Absolutely clueless on how to do this... Geometry Help


by Frogeyedpeas
Tags: absolutely, clueless, geometry
SammyS
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Dec17-11, 12:31 AM
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Quote Quote by Frogeyedpeas View Post
...

Your given a triangle with vertices (-a,0) (a,0) (b,c)

The perpendicular bisector to (-a,0) and (a,0) is met by the equation x = 0

The next side we address has the endpoints (-a,0) and (b,c) it's slope is c/(b+a) so the perpendicular bisector's slope will be -(b+a)/c

The third side has the endpoints (b,c) and (a,0) with a slope of c/(b-a) therefore giving the perpendicular bisector's slope as (b-a)/-c

So now we know that our x coordinate in the intersection point is most definitely going to be equal to 0.

So we now have to solve for the Y coordinate of the intersection point.

The two other perpendicular bisectors take on the form:

y1 = -(b+a)/c + K
y2 = -(b-a)/c + K
...
Find the equation for each of the lines corresponding to these bisectors.

For the perpendicular bisector to the segment with vertices (-a,0) and (b,c):
[itex]\displaystyle y=-\frac{b+a}{c}\,x+k_1[/itex]

Put in the coordinates for the midpoint to find k1
k1 is the y-intercept for this line, so that's where this line intersects the line, x=0.

Do similar for the other bisector.
Frogeyedpeas
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Dec17-11, 11:30 AM
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Okay so lets use the first perpendicular bisector: and the corresponding midpoint

((b-a)/2, c/2)

We know that y = -(b+a)/c*x + K1 So to substitute the x value we get

c/2 = -(b2-a2)/c + K1

multplying the c we recieve:

c2/2 = k -b2 + a2

Which leaves us finally with

K = 1/2(c2+b2-a2 = k

And since that is the Y coordinate of the point of intersection we get the intersection of three perpendicular bisectors as:

(0, 1/2c2 + b2-a2) which unfortunately doesn't match up with my book's answer (I wanted to do this from scratch without working backwards) which is:

(0, (-a2 + b2+2)/2c)

The two expressions are not equal either, yet this one that I have attained seems quite logically deduced
SammyS
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Dec17-11, 07:00 PM
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Quote Quote by Frogeyedpeas View Post
Okay so lets use the first perpendicular bisector: and the corresponding midpoint

((b-a)/2, c/2)

We know that y = -(b+a)/c*x + K1 So to substitute the x value we get

c/2 = -(b2-a2)/c + K1

multplying the c we recieve:

c2/2 = k -b2 + a2

...
Not quite right.

You failed to multiply k by c. Anyway there should have been a 2 in both denominator.

So, after multiplying by 2c you should have:
[itex]\displaystyle c^2=2ck-b^2+a^2[/itex]
Solving for k gives:
[itex]\displaystyle k=\frac{b^2+c^2-a^2}{2c}[/itex]


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