Finding Points in the Third Quadrant Using the Distance Formula

[adillhoff]

Homework Statement

Find the point with coordinates of the form (2a, a) that is in the third quadrant and is a distance 5 from P(1, 3)

Homework Equations

<br /> \begin{distance}<br /> d(P_1, P_2) = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}<br /> \end{distance}<br />

The Attempt at a Solution

To be quite honest I have no idea where to begin. The examples in my book do not cover this specific type of problem. I think I could treat P(1, 3) as the center of a circle with a radius of 5, but at this point in the book we have not covered circles yet.

 

[Mentallic]

It isn't necessary to know about circles, but I'll just teach you a little something quickly about them before moving on.

If you have some point P(a,b) and you want to find a point Q(x,y) which is a distance r from P then by the distance formula we have

r=\sqrt{(x-a)^2+(y-b)^2}

But we don't know in which direction Q is from P so this formula is pretty much giving us a circle with radius r around the point P. This is pretty much the circle formula!

For a circle with radius r and centre (a,b) we have

(x-a)^2+(y-b)^2=r^2

ok so back to the point. We want to find a point Q(2a,a) a distance of 5 from the point P(1,3). Shoving this into the distance (or circle) formula, we get

25=(2a-1)^2+(a-3)^2

Now solve for a. This will be a quadratic in a so you need to turn it into the appropriate form and use the quadratic formula - or factorizing, if that works.

 

[adillhoff]

I wasn't quite sure if I could solve it that way, but it makes a lot of sense. After solving for a using the quadratic formula I get a = 1 - 2 and a = 1 + 2. The problem stated that the point is in the third quadrant which means a < 0. So I am left with a = 1 - 2 = -1.

The answer ends up being (2(-1), -1) or (-2, -1)

Thanks!

 

[Mentallic]

You're welcome