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

 Quote by Frogeyedpeas The two other perpendicular bisectors take on the form: y1 = -(b+a)/c + K y2 = -(b-a)/c + K
The general equation for a straight line is y=mx+c
You seem to be overlooking that x.

As I pointed out, you now have all the information you need to write the equation for every perpendicular bisector.

For each you know its slope, and you know the co-ordinates of a point on the side of the triangle that it passes through. So, considering one perpendicular at a time, substitute for what you know into the straight line equation y=mx+c and hence determine its equation.

Mentor
 Quote by Frogeyedpeas ... 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):
$\displaystyle y=-\frac{b+a}{c}\,x+k_1$

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.
 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

Mentor
 Quote by Frogeyedpeas 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:
$\displaystyle c^2=2ck-b^2+a^2$
Solving for k gives:
$\displaystyle k=\frac{b^2+c^2-a^2}{2c}$