1).

Next, use the slope of the line that connects C to AB and the point C to find the equation of that line also in the form y = mx + b

Set the two equation equal to each other (mx + b = mx +b) and solve for x. The, plug that x value back into either equation to find y. This is the point at which the two pipes meet.

Now, use the distance formula that topsquark provided to find the distance between C and the new point(lets call it D).

Now use Pythagoras to find the distance from D to the refinery.

From C to D + From D to Refinery.

4. Originally Posted by topher0805
Yes.

$\displaystyle a^2 = b^2 + c^2$

So, in the diagram that I drew, find the equation of the line connecting A and B in the form y = mx + b

Then, take the negative reciprocal of the slope of AB. (flip the fraction and multiply it by -1).

Next, use the slope of the line that connects C to AB and the point C to find the equation of that line also in the form y = mx + b

Set the two equation equal to each other (mx + b = mx +b) and solve for x. The, plug that x value back into either equation to find y. This is the point at which the two pipes meet.

Now, use the distance formula that topsquark provided to find the distance between C and the new point(lets call it D).

Now use Pythagoras to find the distance from D to the refinery.

From C to D + From D to Refinery.