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.

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.

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.

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

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

 

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.