Simple Math Problem

[jordan]

I had to start a little work out of the old calculus book today. The teacher's want to make sure we understand how to calculate the slope of a line when given two points before we get into the really tough stuff like the slope of a curve and logarithms this fall. (I know I might not be the best at math, but come on). Ths is a question I came upon in the book that I just couldn't find a way around. Incidently, the only reason I saw it was because the teacher told us to skip over that section. So I naturaly tried some of the problems. One of them has me a little stumped though.

 

It came with a diagram, but since I don't know how to paste pictures on here, I'll just describe it.

 

On a basic graph, we have a point on the x-axis at (a,0) and a symetrical point at (-a,0). To complete the triangle in a point in the first quadrant (doesn't matter where since they're only variables) at point (b,c).

 

Now, the question: "Find the coordinates of the point of intersection of the perpendicular bisectors of the sides?"

 

Explanations are always apreciated.

[NSX]

Any ideas from you so far?

 

Always important in these types of problems are the definition of a slope and that two lines are perpendicular if one's slope is the negative reciprocal of the other.

[bloodhound]

well, its too late for me to work out the solution using the simple equations of the lines, so ill just give u the standard method.

 

Find the equation in the standard form y=mx+c for each line.

 

then for each line the slope of the perpendicular will be -1/m. For the line between (a,0) and (-a,0), the equation of the perpendicular is just y=0

 

find the equations of the perpendicular at midpoint of each side. and then just solve the simultneous equations

 

 

Also the intersection point can be geometrically found, if u draw a circle which passes through these three point. then the point of intersection of the perpendicular bisectors of the sides is the centre of the circle.

[JaKiri]

The easiest way, by far, to find the centre is by vector methods.

[jordan]

Here's a picture of the problem, just for refrence:

 

[jordan]

I wanted that picture to open up in my post, but I only have the link there. Oh well.

 

Now, for L1, the perpedicular bisector is obviously the y-axis. So our answer must lie on the line x=0.

 

 

L2's perp. bisector slope is [math]-(\frac{y2-y1}{x2-x1})^{-1}[/math]

With the other variables and all worked out it's: [math]-(\frac{b-a}{c})[/math] which will yeild a positive slope, which is good.

So the equation is y-y₁=m(x-x₁)

or

[math]y-(c/2)=\frac{-(b-a)}{c}(x-(b-a)/2))[/math]

or

[math]y=\frac{-(b-a)x}{c}*(\frac{(b-a)^2+c^2}{2c})[/math]

 

 

L3:

m=[math]-(\frac{b+a}{c})[/math] which is negative. Good.

Subing in the equation:

[math]y-(c/2)=\frac{-(b+a)}{c}*(x-(b+a)/2)[/math]

or

[math]y=\frac{(-bx-ax)}{c}+(\frac{(b+a)^2+c^2}{2c}[/math]

 

Wow, that took forever and I'm still not sure it's all right. Then I don't know where to go. Any help is apreciated.