I can't seem to get my head around this question, if I had a point P(x,y) that moves along the line y=3x and I have another point Q which is stationary at (2,0), I need to write an expression between points P and Q in terms of x only. I have tried using Pythagorus theorem with D as the distance P to Q, as drawing a diagram shows a right angled triangle but in this case the point at Q the angle changes as x or y changes. Any way my answer doesn't match with the answer given which is D=√(10x²-4x+4) my answer is:

by letting B=the line y=3x and X=the distance along the x-axis (from 0 to 2) and D=the distance between P and Q. So D equals the square root of B²-X², I got the square root of 9x²-4? Please help I know there must be something wrong with my diagram (linear line y=3x with a perpendicular line to x=2 on x=axis with corresponding y-value at that pt of 6, I know that here the line D is changing in length as the point slides along y=3x and also the length of y=3x is changing. Am I remotely on the right track?

Please, tell me, did your diagram look anything like this:
22201

If so, then you are on the right track.
You are trying to solve in terms of x only. You know that as the value of X changes, the value of Y changes three fold. You should also know that, since point P is not stationary, you cannot always solve for D using the pythagoreum theorum.

Try finding another way to solve for the distance between two points!
Chances are, you didn't think of it because you've always gotten a numerical answer, and aren't used to getting a variable answer.

You have to find the distance of P to Q at any point, not the shortest distance. The distance between two points is given by:

\sqrt{(x_2 - x_1)^2+(y_2-y_1)^2}

Just fit in (x,y) and (2,0) and you'll have the answer.

EDIT: oops, didn't see your post HH... :o saw the question, got lunch and posted the answer too late...

It's all good unknown, I've done the same thing before.

I noticed that you gave it away though... lol.
I figured that, by recognizing the difference between two points, and seeing the quadratic equation in the answer given, he (she?) could have gotten that far on his own.
Sometimes, though, it's good to go that far. I trust your judgement here more than my own as far as how big of a push to give, as you seem to be better at answering these than me.

Your answer was good too. I wanted to rep it, but had to spread it... :/

Thanks for the reminder of this formula, I would have never thought of using it, I'm not naturally talented with numbers so many concepts and formulas are forgotten if not used, a few months after they are learnt.


You have to find the distance of P to Q at any point, not the shortest distance. The distance between two points is given by:

\sqrt{(x_2 - x_1)^2+(y_2-y_1)^2}

Just fit in (x,y) and (2,0) and you'll have the answer.

EDIT: oops, didn't see your post HH... :o saw the question, got lunch and posted the answer too late...

I myself got veered away when you posted how you tried it. I made several 'wrong' methods, saying that there must be something wrong here. Then, I found it. It's just a matter of practice.

Yep this is the diagram I was describing I also worked out the gradient function as y=3x+6 but didn't know how to take it from there to relate it to what I need, but the formula given below shows a much better way of reasoning than what I have been trying to do.


Please, tell me, did your diagram look anything like this:
22201

If so, then you are on the right track.
You are trying to solve in terms of x only. You know that as the value of X changes, the value of Y changes three fold. You should also know that, since point P is not stationary, you cannot always solve for D using the pythagoreum theorum.

Try finding another way to solve for the distance between two points!
Chances are, you didn't think of it because you've always gotten a numerical answer, and aren't used to getting a variable answer.

That was what you were doing! I wasn't so much understanding that part! Your line, distance D is not vertical. The line D is perpendicular to your line y=3x. That makes OD the hypotenuse.

Anyway, you now got the answer.

If you had to find D, you would have to work out the point of intersection of the line D and your line y=3x. The gradient of D would be -1/3 since it is perpendicular to y=3x.

Work the equation of D;
\frac{y-0}{x-2} = -\frac13

Then simultaneously solve the two equations to obtain the point of intersection. Finally, finding the length through the formula I gave in my first post here.